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College Algebra

# 1.2Exponents and Scientific Notation

College Algebra1.2 Exponents and Scientific Notation

### Learning Objectives

In this section students will:

• Use the product rule of exponents.
• Use the quotient rule of exponents.
• Use the power rule of exponents.
• Use the zero exponent rule of exponents.
• Use the negative rule of exponents.
• Find the power of a product and a quotient.
• Simplify exponential expressions.
• Use scientific notation.

Mathematicians, scientists, and economists commonly encounter very large and very small numbers. But it may not be obvious how common such figures are in everyday life. For instance, a pixel is the smallest unit of light that can be perceived and recorded by a digital camera. A particular camera might record an image that is 2,048 pixels by 1,536 pixels, which is a very high resolution picture. It can also perceive a color depth (gradations in colors) of up to 48 bits per frame, and can shoot the equivalent of 24 frames per second. The maximum possible number of bits of information used to film a one-hour (3,600-second) digital film is then an extremely large number.

Using a calculator, we enter $2,048×1,536×48×24×3,600 2,048×1,536×48×24×3,600$ and press ENTER. The calculator displays 1.304596316E13. What does this mean? The “E13” portion of the result represents the exponent 13 of ten, so there are a maximum of approximately $1.3× 10 13 1.3× 10 13$ bits of data in that one-hour film. In this section, we review rules of exponents first and then apply them to calculations involving very large or small numbers.

### Using the Product Rule of Exponents

Consider the product $x 3 ⋅ x 4 . x 3 ⋅ x 4 .$ Both terms have the same base, x, but they are raised to different exponents. Expand each expression, and then rewrite the resulting expression.

$x3⋅x4 = x⋅x⋅x3factors⋅x⋅x⋅x⋅x 4factors = x⋅x⋅x⋅x⋅x⋅x⋅x7factors = x7 x3⋅x4 = x⋅x⋅x3factors⋅x⋅x⋅x⋅x 4factors = x⋅x⋅x⋅x⋅x⋅x⋅x7factors = x7$

The result is that $x 3 ⋅ x 4 = x 3+4 = x 7 . x 3 ⋅ x 4 = x 3+4 = x 7 .$

Notice that the exponent of the product is the sum of the exponents of the terms. In other words, when multiplying exponential expressions with the same base, we write the result with the common base and add the exponents. This is the product rule of exponents.

$a m ⋅ a n = a m+n a m ⋅ a n = a m+n$

Now consider an example with real numbers.

$2 3 ⋅ 2 4 = 2 3+4 = 2 7 2 3 ⋅ 2 4 = 2 3+4 = 2 7$

We can always check that this is true by simplifying each exponential expression. We find that $2 3 2 3$ is 8, $2 4 2 4$ is 16, and $2 7 2 7$ is 128. The product $8⋅16 8⋅16$ equals 128, so the relationship is true. We can use the product rule of exponents to simplify expressions that are a product of two numbers or expressions with the same base but different exponents.

### The Product Rule of Exponents

For any real number $a a$ and natural numbers $m m$ and $n, n,$ the product rule of exponents states that

$a m ⋅ a n = a m+n a m ⋅ a n = a m+n$

### Example 1

#### Using the Product Rule

Write each of the following products with a single base. Do not simplify further.

1. $t 5 ⋅ t 3 t 5 ⋅ t 3$
2. $( −3 ) 5 ⋅( −3 ) ( −3 ) 5 ⋅( −3 )$
3. $x 2 ⋅ x 5 ⋅ x 3 x 2 ⋅ x 5 ⋅ x 3$
Try It #1

Write each of the following products with a single base. Do not simplify further.

1. $k 6 ⋅ k 9 k 6 ⋅ k 9$
2. $( 2 y ) 4 ⋅( 2 y ) ( 2 y ) 4 ⋅( 2 y )$
3. $t 3 ⋅ t 6 ⋅ t 5 t 3 ⋅ t 6 ⋅ t 5$

### Using the Quotient Rule of Exponents

The quotient rule of exponents allows us to simplify an expression that divides two numbers with the same base but different exponents. In a similar way to the product rule, we can simplify an expression such as $y m y n , y m y n ,$ where $m>n. m>n.$ Consider the example $y 9 y 5 . y 9 y 5 .$ Perform the division by canceling common factors.

$y9y5 = y⋅y⋅y⋅y⋅y⋅y⋅y⋅y⋅yy⋅y⋅y⋅y⋅y = y⋅y⋅y⋅y⋅y⋅y⋅y⋅y⋅yy⋅y⋅y⋅y⋅y = y⋅y⋅y⋅y1 = y4 y9y5 = y⋅y⋅y⋅y⋅y⋅y⋅y⋅y⋅yy⋅y⋅y⋅y⋅y = y⋅y⋅y⋅y⋅y⋅y⋅y⋅y⋅yy⋅y⋅y⋅y⋅y = y⋅y⋅y⋅y1 = y4$

Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend.

$a m a n = a m−n a m a n = a m−n$

In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents.

$y 9 y 5 = y 9−5 = y 4 y 9 y 5 = y 9−5 = y 4$

For the time being, we must be aware of the condition $m>n. m>n.$ Otherwise, the difference $m−n m−n$ could be zero or negative. Those possibilities will be explored shortly. Also, instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers.

### The Quotient Rule of Exponents

For any real number $a a$ and natural numbers $m m$ and $n, n,$ such that $m>n, m>n,$ the quotient rule of exponents states that

$a m a n = a m−n a m a n = a m−n$

### Example 2

#### Using the Quotient Rule

Write each of the following products with a single base. Do not simplify further.

1. $( −2 ) 14 ( −2 ) 9 ( −2 ) 14 ( −2 ) 9$
2. $t 23 t 15 t 23 t 15$
3. $( z 2 ) 5 z 2 ( z 2 ) 5 z 2$
Try It #2

Write each of the following products with a single base. Do not simplify further.

1. $s 75 s 68 s 75 s 68$
2. $( −3 ) 6 −3 ( −3 ) 6 −3$
3. $( e f 2 ) 5 ( e f 2 ) 3 ( e f 2 ) 5 ( e f 2 ) 3$

### Using the Power Rule of Exponents

Suppose an exponential expression is raised to some power. Can we simplify the result? Yes. To do this, we use the power rule of exponents. Consider the expression $( x 2 ) 3 . ( x 2 ) 3 .$ The expression inside the parentheses is multiplied twice because it has an exponent of 2. Then the result is multiplied three times because the entire expression has an exponent of 3.

$(x2)3 = (x2)⋅(x2)⋅(x2)3factors = (x⋅x︷2factors)⋅(x⋅x ︷2factors)⋅(x⋅x︷2factors)3factors = x⋅x⋅x⋅x⋅x⋅x = x6 (x2)3 = (x2)⋅(x2)⋅(x2)3factors = (x⋅x︷2factors)⋅(x⋅x ︷2factors)⋅(x⋅x︷2factors)3factors = x⋅x⋅x⋅x⋅x⋅x = x6$

The exponent of the answer is the product of the exponents: $( x 2 ) 3 = x 2⋅3 = x 6 . ( x 2 ) 3 = x 2⋅3 = x 6 .$ In other words, when raising an exponential expression to a power, we write the result with the common base and the product of the exponents.

$( a m ) n = a m⋅n ( a m ) n = a m⋅n$

Be careful to distinguish between uses of the product rule and the power rule. When using the product rule, different terms with the same bases are raised to exponents. In this case, you add the exponents. When using the power rule, a term in exponential notation is raised to a power. In this case, you multiply the exponents.

### The Power Rule of Exponents

For any real number $a a$ and positive integers $m m$ and $n, n,$ the power rule of exponents states that

$( a m ) n = a m⋅n ( a m ) n = a m⋅n$

### Example 3

#### Using the Power Rule

Write each of the following products with a single base. Do not simplify further.

1. $( x 2 ) 7 ( x 2 ) 7$
2. $( ( 2t ) 5 ) 3 ( ( 2t ) 5 ) 3$
3. $( ( −3 ) 5 ) 11 ( ( −3 ) 5 ) 11$
Try It #3

Write each of the following products with a single base. Do not simplify further.

1. $( ( 3y ) 8 ) 3 ( ( 3y ) 8 ) 3$
2. $( t 5 ) 7 ( t 5 ) 7$
3. $( ( −g ) 4 ) 4 ( ( −g ) 4 ) 4$

### Using the Zero Exponent Rule of Exponents

Return to the quotient rule. We made the condition that $m>n m>n$ so that the difference $m−n m−n$ would never be zero or negative. What would happen if $m=n? m=n?$ In this case, we would use the zero exponent rule of exponents to simplify the expression to 1. To see how this is done, let us begin with an example.

$t 8 t 8 = t 8 t 8 =1 t 8 t 8 = t 8 t 8 =1$

If we were to simplify the original expression using the quotient rule, we would have

$t 8 t 8 = t 8−8 = t 0 t 8 t 8 = t 8−8 = t 0$

If we equate the two answers, the result is $t 0 =1. t 0 =1.$ This is true for any nonzero real number, or any variable representing a real number.

$a 0 =1 a 0 =1$

The sole exception is the expression $0 0 . 0 0 .$ This appears later in more advanced courses, but for now, we will consider the value to be undefined.

### The Zero Exponent Rule of Exponents

For any nonzero real number $a, a,$ the zero exponent rule of exponents states that

$a 0 =1 a 0 =1$

### Example 4

#### Using the Zero Exponent Rule

Simplify each expression using the zero exponent rule of exponents.

1. $c 3 c 3 c 3 c 3$
2. $−3 x 5 x 5 −3 x 5 x 5$
3. $( j 2 k ) 4 ( j 2 k )⋅ ( j 2 k ) 3 ( j 2 k ) 4 ( j 2 k )⋅ ( j 2 k ) 3$
4. $5 ( r s 2 ) 2 ( r s 2 ) 2 5 ( r s 2 ) 2 ( r s 2 ) 2$
Try It #4

Simplify each expression using the zero exponent rule of exponents.

1. $t 7 t 7 t 7 t 7$
2. $( d e 2 ) 11 2 ( d e 2 ) 11 ( d e 2 ) 11 2 ( d e 2 ) 11$
3. $w 4 ⋅ w 2 w 6 w 4 ⋅ w 2 w 6$
4. $t 3 ⋅ t 4 t 2 ⋅ t 5 t 3 ⋅ t 4 t 2 ⋅ t 5$

### Using the Negative Rule of Exponents

Another useful result occurs if we relax the condition that $m>n m>n$ in the quotient rule even further. For example, can we simplify $h 3 h 5 ? h 3 h 5 ?$ When $m —that is, where the difference $m−n m−n$ is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal.

Divide one exponential expression by another with a larger exponent. Use our example, $h 3 h 5 . h 3 h 5 .$

$h3h5 = h⋅h⋅hh⋅h⋅h⋅h⋅h = h⋅h⋅hh⋅h⋅h⋅h⋅h = 1h⋅h = 1h2 h3h5 = h⋅h⋅hh⋅h⋅h⋅h⋅h = h⋅h⋅hh⋅h⋅h⋅h⋅h = 1h⋅h = 1h2$

If we were to simplify the original expression using the quotient rule, we would have

$h3h5 = h3−5 = h−2 h3h5 = h3−5 = h−2$

Putting the answers together, we have $h −2 = 1 h 2 . h −2 = 1 h 2 .$ This is true for any nonzero real number, or any variable representing a nonzero real number.

A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa.

$a −n = 1 a n and a n = 1 a −n a −n = 1 a n and a n = 1 a −n$

We have shown that the exponential expression $a n a n$ is defined when $n n$ is a natural number, 0, or the negative of a natural number. That means that $a n a n$ is defined for any integer $n. n.$ Also, the product and quotient rules and all of the rules we will look at soon hold for any integer $n. n.$

### The Negative Rule of Exponents

For any nonzero real number $a a$ and natural number $n, n,$ the negative rule of exponents states that

$a −n = 1 a n a −n = 1 a n$

### Example 5

#### Using the Negative Exponent Rule

Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.

1. $θ 3 θ 10 θ 3 θ 10$
2. $z 2 ⋅z z 4 z 2 ⋅z z 4$
3. $( −5 t 3 ) 4 ( −5 t 3 ) 8 ( −5 t 3 ) 4 ( −5 t 3 ) 8$
Try It #5

Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.

1. $( −3t ) 2 ( −3t ) 8 ( −3t ) 2 ( −3t ) 8$
2. $f 47 f 49 ⋅f f 47 f 49 ⋅f$
3. $2 k 4 5 k 7 2 k 4 5 k 7$

### Example 6

#### Using the Product and Quotient Rules

Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.

1. $b 2 ⋅ b −8 b 2 ⋅ b −8$
2. $( −x ) 5 ⋅ ( −x ) −5 ( −x ) 5 ⋅ ( −x ) −5$
3. $−7z ( −7z ) 5 −7z ( −7z ) 5$
Try It #6

Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.

1. $t −11 ⋅ t 6 t −11 ⋅ t 6$
2. $25 12 25 13 25 12 25 13$

### Finding the Power of a Product

To simplify the power of a product of two exponential expressions, we can use the power of a product rule of exponents, which breaks up the power of a product of factors into the product of the powers of the factors. For instance, consider $( pq ) 3 . ( pq ) 3 .$ We begin by using the associative and commutative properties of multiplication to regroup the factors.

$(pq)3 = (pq)⋅(pq)⋅(pq) 3factors = p⋅q⋅p⋅q⋅p⋅q = p⋅p⋅p 3factors⋅ q⋅q⋅q 3factors = p3⋅q3 (pq)3 = (pq)⋅(pq)⋅(pq) 3factors = p⋅q⋅p⋅q⋅p⋅q = p⋅p⋅p 3factors⋅ q⋅q⋅q 3factors = p3⋅q3$

In other words, $( pq ) 3 = p 3 ⋅ q 3 . ( pq ) 3 = p 3 ⋅ q 3 .$

### The Power of a Product Rule of Exponents

For any real numbers $a a$ and $b b$ and any integer $n, n,$ the power of a product rule of exponents states that

$( ab ) n = a n b n ( ab ) n = a n b n$

### Example 7

#### Using the Power of a Product Rule

Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.

1. $( a b 2 ) 3 ( a b 2 ) 3$
2. $( 2t ) 15 ( 2t ) 15$
3. $( −2 w 3 ) 3 ( −2 w 3 ) 3$
4. $1 ( −7z ) 4 1 ( −7z ) 4$
5. $( e −2 f 2 ) 7 ( e −2 f 2 ) 7$
Try It #7

Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.

1. $( g 2 h 3 ) 5 ( g 2 h 3 ) 5$
2. $( 5t ) 3 ( 5t ) 3$
3. $( −3 y 5 ) 3 ( −3 y 5 ) 3$
4. $1 ( a 6 b 7 ) 3 1 ( a 6 b 7 ) 3$
5. $( r 3 s −2 ) 4 ( r 3 s −2 ) 4$

### Finding the Power of a Quotient

To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let’s look at the following example.

$( e −2 f 2 ) 7 = f 14 e 14 ( e −2 f 2 ) 7 = f 14 e 14$

Let’s rewrite the original problem differently and look at the result.

$(e−2f2)7 = ( f2 e2)7 = f14e14 (e−2f2)7 = ( f2 e2)7 = f14e14$

It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.

$( e −2 f 2 ) 7 = ( f 2 e 2 ) 7 = ( f 2 ) 7 ( e 2 ) 7 = f 2⋅7 e 2⋅7 = f 14 e 14 ( e −2 f 2 ) 7 = ( f 2 e 2 ) 7 = ( f 2 ) 7 ( e 2 ) 7 = f 2⋅7 e 2⋅7 = f 14 e 14$

### The Power of a Quotient Rule of Exponents

For any real numbers $a a$ and $b b$ and any integer $n, n,$ the power of a quotient rule of exponents states that

$( a b ) n = a n b n ( a b ) n = a n b n$

### Example 8

#### Using the Power of a Quotient Rule

Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.

1. $( 4 z 11 ) 3 ( 4 z 11 ) 3$
2. $( p q 3 ) 6 ( p q 3 ) 6$
3. $( −1 t 2 ) 27 ( −1 t 2 ) 27$
4. $( j 3 k −2 ) 4 ( j 3 k −2 ) 4$
5. $( m −2 n −2 ) 3 ( m −2 n −2 ) 3$
Try It #8

Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.

1. $( b 5 c ) 3 ( b 5 c ) 3$
2. $( 5 u 8 ) 4 ( 5 u 8 ) 4$
3. $( −1 w 3 ) 35 ( −1 w 3 ) 35$
4. $( p −4 q 3 ) 8 ( p −4 q 3 ) 8$
5. $( c −5 d −3 ) 4 ( c −5 d −3 ) 4$

### Simplifying Exponential Expressions

Recall that to simplify an expression means to rewrite it by combing terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions.

### Example 9

#### Simplifying Exponential Expressions

Simplify each expression and write the answer with positive exponents only.

1. $( 6 m 2 n −1 ) 3 ( 6 m 2 n −1 ) 3$
2. $17 5 ⋅ 17 −4 ⋅ 17 −3 17 5 ⋅ 17 −4 ⋅ 17 −3$
3. $( u −1 v v −1 ) 2 ( u −1 v v −1 ) 2$
4. $( −2 a 3 b −1 )( 5 a −2 b 2 ) ( −2 a 3 b −1 )( 5 a −2 b 2 )$
5. $( x 2 2 ) 4 ( x 2 2 ) −4 ( x 2 2 ) 4 ( x 2 2 ) −4$
6. $( 3 w 2 ) 5 ( 6 w −2 ) 2 ( 3 w 2 ) 5 ( 6 w −2 ) 2$
Try It #9

Simplify each expression and write the answer with positive exponents only.

1. $( 2u v −2 ) −3 ( 2u v −2 ) −3$
2. $x 8 ⋅ x −12 ⋅x x 8 ⋅ x −12 ⋅x$
3. $( e 2 f −3 f −1 ) 2 ( e 2 f −3 f −1 ) 2$
4. $( 9 r −5 s 3 )( 3 r 6 s −4 ) ( 9 r −5 s 3 )( 3 r 6 s −4 )$
5. $( 4 9 t w −2 ) −3 ( 4 9 t w −2 ) 3 ( 4 9 t w −2 ) −3 ( 4 9 t w −2 ) 3$
6. $( 2 h 2 k ) 4 ( 7 h −1 k 2 ) 2 ( 2 h 2 k ) 4 ( 7 h −1 k 2 ) 2$

### Using Scientific Notation

Recall at the beginning of the section that we found the number $1.3× 10 13 1.3× 10 13$ when describing bits of information in digital images. Other extreme numbers include the width of a human hair, which is about 0.00005 m, and the radius of an electron, which is about 0.00000000000047 m. How can we effectively work read, compare, and calculate with numbers such as these?

A shorthand method of writing very small and very large numbers is called scientific notation, in which we express numbers in terms of exponents of 10. To write a number in scientific notation, move the decimal point to the right of the first digit in the number. Write the digits as a decimal number between 1 and 10. Count the number of places n that you moved the decimal point. Multiply the decimal number by 10 raised to a power of n. If you moved the decimal left as in a very large number, $n n$ is positive. If you moved the decimal right as in a small large number, $n n$ is negative.

For example, consider the number 2,780,418. Move the decimal left until it is to the right of the first nonzero digit, which is 2.

We obtain 2.780418 by moving the decimal point 6 places to the left. Therefore, the exponent of 10 is 6, and it is positive because we moved the decimal point to the left. This is what we should expect for a large number.

$2.780418× 10 6 2.780418× 10 6$

Working with small numbers is similar. Take, for example, the radius of an electron, 0.00000000000047 m. Perform the same series of steps as above, except move the decimal point to the right.

Be careful not to include the leading 0 in your count. We move the decimal point 13 places to the right, so the exponent of 10 is 13. The exponent is negative because we moved the decimal point to the right. This is what we should expect for a small number.

$4.7× 10 −13 4.7× 10 −13$

### Scientific Notation

A number is written in scientific notation if it is written in the form $a× 10 n , a× 10 n ,$ where $1≤| a |<10 1≤| a |<10$ and $n n$ is an integer.

### Example 10

#### Converting Standard Notation to Scientific Notation

Write each number in scientific notation.

Distance to Andromeda Galaxy from Earth: 24,000,000,000,000,000,000,000 m

Diameter of Andromeda Galaxy: 1,300,000,000,000,000,000,000 m

Number of stars in Andromeda Galaxy: 1,000,000,000,000

Diameter of electron: 0.00000000000094 m

Probability of being struck by lightning in any single year: 0.00000143

#### Analysis

Observe that, if the given number is greater than 1, as in examples a–c, the exponent of 10 is positive; and if the number is less than 1, as in examples d–e, the exponent is negative.

Try It #10

Write each number in scientific notation.

1. U.S. national debt per taxpayer (April 2014): $152,000 2. World population (April 2014): 7,158,000,000 3. World gross national income (April 2014):$85,500,000,000,000
4. Time for light to travel 1 m: 0.00000000334 s
5. Probability of winning lottery (match 6 of 49 possible numbers): 0.0000000715

#### Converting from Scientific to Standard Notation

To convert a number in scientific notation to standard notation, simply reverse the process. Move the decimal $n n$ places to the right if $n n$ is positive or $n n$ places to the left if $n n$ is negative and add zeros as needed. Remember, if $n n$ is positive, the value of the number is greater than 1, and if $n n$ is negative, the value of the number is less than one.

### Example 11

#### Converting Scientific Notation to Standard Notation

Convert each number in scientific notation to standard notation.

1. $3.547× 10 14 3.547× 10 14$
2. $−2× 10 6 −2× 10 6$
3. $7.91× 10 −7 7.91× 10 −7$
4. $−8.05× 10 −12 −8.05× 10 −12$

### Try It #11

Convert each number in scientific notation to standard notation.

1. $7.03× 10 5 7.03× 10 5$
2. $−8.16× 10 11 −8.16× 10 11$
3. $−3.9× 10 −13 −3.9× 10 −13$
4. $8× 10 −6 8× 10 −6$

#### Using Scientific Notation in Applications

Scientific notation, used with the rules of exponents, makes calculating with large or small numbers much easier than doing so using standard notation. For example, suppose we are asked to calculate the number of atoms in 1 L of water. Each water molecule contains 3 atoms (2 hydrogen and 1 oxygen). The average drop of water contains around $1.32× 10 21 1.32× 10 21$ molecules of water and 1 L of water holds about $1.22× 10 4 1.22× 10 4$ average drops. Therefore, there are approximately $3⋅( 1.32× 10 21 )⋅( 1.22× 10 4 )≈4.83× 10 25 3⋅( 1.32× 10 21 )⋅( 1.22× 10 4 )≈4.83× 10 25$ atoms in 1 L of water. We simply multiply the decimal terms and add the exponents. Imagine having to perform the calculation without using scientific notation!

When performing calculations with scientific notation, be sure to write the answer in proper scientific notation. For example, consider the product $( 7× 10 4 )⋅( 5× 10 6 )=35× 10 10 . ( 7× 10 4 )⋅( 5× 10 6 )=35× 10 10 .$ The answer is not in proper scientific notation because 35 is greater than 10. Consider 35 as $3.5×10. 3.5×10.$ That adds a ten to the exponent of the answer.

$( 35 )× 10 10 =( 3.5×10 )× 10 10 =3.5×( 10× 10 10 )=3.5× 10 11 ( 35 )× 10 10 =( 3.5×10 )× 10 10 =3.5×( 10× 10 10 )=3.5× 10 11$

### Example 12

#### Using Scientific Notation

Perform the operations and write the answer in scientific notation.

1. $( 8.14× 10 −7 )( 6.5× 10 10 ) ( 8.14× 10 −7 )( 6.5× 10 10 )$
2. $( 4× 10 5 )÷( −1.52× 10 9 ) ( 4× 10 5 )÷( −1.52× 10 9 )$
3. $( 2.7× 10 5 )( 6.04× 10 13 ) ( 2.7× 10 5 )( 6.04× 10 13 )$
4. $( 1.2× 10 8 )÷( 9.6× 10 5 ) ( 1.2× 10 8 )÷( 9.6× 10 5 )$
5. $( 3.33× 10 4 )( −1.05× 10 7 )( 5.62× 10 5 ) ( 3.33× 10 4 )( −1.05× 10 7 )( 5.62× 10 5 )$

### Try It #12

Perform the operations and write the answer in scientific notation.

1. $( −7.5× 10 8 )( 1.13× 10 −2 ) ( −7.5× 10 8 )( 1.13× 10 −2 )$
2. $( 1.24× 10 11 )÷( 1.55× 10 18 ) ( 1.24× 10 11 )÷( 1.55× 10 18 )$
3. $( 3.72× 10 9 )( 8× 10 3 ) ( 3.72× 10 9 )( 8× 10 3 )$
4. $( 9.933× 10 23 )÷( −2.31× 10 17 ) ( 9.933× 10 23 )÷( −2.31× 10 17 )$
5. $( −6.04× 10 9 )( 7.3× 10 2 )( −2.81× 10 2 ) ( −6.04× 10 9 )( 7.3× 10 2 )( −2.81× 10 2 )$

### Example 13

#### Technology

For the following exercises, use a graphing calculator to simplify. Round the answers to the nearest hundredth.

51.

$( 12 3 m 33 4 −3 ) 2 ( 12 3 m 33 4 −3 ) 2$

52.

$17 3 ÷ 15 2 x 3 17 3 ÷ 15 2 x 3$

#### Extensions

For the following exercises, simplify the given expression. Write answers with positive exponents.

53.

$( 3 2 a 3 ) −2 ( a 4 2 2 ) 2 ( 3 2 a 3 ) −2 ( a 4 2 2 ) 2$

54.

$( 6 2 −24 ) 2 ÷ ( x y ) −5 ( 6 2 −24 ) 2 ÷ ( x y ) −5$

55.

$m 2 n 3 a 2 c −3 ⋅ a −7 n −2 m 2 c 4 m 2 n 3 a 2 c −3 ⋅ a −7 n −2 m 2 c 4$

56.

$( x 6 y 3 x 3 y −3 ⋅ y −7 x −3 ) 10 ( x 6 y 3 x 3 y −3 ⋅ y −7 x −3 ) 10$

57.

$( ( a b 2 c ) −3 b −3 ) 2 ( ( a b 2 c ) −3 b −3 ) 2$

58.

Avogadro’s constant is used to calculate the number of particles in a mole. A mole is a basic unit in chemistry to measure the amount of a substance. The constant is $6.0221413× 10 23 . 6.0221413× 10 23 .$ Write Avogadro’s constant in standard notation.

59.

Planck’s constant is an important unit of measure in quantum physics. It describes the relationship between energy and frequency. The constant is written as $6.62606957× 10 −34 . 6.62606957× 10 −34 .$ Write Planck’s constant in standard notation.

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# Binomial Coefficient Challenge 11!

Find a closed form of -

$\large \displaystyle \sum_{k=1}^n{\binom{r}{k} {\left(-1\right)}^k H_k}$

where $$H_k$$ denotes $$k$$th Harmonic number.

Note by Kartik Sharma
1 month, 4 weeks ago

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Theorem : $\sum_{r=1}^{m} (-1)^r \dbinom{n}{r} H_{r} = (-1)^m \dbinom{n-1}{m} \left( H_{m} + \dfrac{1}{n} \right) - \dfrac{1}{n}$

Let,

$\text{S} = \sum_{r=1}^{m} (-1)^r \dbinom{n}{r} H_{r}$

From Binomial Coefficient Challenge 4, we have,

$\sum_{k=0}^{n-1} \dbinom{k}{r} \dfrac{1}{n-k} = \dbinom{n}{r} (H_{n} - H_{r})$

For $$r \geq 1$$,

$\sum_{k=1}^{n-1} \dbinom{k}{r} \dfrac{1}{n-k} = \dbinom{n}{r} (H_{n} - H_{r})$

$\implies \sum_{k=1}^{n-1} \sum_{r=1}^{m} (-1)^r \dbinom{k}{r} \dfrac{1}{n-k} = H_{n} \sum_{r=1}^{m} (-1)^r \dbinom{n}{r} - \sum_{r=1}^{m} (-1)^r \dbinom{n}{r} H_{r}$

$\implies \sum_{k=1}^{n-1} \left((-1)^m \dbinom{k-1}{m} - 1 \right)\dfrac{1}{n-k} = H_{n} \left[ (-1)^m \dbinom{n-1}{m} - 1 \right] - \text{S}$

$\implies (-1)^m \sum_{k=0}^{n-2} \dbinom{k}{m} \dfrac{1}{n-k-1} - H_{n-1} = (-1)^m H_{n} \dbinom{n-1}{m} - H_{n} - \text{S}$

$\implies \text{S} = (-1)^m \dbinom{n-1}{m} \left( H_{m} + \dfrac{1}{n} \right) - \dfrac{1}{n}$

$\therefore \sum_{r=1}^{m} (-1)^r \dbinom{n}{r} H_{r} = (-1)^m \dbinom{n-1}{m} \left( H_{m} + \dfrac{1}{n} \right) - \dfrac{1}{n} \quad \square$

For $$m \geq n$$, we see that the first term vanishes and we are left with the special case,

$\sum_{r=1}^{n} (-1)^r \dbinom{n}{r} H_{r} = - \dfrac{1}{n}$ · 1 month, 4 weeks ago

There is no typo. It is a finite sum and the answer should be valid for any $$r$$ and $$n$$. · 1 month, 4 weeks ago

I have answered for a general case. I have used $$n$$ and $$m$$ instead of $$r$$ and $$n$$. Btw, what's your approach? · 1 month, 4 weeks ago

If we define $a_n \; = \; \left\{ \begin{array}{lll} -\tfrac{1}{n} & \hspace{1cm} & n \ge 1 \\ 0 & & n = 0 \end{array}\right. \hspace{2cm} b_n \; = \; \left\{ \begin{array}{lll} H_n & \hspace{1cm} & n \ge 1 \\ 0 & & n = 0 \end{array} \right.$ then $H_n \; = \; -\sum_{k=1}^n \binom{n}{k} (-1)^k \frac{1}{k} \hspace{2cm} n \ge 1$ so that $b_n \; = \; \sum_{k=0}^n \binom{n}{k} (-1)^k a_k$ and so, since we are using the Binomial Transform here, $a_n \; = \; \sum_{k=0}^n \binom{n}{k} (-1)^k b_k$ and hence $\sum_{k=1}^n \binom{n}{k}(-1)^k H_k \; = \; -\frac{1}{n} \hspace{2cm} n \ge 1$ I presume that there is a slight typo in the question - $$\binom{n}{k}$$ instead of $$\binom{r}{k}$$. Of course, this makes no difference if $$n > r$$. · 1 month, 4 weeks ago

There is no typo. It is a finite sum and the answer should be valid for any $$r$$ and $$n$$. · 1 month, 4 weeks ago

Surprisingly, there is a general closed form too! I discovered it using challenge 4. BCC 4 is indeed an intriguing problem. · 1 month, 4 weeks ago

After some experience in finite calculus, it is not difficult to observe

$$\displaystyle \Delta\left({(-1)}^{k-1}\binom{r-1}{k-1}\right) = \binom{r-1}{k} {(-1)}^k - \binom{r-1}{k-1}{(-1)}^{k-1} = {(-1)}^k \binom{r}{k}$$

And we would be using SBP -

$\displaystyle \sum {u(x) \Delta(v(x))} = u(x)v(x) - \sum {v(x+1) \Delta(u(x)}$

Therefore, our finite sum -

$\displaystyle \sum {\binom{r}{k}{(-1)}^k H_k \delta k}$

$\displaystyle = {(-1)}^{k-1}\binom{r-1}{k-1} H_k - \sum {\binom{r-1}{k}{(-1)}^k \frac{\delta k}{k+1}}$

$\displaystyle = {(-1)}^{k-1}\binom{r-1}{k-1} H_k - \frac{1}{r}\sum {\frac{r}{k+1}\binom{r-1}{k}{(-1)}^k \delta k}$

$\displaystyle = {(-1)}^{k-1}\binom{r-1}{k-1} H_k + \frac{1}{r}\sum {\binom{r}{k+1}{(-1)}^{k+1} \delta k}$

Using our first difference in the sum form,

$\displaystyle = {(-1)}^{k-1}\binom{r-1}{k-1} H_k + \frac{1}{r} {(-1)}^k\binom{r-1}{k}$

Now using the limits $$1$$ and $$n+1$$,

$\displaystyle = {(-1)}^{n}\binom{r-1}{n} H_{n+1} - \frac{1}{r} {(-1)}^n\binom{r-1}{n+1} - 1 + \frac{r-1}{r}$

$\displaystyle = {(-1)}^{n}\binom{r-1}{n} \left(H_n + \frac{1}{r}\right) - \frac{1}{r}$ · 1 month, 3 weeks ago

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## Wednesday, March 05, 2008

### [TECH] Sorting partially sorted sequences.

Partially Sorted Sequence: A sequence k1,k2,...kn of n keys such that any key kj differs from its sorted position by atmost d as an example for d=3 we have 3,5,6,1,2,4 I came across this problem recently, turns out that we can solve this problem in several ways in O(nlog(d)) time. I found an interesting and simple way to solve this problem, a simple observation reveals that the smallest element in the entire sequence of n keys exists in the first d keys.

/*Build a heap(min) H with first d elements
*in O(d) time
*/
for(i=d;i<n;i++){
DeleteMin(H); /*output this key*/
InsertHeap(H,ki);
}
/*Note the Heap will have atmost d keys in it
at any time so the Insert and Delete can be done in
O(log(d)) worst case.
*/

(n-d)log(d) = O(nlog(d))


text:
# Why is denser air heavier than less dense air?

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It does not become lighter in the sense of less massive, rather it expands to take up more volume. The principle of buoyancy (since we are immersed in an atmosphere) then makes it appear lighter.

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The air expands as heated becauses the average separation of the gas molecules get larger at higher temperature. Hence, there are less gas molecules per unit volume, that is the reason why the it is lighter or density is smaller.

The website is trying to explain the process of the mechanism of vertical motion of atmosphere caused by adiabatic expansion or shrinkage. There are basically two modes:

When the air is near the ground, it is heated and it rise up because it is less dense than that of cold air. During the process of rising, it expands more as the pressure is smaller, then it gets cooled. This called adiabatic cooling. The reverse process occurs on the cold air.

You can refer to the ideal gas law and the for basic process of the ideal gas: isothermal(constant T),isobaric(constant P),adiabatic/isotropic(Q=0/constant S) and isochoric(constant V) process. The following link may help you.

http://hyperphysics.phy-astr.gsu.edu/hbase/HFrame.html

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It does not become lighter in the sense of less massive, rather it expands to take up more volume. The principle of buoyancy (since we are immersed in an atmosphere) then makes it appear lighter.

So if there was no atmosphere like on the Moon then denser air will not be lighter than less dense air?

Merged post follows:

Consecutive posts merged
The air expands as heated becauses the average separation of the gas molecules get larger at higher temperature. Hence, there are less gas molecules per unit volume, that is the reason why the it is lighter or density is smaller.

What I don't get is that by density we mean mass per unit volume but when we say 'lighter' we actually mean 'weights less or of less weight'. So why does less mass per unit volume mean less weight?

Edited by Uri
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So if there was no atmosphere like on the Moon then denser air will not be lighter than less dense air?

Assuming the same mass of air they would weigh the same on the moon.

Assuming the same volume the denser air would weigh more due to the additional mass in the same volume.

No atmosphere means no buoyant forces to complicate matters.

Most would say less dense air is lighter generally, but it depends on what you are comparing and what you mean by lighter.

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Less dense air is going to have less mass (and therefore less weight) per unit volume, yeah. That's what "less dense" means. A mass of air that is less dense than the air around it is going to rise up, because the denser, heavier air around it is going to be pulled down more and force the less dense air up and out of the way. That's buoyancy.

Think about a rock dropped into the ocean. Sure, obviously, the ocean is heavier than the rock. But the rock is heavier per unit volume, so it sinks.

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Assuming the same mass of air they would weigh the same on the moon.

Assuming the same volume the denser air would weigh more due to the additional mass in the same volume.

I still don't get it. Why is mass always compared to volume? Mass is just the quantity of matter, regardless of the volume. So shouldn't two objects of two different volumes still have the same weight regardless of their volume if their mass is the same or is mass always a function of volume?

That is what I have been taught but it could be wrong.

Edited by Uri
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I still don't get it. Why is mass always compared to volume? Mass is just the quantity of matter, regardless of the volume. So shouldn't two objects of two different volumes still have the same weight regardless of their volume if their mass is the same or is mass always a function of volume?

That is what I have been taught but it could be wrong.

Which has more weight, 10 kg of lead or 10 kg of feathers? What about if you weighed them on the moon?

Which has more mass, 10 lb of lead or 10 lb of feathers?

A lot depends on how you define your terms. Does a boat weigh nothing when floating in water - or just enough to displace an equal weight of water?

Edited by J.C.MacSwell
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So shouldn't two objects of two different volumes still have the same weight regardless of their volume if their mass is the same or is mass always a function of volume?

The answer depends on what you mean by "weight". Weight has multiple meanings.

• Legal weight. Legally (and colloquially), weight is a synonym for mass. A one pound can of beans weighs 10 pounds on the surface of the Earth, on the Moon, and on the space station.

• Gravitational weight: $W_g=m*g$, where m is the object's mass and g is the gravitational acceleration of the object. This is the definition of weight used in most introductory physics texts. One problem with this definition: It is unmeasurable.

• Scale weight: $W_s = \sum F_{\text{ext}}-W_g$, where Fext are the external forces (including gravity) acting on the object. This is the quantity measured by an ideal scale. This is the definition of weight used in some undergraduate physics texts very and is closely aligned with the definition of weight used in general relativity.

So, if you use gravitational weight as your definition, two objects with the same mass have the same weight. If you use scale weight as your definition, two objects with the same mass can have vastly different weights. One of the external forces acting on an object is buoyancy, and buoyancy is a function of the object's volume.

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Which has more weight, 10 kg of lead or 10 kg of feathers? What about if you weighed them on the moon?

Actually they should weigh the same in a similar gravitational field. Their mass should also be same. Unlike weight, mass is the amount of material an object contains and does not change under a different gravitational field.

A lot depends on how you define your terms. Does a boat weigh nothing when floating in water - or just enough to displace an equal weight of water?

If the boat is floating it's weight is equal to the weigth of the volume of water displaced.

Merged post follows:

Consecutive posts merged
Most would say less dense air is lighter generally, but it depends on what you are comparing and what you mean by lighter.

But why is less dense air lighter? By 'lighter' I mean has less weight or weights less and that is also the dictionary's definition of 'lighter'.

What I mean is that even if air expands in volume it's mass and hence it's weight should stay the same because mass is simply the quantity of material which a body contains, irrespective of its volume.

Edited by Uri
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Actually they should weigh the same in a similar gravitational field. Their mass should also be same. Unlike weight, mass is the amount of material an object contains and does not change under a different gravitational field.

If the boat is floating it's weight is equal to the weigth of the volume of water displaced.

Fair enough, you have used pretty much the same definition of weight in each case.

However, if you used a scale, 10 kg of lead would weigh more than 10 kg of feathers due to the greater buoyant force of the atmosphere on the feathers. That is more or less the definition some others have used in this thread. They have allowed buoyancy to be factored into the terms heavier or lighter. Still valid, but they are weighing things differently.

OTOH, mass is always defined the same way. So on the moon, with no buoyant forces from the atmosphere 10 kg of lead or feathers weigh the same, even on a scale.

The difference might be subtle, but become quite significant with respect to something as "light as air".

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Most would say less dense air is lighter generally, but it depends on what you are comparing and what you mean by lighter.

But why is less dense air less massive? This is what I don't understand.

By 'lighter' I mean has less weight or weights less and that is also the dictionary's definition of 'lighter'.

What I mean is that even if air expands in volume its mass and hence its weight should stay the same because mass is simply the quantity of material which a body contains, irrespective of its volume.

Edited by Uri
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By 'lighter' I mean has less weight or weights less and that is also the dictionary's definition of 'lighter'.

The problem here, Uri, is that you are stuck on the idea of weight being mass times gravitational acceleration. That is not the only definition. See post #9.

From http://www.thefreedictionary.com/weight (emphasis mine):

1. A measure of the heaviness of an object.

4. The measured heaviness of a specific object: a two-pound weight.

Note well: Gravitational weight cannot be measured. These definitions are alluding about some other aspect of the word "weight". They are in fact alluding to what I called in post #9 "scale weight".

Gravitational weight obviously is not the meaning of weight used in the article you cited in the original post.

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The word "lighter" has several meanings.

When comparing two masses, the lighter mass has less mass than the heavier mass. Eg, 1 kg of lead is lighter than 2 kg of styrofoam.

When comparing two substances, the lighter substance has a lower density than the heavier substance. Eg, styrofoam is lighter than lead, or your example, hot air is lighter than cold air.

As well, lighter can simply mean the weight is smaller. Eg, scuba gear is lighter when you are underwater (due to buoyancy), or you're lighter on the moon (due to lower gravity).

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What I mean is that even if air expands in volume its mass and hence its weight should stay the same because mass is simply the quantity of material which a body contains, irrespective of its volume.

The total mass of air has the same mass, yes. But any given volume of that air has less mass. That's what matters. It's like replacing 100 cubic meters of "heavy air" with 200 cubic meters of "light air." The light air gets forced upwards, to make room for heavier air. Yes, it contains all the same molecules. Yes, the total mass of each total volume is the same. But a cubic meter of light air is lighter than a cubic meter of heavy air, and that's what matters. That's all that's being said. Similarly, the whole weight of the Earth's atmosphere is a whole lot more than a cannonball, and yet you drop a cannonball and it falls right through the air, because, we say, it's "heavier than air." And by that we mean heavier than a mass of air occupying the same volume, i.e denser.

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But why is less dense air less massive? This is what I don't understand.

By 'lighter' I mean has less weight or weights less and that is also the dictionary's definition of 'lighter'.

What I mean is that even if air expands in volume its mass and hence its weight should stay the same because mass is simply the quantity of material which a body contains, irrespective of its volume.

This is a matter of definition. "Density" is the amount of mass per unit volume; for example, water (at standard temperature and pressure, "STP") has a density of 1 g per cm3. In contrast, gasoline has a density of around 0.75 g/cm3, and mercury has a density of 13.5 g/cm3. Gasoline is less dense than water; mercury is more dense. 100 mL of water will weigh 100 g, while 100 mL of gasoline will weigh about 75 g. It weighs less by definition.

The density of air at STP is about 0.0012 g/cm3. Air density varies with pressure (higher pressure raises density), temperature (density decreases with temperature), humidity (moist air is more dense than dry air), and composition (gases like CO2 increase the density).

This is the principle upon which hot air balloons rely. Hot air is less dense than cold air. So, if you heat the air inside your balloon, it becomes less dense. Since the pressure has to remain the same as the surrounding atmosphere (hot air balloons are not sealed), and the volume remains constant, raising the temperature means that the number of moles of gas included must decrease: PV = nRT, or n = PV/(RT). "R" is the gas constant, and does not change. Gas (air) leaves the balloon, so that the whole system is less massive -- therefore, the air in the balloon is less dense, and the whole thing rises.

Make sense?

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A few corrections.

This is a matter of definition. "Density" is the amount of mass per unit volume; for example, water (at standard temperature and pressure, "STP") has a density of 1 g per cm3.

Actually, that's what it is at 4 C, the temperature at which it is densest at standard pressure. However, it is usually very close to 1 g/cm3 at other temperatures.

humidity (moist air is more dense than dry air),

Actually, moist air is less dense than dry air (probably a typo on your part considering you know the ideal gas law). This is counterintuitive to people who note that liquid water is heavier than air. However, the molecular weight of water is 18 g/mol, compared to the heavier 28 g/mol for N2 and 32 g/mol for O2. Since the density of gases is related to the molecular weight of the gas, this makes moist air less dense than dry air.

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A few corrections.

Actually, that's what it is at 4 C, the temperature at which it is densest at standard pressure. However, it is usually very close to 1 g/cm3 at other temperatures.

Actually, moist air is less dense than dry air (probably a typo on your part considering you know the ideal gas law). This is counterintuitive to people who note that liquid water is heavier than air. However, the molecular weight of water is 18 g/mol, compared to the heavier 28 g/mol for N2 and 32 g/mol for O2. Since the density of gases is related to the molecular weight of the gas, this makes moist air less dense than dry air.

Moist air is heavier though, when it has liquid water suspended in the air. I think a lot of people consider that moist air.

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I stand corrected: humid air just feels twice as dense as dry air

I've been to Texas: I know

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Originally posted by GDG:

Since the pressure has to remain the same as the surrounding atmosphere (hot air balloons are not sealed), and the volume remains constant, raising the temperature means that the number of moles of gas included must decrease: PV = nRT, or n = PV/(RT). "R" is the gas constant, and does not change. Gas (air) leaves the balloon, so that the whole system is less massive -- therefore, the air in the balloon is less dense, and the whole thing rises.

Here, V must increase to keep P and n constant. n is the mole of the air inside the balloon, it cannot be changed as hot air doesn't leak out. With the constant n and larger V, its density is smaller, lighter.

Originally posted by Mr Skeptic:

However, the molecular weight of water is 18 g/mol, compared to the heavier 28 g/mol for N2 and 32 g/mol for O2. Since the density of gases is related to the molecular weight of the gas, this makes moist air less dense than dry air.

I doubt about this statement. What you talk about here is the molar mass, it is rather different with density. If we consider the moist as a mixture of water(liquid) and dry air, apparently, the density of moist air will be larger.

I can also explain in your point, 1 mole of dry air occupies much larger volume than 1 mole of water. The differences in volume is much larger than the difference in mass when we consider the same quantity of dry air and water. Overall result is that density will be increased when water droplet is added.

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Here, V must increase to keep P and n constant. n is the mole of the air inside the balloon, it cannot be changed as hot air doesn't leak out. With the constant n and larger V, its density is smaller, lighter.

I'm pretty sure GDG had it right for a hot air balloon. Your right that it doesn't or shouldn't leak out, but it certainly is free to get out, as it is open at the bottom.

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Overall result is that density will be increased when water droplet is added.

But you don't just "add" a water droplet. The water molecules in the air effectively "knock out" some of the other molecules in the air. If they didn't, the pressure would go up because there would be more molecules per unit volume, causing more impacts per unit area, resulting in higher pressure. But, we are assuming constant pressure and temperature when comparing the two.

And, since most of the molecules in air (oxygen and nitrogen being the overwhelming majority) are heavier than water, the water results in less overall mass per unit volume for a given pressure and temperature of humid air.

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I think that what he's saying is that air plus liquid water (eg mist or fog) is more dense then the air without that liquid water. That's true, but most people consider moist air to be air with lots of water vapor, in which case it will be less dense than dry air.

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Here, V must increase to keep P and n constant. n is the mole of the air inside the balloon, it cannot be changed as hot air doesn't leak out. With the constant n and larger V, its density is smaller, lighter.

Once the balloon is inflated, V does not increase. Hot air balloons do not work the same way as toy rubber balloons. The bottom of a hot air balloon is open: it has a burner suspended below the opening, which the operator (pilot?) flicks on every few minutes to heat the air inside the balloon. (Believe me, riding in a hot air balloon is a much noisier experience than you would expect. The burners are surprisingly noisy.) The balloon fabric does not stretch. Since P and V are constant, n must decrease with increasing T. Thus, hot air escapes from the bottom of the balloon, leaving a smaller mass of (hot) air inside the balloon.

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# Re: Rain droplets falling on a table

This questions is almost exactly similiar to the the following question, with an extra condition :

Rain droplets falling on a table

Suppose you have a circular table of radius R. This table has been left outside, and it begins to rain at a constant rate of one droplet per second. The drops, which can be considered points as they fall, can only land in such a way such that they impact the surface of the table. Once they strike the table, they form a puddle of radius r, centered at their point of impact. What is the expected number of droplets it takes to cover the table in water?

Now, suppose every water droplet that falls on the table dries out after $(R/r)^2$ seconds. What will the graph of the propability $P(N)$ that the table will be covered with N droplets versus N look like? Can someone please help me with this?

-
after the drop hits the table, should we assume the drop spreads out to cover the area $\pi r^2$ in addition to that which is already covered? For example, if the first drop is hit by the second drop then is the total area covered by those two drops $2\pi r^2$? Also, is the drying event discontinuous or should we assume some linear evaporation to zero taking $(R/r)^2$ seconds? I need to know the rules before I play. –  James S. Cook Sep 1 '12 at 4:47
" if the first drop is hit by the second drop then is the total area covered by those two drops $2πr^2$ ?", i that that is why we are talking about probability, the 2nd drop can fall in any way, the area covered need not always be $2πr^2$. The water droplets evaporate after $(R/r)^2$ seconds. Thats all. I don't think the mechanism will matter. Sorry for late reply. –  Swapnanil Saha Sep 1 '12 at 13:40
You're standing on a flat floor. Think about pouring a bucket of water on your feet. Pour another bucket on your feet. Surely the puddle is larger for two buckets than it would be for a single bucket. If we apply this idea to raindrops then it doesn't matter if the new drop hits where an old drop already hit. The new drop will just spread out to cover $\pi r^2$-additional area. The number of drops should be measured by area of water divided by $\pi r^2$. –  James S. Cook Sep 1 '12 at 16:00
I dont think this is a math question in its current form because it requires the knowledge of physical behavior of water. Specifically, we need to be given information on the area covered by two drops of water that fall within distance $r$ of each other. Physically, these drops will spread over $2\pi r^2$. However if one thinks of the covered areas as sets then the collectively covered area is the area of union of the two sets, which is less than $2\pi r^2$. –  Ankur Sep 2 '12 at 0:29


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## Saturday, March 28, 2020

### Putting Humanity First

When you unfortunately need to go to a friend’s funeral, you also have to figure out how to get there.  You need to make arrangements for hotel stay, travel, find out exactly when and where the services will be.  Maybe travel won’t go to plan, and there are things you can rightly complain about. But the main reason why you go to the funeral is for your friend you lost. It’s about the celebration of life, not the travel details.

At present, we have a lot on our plates. Converting an in-person class to online/virtual format is a huge task, and all the more challenging under the weight and stress of a crisis. But in the big picture how we deliver our content online is like the travel details to attend a friend's funeral. We have a lot of huge issues now that we never had to deal with before.

Lots of things don’t compute right now, since coronavirus arrived. I don't have a lot of answers. I do have some thoughts. Our whole world, our daily lives, even potentially our very own health and the health of loved ones are threatened. We are witnessing leaders call for trading lives to save the Dow Jones average. We are hearing reports of healthcare workers on the front lines, without proper equipment and support. Family businesses are going out of business. We are also seeing racism against Asians and Asian Americans.

In light of this, going on with business as usual with our teaching is for me something I am struggling with. I know this is a personal viewpoint. Some may need to dive into the details of a class to cope and distract. I get that, and I’m not arguing against what others are doing in this same situation.  What I am saying is that I personally can’t just sit back and say nothing or just teach my spring quarter class just like I did the last time. The world has changed, and so have I. In April, May and possibly beyond, we’ll be in the midst of toughest part of the first wave of the coronavirus. You can’t just ignore all the bodies in the corner, while going on about integration by parts without acknowledging what’s happening.

At the same time, we can’t lose sight of the goals of our classes. We still need to educate. Students still need to learn and build up for their lives and careers.  I’m not advocating for not teaching content either.

The world has far too many damaging, false dichotomies. Choosing between lives and the economy is the latest, terribly inhuman example. Others suggest we must choose between skills and conceptual understanding. Some suggest, falsely, that the choice is between helping underrepresented students or white students, as if kindness and inclusion is like pie. False dichotomies are red flags. It signals an agenda, bias, lazy thinking, or unwillingness to dig into the details of a complex issue.

The question for me at least is whether we will choose to be responsible to each other and ourselves and behave like grown ups.  Responsible adults study the data and evidence, and get on with solving problems in this new normal.

Specific to education, teachers can inspire and attend to basic skills, as well as ask students in earnest, “How are you doing?” Teachers can be flexible with lessons because the news we all just received is truly bad. Teachers can keep the door open, create space in class and in office hours (even virtually), check in on students, and show their own humanity, reasoning, and maturity in the face of a global crisis.  We are all scared. We all feel a range of emotions, and none of us have it all figured out or know what the future holds.

I’m teaching integral calculus online starting in about a week, and I plan to create spaces to build some community and have time set aside for us to discuss the global pandemic and how students are dealing with it. We (students and instructors) all have smart things we can share that will help others, and we can teach and learn to be better to one another. I'll also spend a lot of time developing an assessment plan that is standards based/mastery based, so that the stress of testing is taken off. The focus should be on learning (as always and even more important now).  That in a nutshell is how I'll try to be a responsible adult in my role as a math instructor. I teach students first and foremost, with Math as the context.

Yes, we can teach the usual things, all the while putting humanity first.

## Wednesday, March 18, 2020

### Thoughts On Human-Centered Virtual Teaching (Coronavirus)

My head is still spinning. We're living through the start of a global pandemic. I've had to do so much at home and work to prepare, that's it's been tough to even get enough sleep and take care of myself. Or fold that giant pile of laundry that's been sitting in the corner for days.

On to teaching... I've read many posts on twitter, in email, and articles about what to do to go virtual with our teaching. These discussions are great, and I've already learned a lot. People are stepping up and sharing really useful, real-world advice.  In this post I'm going to focus on one aspect of all this. I call it human-centered teaching.

When we teach in person, it's easier to get to know your students, mentor during office hours, run into students on campus and have a conversation. All these things are part of the package of student-centered instruction and being on campus, and many of these things are harder or nonexistent or must be recreated in different way in online settings. I have not figured this all out by any stretch. I don't think it's something we should feel bad about, and I know over time we will develop and figure some of this out.

I'm on quarter system, so we actually start spring term in early April. But even if we were on semesters I'd do the same thing I'm going to mention.  On the first virtual meeting day, I have two main goals. One is to meet and set virtual class norms (e.g. everyone turns on video, raising hands, muting), and practice using the technology. All of this is new, so we're going to need to get comfortable with our new normal routines.

The second goal is to open up a space for students to share some of their thoughts and feelings and make human connections with their classmates and me. This is the most important thing, in my opinion, to accomplish on day 1 (and keep going). The context surrounding us is a global pandemic. Millions will get sick, and globally many will die or lose someone close. This is hard stuff, and it's desperately real. Students who are young and just at the beginning of their adult lives are dealing with the uncertainty and turmoil of an enormous natural disaster, all the while going through the disruption of having to move home and be separated from their friends. In light of this, it's even more important to try and create a team spirit in our classes.

How will I try to do this? I plan to set aside class time for students to share their thoughts on the coronavirus pandemic. One specific strategy I used to get people to participate online (zoom conference calls) is to use a google doc with a 3xN table (N= number of students +1).  Here's a hypothetical example:

Students pick a row and are paired with another student with the same number.  I give them a pronpt, and they type at the same time, while I observe and make comments. (Using this strategy also needs to be normed.)

Further, I plan to create a chat space on Canvas for discussion strategies for how to study, self care, random questions or comments. I'll send out regular emails to keep the door open for students, and invite students to virtual office hours (or 1-1 meetings) to discuss math, learning, be a mentor, just be there to listen. I'm sure I'll have to adapt and add more things, but these are my initial plans to setup a framework for human interaction.

Another important mindset I'll focus on is keeping a healthy perspective for myself. We are not trying to recreate our in-person classes. Almost all of us are going online for the first time. Trying to cover everything just like before and in the same way is not going to happen. My focus will be on being present in each moment, trying my best, and working to improve each and everyday (Shokunin Spirit).

Lastly, I'd like to mention that we should be gentle on ourselves. We all going through a lot, and we should allow ourselves to be human, to make mistakes or not have the "best" class session. What truly matters is that we care and our students know we care about them. If it doesn't go like we planned, it's ok.

That's my initial plan for implementing human-centered virtual teaching. Sending positive thoughts. Stay safe and healthy!

## Thursday, March 5, 2020

### Draft Plans for Running a Virtual Class

Let me frame this post first with some initial thoughts... Sometimes life throws us all a curveball. Coronavirus is here, and CDC and others have warned us that major disruptions are coming. This is a very serious situation with a lot of levels, nuances, emotions, and sadly tragedy. People have and will suffer, and first and foremost I am wishing you all the best and hope you can stay strong and safe.

Some basic things I'm doing now to help keep my students and myself safe:
1. Inform my students about what they can do.
2. I bring cleaning supplies to class daily to clean desks.
3. I also bring tissues and will bring hand sanitizer when it's in stock again.
4. I've adjusted my attendance policy, and am asking students to stay home if they feel sick.
School closure is a strategy used in country for social distancing. This is used to flatten the epidemic curve to slow spread, provide necessary time and capacity for medical care, and a necessary part of battling a disease to keep people safe.  I'm for whatever keeps us the safest.

What this means is many of us may have to teach virtual classes. I'm on quarter system, and we are wrapping up winter term. Spring term starts at the end of March, and that could be when things change significantly. People on semester system might need to switch in the middle of the term, so that brings its own set of issues.

A bit about some of my relevant background. Many of us have not taught online. I haven't, at least not a fully online class. I have taught hybrid online here and there, and have run faculty development workshops using zoom and a couple other tools.  Tools like zoom are going to make this easier, and I think the tech is there to make this work well enough.

I'm sharing my own draft plans via google slides (linked HERE and embedded below). These are my draft thoughts on how to organize class via a simple framework, which others might find useful as a starting point.  I'll update this regularly as things develop. I know there are much more experienced people on this issue, so find and share resources you have.

At the bare minimum, an instructor can give lectures via zoom at the scheduled class time, and setup office hours via zoom. I personally will do much more, since I use IBL methods, but I think it's worth sharing that switching to virtual is not that scary, and certainly worth it given the situation.

## Wednesday, January 15, 2020

### Summer 4-Day IBL Workshops 2020

AIBL is offering a range of IBL Workshops in 2020. This post is about one of the types. Other types of workshops will be posted here soon.

Our "classic" IBL Workshop is a 4-day summer intensive workshop. We've been running these since 2006, and have developed these over time to make them as good as we can. A summer intensive workshop is one of the best ways to get started with IBL or sharpen up your IBL skills. Our main target audiences are math instructors at 2-year, 4-year, other graduate degree granting institutions. We especially encourage faculty in the early stages of their career to attend one of our workshops. Community college math instructors are also welcome, and I want to point this out specifically.

Why attend a workshop? The main benefit is that it supercharges your teaching. You get to work with experienced facilitators and join a community of math instructors going through the same type of transition to active, student-centered teaching. It's fun, you make new friends and colleagues, and you could teach in ways that transforms the learning experience for your students.

Our motto is People First Professional Development. We don't have specific curricula or a set model of IBL. We work with individuals and put at the center of our work you, the participant, and your students. We work from that starting point to help instructors find a good solution within the IBL framework.

Workshops are funded generously by the National Science Foundation (NSF DUE 1525058). Most of the cost of attending is paid for by the grant. Participants are responsible for paying the registration fee, some meals, and travel costs. Early-career faculty (e.g. postdocs, new instructors, assistant professors) are eligible for travel scholarships of up to $500. We work with you and your department to figure out how to get you there. The LA Workshop is June 16-19, 2020 at the Staybridges Suites in Torrance, CA. The DC Workshop is at the MAA Carriage House, Washington DC, June 23-26, 2020. For those who cannot travel, we are planning a virtual IBL Workshop in July. More details are forthcoming. We realize that not everyone can travel and that work travel increases significantly our carbon footprint. We are working hard to develop and provide "next gen," sustainable professional development. More details about traveling workshops and workshops for courses for future elementary teachers are coming soon. Hope to see you at a workshop this year! Links: ## Tuesday, November 12, 2019 ### Interview: Professor Matthew Boelkins, Grand Valley State University and Active Calculus Hi everyone! A massive thanks to Professor Matthew Boelkins, Grand Valley State University, for taking the time to share some info about his project, Active Calculus. -SY 1. Please tell us about yourself. I've had the good fortune to serve on the math faculty at Grand Valley State University in Allendale, MI, for more than 20 years. I'm also one of the two editors-in-chief of PRIMUS (Problems, Resources, and Issues in Mathematics Undergraduate Studies), and have held that position for 5 years, following 5 years as associate editor. My interests span a wide range of undergraduate teaching and learning of mathematics, but much of my recent creative energy has been focused on developing textbooks that encourage active learning. 2. Briefly describe “Active Calculus”? How much does it cost to the student? Active Calculus (there are both single and multivariable texts; I'm the lead author of the single variable one, and will focus on that in my responses) is a free, open-source textbook designed for a standard calculus sequence taught from an active learning perspective. Rather than lots of worked examples, most of the text is structured around activities for students that are designed to be completed before or during class, the latter with encouragement and feedback from peers and instructor. Interested people can learn more at https://activecalculus.org/. 3. What are some of the reasons why you decided to write “Active Calculus?” There were two main reasons. First, I read an article in MAA FOCUS about the \$21M home built by the author of a popular calculus textbook.  I figured that if the author had earned that much in royalties, the publisher had likely made \$210M. It was no longer tenable for me to ask my students to pay \$150 for a textbook that had ideas in it that had been well understood by humankind for decades, even centuries.  I wanted to write something that would be free for them.

In addition, while at the time I began writing (around 2010) I didn't fully know the scholarship that tells us why active learning is better,  I used a lot of active learning in my own teaching and had found my students to be more successful than when I used to lecture.  Having developed a collection of activities for calculus to use in my own teaching, I had the idea to use those as the foundation for the textbook, and that's what led to Active Calculus.  When the Freeman report came out in 2014 (https://www.pnas.org/content/111/23/8410), that provided added motivation for making the text a good one.

4. What is a typical day like in a course using “Active Calculus” when you teach using it? Can you use it with flipped learning or with IBL methods?
For each class meeting, I prepare a written script with estimated times.  In advance of most classes, students complete a "daily prep assignment" that normally consists of a short reading, a preview activity from the text, and 1-2 additional questions; students spend 30-45 minutes completing these, and their work is graded on effort and completeness for 5\% of their semester mark.  Most classes then begin with a "daily prep debrief & discuss" (6-8 minutes) where students check in with one another and see what questions they might want to discuss as a class.

From there, we usually engage in some brief (5-7 minutes) lecture & discussion to build on daily prep and set stage for an activity.  Students then work in groups of 3-4 on an activity from the text for 15-20 minutes, followed by or including some discussion for closure, transition, and new ideas (adding another 5-10 minutes), and then we are on to the next activity for 15-20 minutes.  Often I teach our 4-credit calculus class on a schedule with two 2-hour meetings a week, so we basically rinse and repeat this schedule for a second hour, but without a daily prep assignment to start the 2nd hour.

I've written a blog post that has some more information and detail, including references to key preface sections of the text for students and instructors, as well as a video for students on how to use the text: https://opencalculus.wordpress.com/2019/08/05/how-to-use-active-calculus/

I think the text is particularly well-suited to a flipped learning setup:  when I have students complete daily prep assignments, they are doing some key basic learning on their own outside of class.  The in-class activity-driven style also fits with engaging students during class in the some of the most important and demanding work of the course.  Some of my GVSU colleagues have created screencasts that accompany the text, and these would work especially well for a flipped experience:

For IBL practitioners who have worked with that approach using more traditional texts like Stewart or Hughes-Hallett, I think you'd find Active Calculus to be a suitable companion for such a course.

5. What are some of the responses by faculty and students after using “Active Calculus”?
For views of some faculty who have reviewed or used the text, see https://open.umn.edu/opentextbooks/textbooks/active-calculus-2-0.  I get a lot of email traffic about the text, and many of those responses express gratitude for the text.  Recently, an instructor who is a first-time user this fall sent me a very kind thank-you note: "I wanted to write to say how much I've appreciated Active Calculus.  I was a bit suspect (I've been using Stewart), but I am so impressed with it.  I like the problems, the students working in class, the organization ... it's really quite good.  I'm excited to be teaching this class again.  And thanks for saving our students some money as well."

This fall, I also surveyed Active Calculus users via my email list and Twitter, and in an open comment part of the survey, I got additional feedback such as:

"I appreciate you and your work on this so much. It's really let me teach calculus in the way I want to without having to create the material from scratch. A million thanks."

"The book is well thought out with great examples. When I teach the class, I really feel like I am actively working through the book with the students. One comment our Quantitative & Symbolic Reasoning Center director told me was that students in Active Calculus sections are asking why a concept is true and being stuck on the theory, while students not using AC are asking questions on the algebra and not thinking about the concepts of the course. "

"I really appreciate the structure and the way that the text works. It has been a game changer for my teaching of Single Variable Calculus! Our pre-calc teacher is now using Active Prelude as one of her core texts and we are excited to see how that influences students being prepared for my class."

"I've been using Active Calculus since my first year of teaching AP Calculus AB and I can't imagine using any other curriculum.  I cannot thank you enough for the time and energy that you have put into your work and my hundreds of students over the years thank you as well."

As to what students say, I think that many students find the book different at first, and thus at times frustrating.  They have been trained to expect a book that has lots of fully-worked examples, followed by exercises that are similar to the examples.  Active Calculus is not that way.

Some of the instructors who responded to my survey commented on what students say:

"It's an excellent book. Students and I both really appreciate its clarity and accessibility. Just yesterday I was recommending to a student the very nice "Summary" bits at the end of each section, and they said, "oh, that's exactly what I was looking for.""

"I really like the text and got mostly positive comments from the students on the course evaluations last year.  In the past, the comments were almost all negative or neutral."

"I asked my students and they expressed appreciation for how easy it is to read. Some students also said they wished there were more examples, which gave me an opportunity to remind them why there aren't more examples. :)"

6. How widespread is “Active Calculus”?
One of the challenges of having a text that's free and available online is knowing exactly where it is being used.

I'm aware of at least 17 4-year universities, 2 2-year colleges, and 5 high schools that have formally adopted Active Calculus as their required textbook.  People from at least another 30 institutions have responded to my recent survey to say they are using it fully in their own course, even though the text hasn't been adopted by all of their peers.

Here's the list of adopters I have at present:
California State University, Monterey Bay
Carroll College
Doane University
Dordt University
Lane Community College
Laurentian University (@ St. Lawrence College)
Lebanon Valley College
Lenoir-Rhyne University
Scottsdale Community College
Sonoma State University
St. Mary's College of Maryland
Texas Lutheran University
The College of Idaho
Vermont Commons School
Vernon Hills High School
Westfield State University
Westminster College (SLC)
Westmont College
Wyoming Seminary

7. Could this book be used in high school?
Absolutely, and several of them have publicly shared their experiences.  For instance, Dave Sabol of St. Ignatius High School has used AC for several years and writes about corresponding activities he has developed on his blog, https://therationalradical.wordpress.com/calculus-resources/.  Jim Pardun and Steve Korney of Vernon Hills High School recently presented on their work teaching AP Calulus using AC at the regional NCTM conference in Nashville.

The overall goal is to keep making both Active Calculus and Active Prelude to Calculus better.

For Active Calculus, the original text was written in LaTeX and later converted to PreTeXt, which allows the HTML output.  I'm realizing that I haven't yet taken advantage of many of the features PreTeXt offers, such as having better cross-referencing, a better index, and more interactive features.  A first goal is to make revisions to take advantage of those features.  I am also considering adding Sage cells to the text to offer some embedded computation and experimentation for students; having some interactive computational opportunities is a second thing I hope to incorporate in the not distant future.  And a third significant goal is to have some additional exercises, ideally with many of them focused on modeling and applications.

For Active Prelude, this is the first year the text has been public, so I'm waiting on some user feedback to see where to focus energies next.

9. Anything else?
People can learn more at https://activecalculus.org and https://opencalculus.wordpress.com/, and I always appreciate hearing directly from users or people interested in the text by email at boelkinm at gvsu dot edu.

## Thursday, October 24, 2019

Even Mr. Miyagi had student buy-in issues. If you've seen the movie, The Karate Kid, you know these "wax on, wax off" training scenes, which is clearly active, Daniel-san-centered teaching. You wouldn't just want to only watch videos or watch demos, if you need to face real humans in a karate tournament. It seems like avoiding a kick to the head might require more than factual knowledge of what a kick is and knowing about the existence of a block or an avoidance move.

Let's take a look at a key segment, where Daniel doesn't see the point of all his practice, and has a blow up.

When viewing this scene from a teaching point of view, it comes down to Mr. Miyagi not framing and signposting the activities.  "You're doing this to get stronger at ..."  Instead, he assigned Daniel exercises, and did not tell him what these exercises were for. So naturally Daniel was upset, because he could not see the connection to learning karate. Finally, Mr. Miyagi does some student buy-in work, and has Daniel demonstrate that he actually has strengthened his defense abilities. At that point, Daniel made a big step forward in student buy-in, after getting some positive feedback from this teacher.

(Now of course this a movie, and drama is needed. I wouldn't change the movie, nor is this a film critique.)

If we boil things down to their essence, the situation is (a) instructors know what the connections are and the purpose of the work, however (b) students don't always see (or can't see) these connections, since they lack expert insight.  Hence, it's vital for teachers to signpost, "We are learning in this way, because..."

Returning to The Karate Kid, at some point, there has to be authentic meaning to the work. Daniel fully buys in after the "anniversary scene", where it is revealed that Mr. Miyagi had suffered tragedy in his family life, while earning a congressional medal of honor for his military service to the United States. This experience provides much needed perspective for the young man. The next scene is Daniel practicing with intent on his own - a sign of full buy-in.  Game on!

This suggests that buy-in has multiple levels. It's one thing to see why you are doing an exercise. It's another to be fully committed to the path forward. This is where assignments and activities focused on the meaning and value of the work can be useful. This can include but is not limited to the value of problem solving, growth mindset, and inspirational content, where students have an opportunity to learn universal lessons about what their education is for.

Wax on, wax off,
Wax on, wax off...

## Friday, September 27, 2019

### Cold Calling Pitfalls

This blog post is about cold calling. Randomly calling on students can go well, but it's not so easy as pulling a name and asking, "What's the answer!?"

The pitfalls of basic random cold calling is that it's not an equitable practice in its barebones form. Some issues include (but are not limited to) the following.

1. Making a mistake in front of the whole class can be stressful. This is a high risk situation.
2. Not everyone thinks fast, so they may be in the process of figuring something out, and then called on the spot to have the goods, right then and there.
3. Thinking fast is not necessarily smart, and thinking slowly is not necessarily not-smart. But having to be quick on your feet biases the definition of smart as having the answer quickly in class on the spot.
4. Stereotype threat can be triggered. Some students are the only one of their identity in your class. If this student makes a mistake, then the stereotypes could be "validated."
What can we do?
Randomizing who contributes does have a positive facet. It spreads the chances around. That's good. The key is to do it in an inclusive way.

Here's one example:
Instead of going with something like "What's the answer!?"you can add think-pair-share into the mix.  First, state the question or task, and ask students to think first and the talk to their partner. Then when you select a person at random, the prompt could be, "Share with us what you discussed with your partner." What students share could be a partial answer, a question, a comment, the whole answer, or perhaps asking for another minute to finish their conversation. If framed as a way to generate discussion and engagement, rather than having to be right, this can reduce anxiety and be a more inviting experience.

Riffs: We can think of think-pair-share as the base layer, and add on or adapt layers above it. Instead of randomly calling on a pair, you could visit a few groups and see what they tried. You can usually very quickly find a pair willing to share. Quiet students who have a useful idea can be encouraged to share, and this specific teacher move is where you enact inclusion! The quiet student is invited in by the instructor and is validated for having something worthy of being shared.

Keeping track of who has shared will then help you spread chances equitably.  This riff works well earlier in the term, as you are developing student buy-in and comfort with talking about and discussing math. It's lower stakes, and doesn't put people on the spot.

Another riff is for classes that include student presentations. Assigning randomly the first chimers is a way to spread who makes the comments. Otherwise, it's the usual people raising their hands again and again. This can be done by selecting two students, then asking everyone to talk to their partner about the presented work. The selected students chime in first. They can ask a question, comment, or give a compliment. Again the contributions are done in a way, where students have time to think, discuss, and prepare remarks. Once the first chimers have made their comments, you can open the floor for further comments.

Cold calling can be warmed up with some additions and tweak. Using a small amount of time, allowing for thinking and talking, and taking perceived judgment off the table, creates an environment where students are less concerned about how they might appear and more focused on the Math.


text:
होम Alligation

# Alligation

### A vessel contains 60 litres of milk..

0
A vessel contains 60 litres of milk. 12 litres of milk taken out from it and replaced with water. Then again from mixture, 12 litres is taken out and replaced with water. The ratio of milk and...

### A trader has 50 kg of pulses, part of which he sells

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A trader has 50 kg of pulses, part of which he sells at 14% profit and rest at 6% loss. On the whole his loss is 4%. How much quantity is sold at...

### An amount of Rs. 680 was invested at 6%..

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An amount of Rs. 680 was invested at 6% rate of interest and another sum of money was invested at 10% interest. If the average interest on the total at the end of...

### In a zoo, there are deers and ducks…

0
In a zoo, there are deers and ducks. If the heads are counted, there are 180, while the legs are 448. What will be the number of deers in the zoo? A. 136 B. 68 C....

### Three types of wheat of Rs. 1.27, Rs. 1.29..

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Three types of wheat of Rs. 1.27, Rs. 1.29 and Rs. 1.32 per kg are mixed together to be sold at Rs. 1.30 per kg. In what ratio should this wheat be mixed? A....

### To gain 10% on selling sample of milk..

0
To gain 10% on selling sample of milk at the cost price of pure milk, the quantity of water to be mixed with 50 kg. of pure milk is: A. 2.5 kg B. 5 kg C....

### In what ratio must a mixture of 30% alcohol

0
In what ratio must a mixture of 30% alcohol strength be mixed with that of 50% alcohol strength so as to get a mixture of 45% alcohol strength? A. 1 : 2 B. 1 :...

### The marks of six boys in a group are 48, 59

0
The marks of six boys in a group are 48, 59, 87, 37, 78 and 57. What are the average marks...


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

5.1: Midsegments

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Define midsegment.
• Use the Midsegment Theorem.

Review Queue

Find the midpoint between the given points.

1. (-4, 1) and (6, 7)
2. (5, -3) and (11, 5)
3. Find the equation of the line between (-2, -3) and (-1, 1).
4. Find the equation of the line that is parallel to the line from #3 through (2, -7).

Know What? A fractal is a repeated design using the same shape (or shapes) of different sizes. Below, is an example of the first few steps of a fractal. Draw the next figure in the pattern.

Defining Midsegment

Midsegment: A line segment that connects two midpoints of the sides of a triangle.

is the midsegment between and .

The tic marks show that and are midpoints.

and

Example 1: Draw the midsegment between and for above.

Solution: Find the midpoints of and using your ruler. Label these points and . Connect them to create the midsegment.

Example 2: You now have all three midpoints of . Draw in midsegment and .

Solution:

For every triangle there are three midsegments.

Midsegments in the Plane

Let’s transfer what we know about midpoints in the plane to midsegments in the plane. We will need to use the midpoint formula, .

Example 3: The vertices of are and . Find the midpoints of all three sides, label them and . Then, graph the triangle, plot the midpoints and draw the midsegments.

Solution: Use the midpoint formula 3 times to find all the midpoints.

and point

and , point

and , point

The graph is to the right.

Example 4: Find the slopes of and .

Solution: The slope of is . The slope of is .

From this we can conclude that . If we were to find the slopes of the other sides and midsegments, we would find and .

Example 5: Find and .

Solution: Now, we need to find the lengths of and . Use the distance formula.

From this we can conclude that is half of . If we were to find the lengths of the other sides and midsegments, we would find that is half of and is half of .

The Midsegment Theorem

The conclusions drawn in Examples 4 and 5 can be combined into the Midsegment Theorem.

Midsegment Theorem: The midsegment of a triangle is half the length of the side it is parallel to.

If is a midsegment of , then and .

Example 6a: Mark all the congruent segments on with midpoints , and .

Solution: Drawing in all three midsegments, we have:

Also, this means the four triangles are congruent by SSS.

Example 6b: Mark all the parallel lines on , with midpoints , and .

Solution:

To play with the properties of midsegments, go to http://www.mathopenref.com/trianglemidsegment.html.

Example 7: , and are the midpoints of the sides of the triangle.

Find

a)

b)

c) The perimeter of

Solution: Use the Midsegment Theorem.

a)

b)

c) Add up the three sides of to find the perimeter.

Remember: No line segment over means length or distance.

Example 8: Algebra Connection Find the value of and . and are midpoints.

Solution: . To find , set equal to 17.

Know What? Revisited To the left is a picture of the figure in the fractal pattern.

Review Questions

• Questions 1-5 use the definition of a midsegment and the Midsegment Theorem.
• Questions 6-9 and 18 are similar to Example 7.
• Questions 10-17 are similar to Example 8.
• Questions 19-22 are similar to Example 3.
• Questions 23-30 are similar to Examples 3, 4, and 5.

Determine if each statement is true or false.

1. The endpoints of a midsegment are midpoints.
2. A midsegment is parallel to the side of the triangle that it does not intersect.
3. There are three congruent triangles formed by the midsegments and sides of a triangle.
4. If a line passes through two sides of a triangle and is parallel to the third side, then it is a midsegment.
5. There are three midsegments in every triangle.

, and are midpoints of the sides of and .

1. If , find and .
2. If , find .
3. If , and , find and .
4. If and , find .

For questions 10-17, find the indicated variable(s). You may assume that all line segments within a triangle are midsegments.

1. The sides of are 26, 38, and 42. is formed by joining the midpoints of .
1. What are the lengths of the sides of ?
2. Find the perimeter of .
3. Find the perimeter of .
4. What is the relationship between the perimeter of a triangle and the perimeter of the triangle formed by connecting its midpoints?

Coordinate Geometry Given the vertices of below find the midpoints of each side.

1. and
2. and
3. and
4. and

Multi-Step Problem The midpoints of the sides of a triangle are , and . Answer the following questions. The graph is below.

1. Find the slope of , and .
2. The side that passes through should be parallel to which midsegment? ( are all midsegments of a triangle).
3. Using your answer from #24, take the slope of and use the “rise over run” in either direction to create a parallel line to that passes through . Extend it with a ruler.
4. Repeat #24 and #25 with and . What are coordinates of the larger triangle?

Multi-Step Problem The midpoints of the sides of are , and . Answer the following questions.

1. Find the slope of , and .
2. Plot the three midpoints and connect them to form midsegment triangle, .
3. Using the slopes, find the coordinates of the vertices of . (#22 above)
4. Find using the distance formula. Then, find the length of the sides it is parallel to. What should happen?

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