text:
Looking for Geometry worksheets?
Check out our pre-made Geometry worksheets!
 Tweet

##### Browse Questions

You can create printable tests and worksheets from these Grade 7 Geometry and Measurement questions! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page.

1 2 3 4 ... 23
Arrange the following polygons in order from least to greatest number of sides:

Hexagon, triangle, decagon, pentagon
1. decagon, hexagon, pentagon, triangle
2. triangle, hexagon, pentagon, decagon
3. hexagon, triangle, decagon, pentagon
4. triangle, pentagon, hexagon, decagon
Grade 7 Metric System and SI
A thousand grams is called a
1. Milligram
2. Decagram
3. Gram
4. Kilogram
Find the area of a circle where r = 7 m. Use pi = 3.14.
1. $153.86m^2$
2. $43.96m^2$
3. $13.28m^2$
4. $153m^2$
Which one has the largest area:
a circle with a diameter of 10 inches
a rectangle that measures 12 inches by 11 inches
a circle with a radius of 6 inches
a square that is 11 inches on each side
1. a circle with a diameter of 10 inches
2. rectangle that measures 12 inches by 11 inches
3. circle with a radius of 6 inches
4. square that is 11 inches on each side
The distance around a circle
1. Circle
2. Circumference
3. Complementary
4. Coefficient
Write as a fraction.

12oz.:3lbs.
1. $12/3$
2. $4/1$
3. $1/4$
4. $3/4$
5. None of these
Grade 7 Metric System and SI
Metric measurements of mass are done in
1. Meters
2. Liters
3. Newtons
4. Grams
Grade 7 Metric System and SI
Metric measurements of length are done in
1. Meters
2. Liters
3. Newtons
4. Grams


text:
# Absolutely Closed Metric Spaces.

1. Aug 30, 2009

### qspeechc

Hi.

An absolutely closed metric space M is such that: If N is a meric space containing M, then M is closed in N.

I would like to show that an absolutely closed metric space is complete, how do I do this? I know the proof of the converse but that's no help obviously.
I know intuitively that an absolutely closed space should be complete: if you "imbed" the space in its completion, it will be a closed subset, and hence complete itself. But! We only know that any space is isometric to a subset of its completion, not that a completion containing the space exists (am I making sense?).
The problem seems to me that for a given metric space M, you do not know if there are any other metric spaces containing it.
I tried to show that the image of an absolutely closed metric space under an isometric function is itself absolutely closed, but I couldn't. If I could show that, then the image of M would be a closed subset of M's completion, and hence complete, thus M would be complete.

Any help?

2. Aug 30, 2009

### quasar987

But isn't that isometry is also a bijection? Call a bijective isometry an isomorphism of metric spaces.

If two metrics spaces are isomorphic, and one of them is complete, then surely the other is too, since Cauchy sequences are preserved by isomorphisms.

3. Aug 30, 2009

### qspeechc

Yes, but then we would have to show that the image of the absolutely closed space is also absolutely closed, or at least a closed subset of its completion. But it is not possible to show that the image of M as a subset of its completion is closed.

Lets say our isometry from M to a subset of its completion is f. We want to show f(M) is a closed subset of the completion of M, let's call it C.
Take a sequence in f(M) converging to some point in C...how do we show this limit is in f(M)?

4. Aug 30, 2009

### quasar987

I see...

But suppose we have constructed the completion (M',d') of (M,d) and h:M-->M' is an injective isometry. Can't we define M'' as $M\cup (M'-h(M))$ with the obvious metric, so that M'' is a complete metric space effectively containing M?

5. Aug 30, 2009

### qspeechc

Yes, that looks like it will work.

The metric we will use on M'' takes on the values of the corresponding points in M'. We must first check M'' is complete: obvious. If x_n is Cauchy in M'', the corresponding sequence is Cauchy in M' and hence converges in M', so x_n converges to the corresponding point in M''. M is absolutely closed, so is a closed subset of M'', hence it is complete. Is this correct?

format: plain


text:
# Revision history [back]

### ODE solving question

I'm stumped here. I am implementing a parachute model for use in my modeling class and am using a nice ODE class devised by oddrobot and modified by Jason Grout. To model the opening of the parachute, I need to make a parameter's value depend on the height of the parachutist. But, the if..then statements in my code appear to be getting ignored. What am I missing here?

Here is what I'm doing. (Note that DESolution is a command in the ODE solving routine that I'm using.)

var('x,v')
g1=9.81

def g(x,v):
if x>50:
return(-g1-0.3*abs(v)*v)
elif x>0:
return(-g1-2*abs(v)*v)
else:
return(0)
def f(x,v):
if x>0:
return(v)
else:
return(0)
F=[f(x,v),g(x,v)]
solution=DESolution(F,[t,0,40],[(x,100),(v,0)])
solution.coordinates(['red','blue'])

The complete code is published at:

http://sage.maa.org/home/pub/104/


text:
## Annuities

After reading this section, you will know how to identify, define, and calculate an annuity's present and future value. An annuity is the structure of a financial instrument that is a finite series of level payments that have a definite end. When you are finished, you will be able to recognize the two types of annuities: an ordinary annuity and an annuity due, and explain how they are different. You will also be able to calculate each of these types of annuities and contrast them to their opposites: perpetuities. Annuities are key to understanding because they mimic the payment structure of a bond's coupon payment. This section is foundational for being able to calculate bond prices.

### Present Value of Annuity

The $PV$ of an annuity can be found by calculating the $PV$ of each individual payment and then summing them up.

#### LEARNING OBJECTIVE

• Calculate the present value of annuities

#### KEY TAKEAWAYS

##### Key Points
• The $PV$ for both annuities-due and ordinary annuities can be calculated using the size of the payments, the interest rate, and number of periods.
• The $PV$ of a perpetuity can be found by dividing the size of the payments by the interest rate.
• Payment size is represented as $p$, $pmt$, or $A$; interest rate by $i$ or $r$; and number of periods by $n$ or $t$.

##### Key Terms
• perpetuity: An annuity in which the periodic payments begin on a fixed date and continue indefinitely.

The Present Value ($PV$) of an annuity can be found by calculating the $PV$ of each individual payment and then summing them up . As in the case of finding the Future Value ($FV$) of an annuity, it is important to note when each payment occurs. Annuities-due have payments at the beginning of each period, and ordinary annuities have them at the end.

Recall that the first payment of an annuity-due occurs at the start of the annuity, and the final payment occurs one period before the end. The $PV$ of an annuity-due can be calculated as follows:

$\mathrm{P}_{0}=\frac{\mathrm{P}_{\mathrm{n}}}{(1+\mathrm{i})^{\mathrm{n}}}=\mathrm{P} \frac{1-(1+\mathrm{i})^{-\mathrm{n}}}{\mathrm{i}} \cdot(1+\mathrm{i})$

where $P$ is the size of the payment (sometimes $A$ or $pmt$), i is the interest rate, and $n$ is the number of periods.

An ordinary annuity has annuity payments at the end of each period, so the formula is slightly different than for an annuity-due. An ordinary annuity has one full period before the first payment (so it must be discounted) and the last payment occurs at the termination of the annuity (so it must be discounted for one period more than the last period in an annuity-due). The formula is:

$\mathrm{P}_{0}=\frac{\mathrm{P}_{\mathrm{n}}}{(1+\mathrm{i})^{\mathrm{n}}}=\mathrm{P} \cdot \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{(1+\mathrm{i})^{\mathrm{n}+\mathrm{k}-1}}=\mathrm{P} \cdot \frac{1-\left[\frac{1}{(1+\mathrm{i})^{\mathrm{n}}}\right]}{\mathrm{i}}$

where, again, $P$, $i$, and $n$ are the size of the payment, the interest rate, and the number of periods, respectively.

Both annuities-due and ordinary annuities have a finite number of payments, so it is possible, though cumbersome, to find the $PV$ for each period. For perpetuities, however, there are an infinite number of periods, so we need a formula to find the $PV$. The formula for calculating the $PV$ is the size of each payment divided by the interest rate.

#### Example 1

Suppose you have won a lottery that pays $1,000 per month for the next 20 years. But, you prefer to have the entire amount now. If the interest rate is 8%, how much will you accept? Consider for argument purposes that two people, Mr. Cash, and Mr. Credit, have won the same lottery of$1,000 per month for the next 20 years. Now, Mr. Credit is happy with his $1,000 monthly payment, but Mr. Cash wants to have the entire amount now. Our job is to determine how much Mr. Cash should get. We reason as follows: If Mr. Cash accepts x dollars, then the x dollars deposited at 8% for 20 years should yield the same amount as the$1,000 monthly payments for 20 years. In other words, we are comparing the future values for both Mr. Cash and Mr. Credit, and we would like the future values to be equal.

Since Mr. Cash is receiving a lump sum of x dollars, its future value is given by the lump sum formula:

$x\left(\frac{1+.08}{12}\right)^{240}$

Since Mr. Credit is receiving a sequence of payments, or an annuity, of 1,000 per month, its future value is given by the annuity formula: $\frac{1000\left[\left(\frac{1+0.08}{12}\right)^{240}-1\right]}{\frac{0.08}{12}}$ The only way Mr. Cash will agree to the amount he receives is if these two future values are equal. So we set them equal and solve for the unknown: \begin{aligned}&\mathrm{x}\left(\frac{1+.08}{12}\right)^{240}=\frac{1000\left[\left(\frac{1+0.08}{12}\right)^{240}-1\right]}{\frac{0.08}{12}} \\&\mathrm{x} \cdot(4.9268)=1,000 \cdot(589.02041) \\&\mathrm{x} \cdot 4.9268=589,020.41 \\&\mathrm{x}=119,554.36\end{aligned} The reader should also note that if Mr. Cash takes his lump sum of119,554.36 and invests it at 8% compounded monthly, he will have $589,020.41 in 20 years. #### Example 2 Find the monthly payment for a car costing$15,000 if the loan is amortized over five years at an interest rate of 9%.

Again, consider the following scenario: Two people, Mr. Cash and Mr. Credit, go to buy the same car that costs $15,000. Mr. Cash pays cash and drives away, but Mr. Credit wants to make monthly payments for five years. Our job is to determine the amount of the monthly payment. We reason as follows: If Mr. Credit pays x dollars per month, then the x dollar payment deposited each month at 9% for 5 years should yield the same amount as the$15,000 lump sum deposited for 5 years. Again, we are comparing the future values for both Mr. Cash and Mr. Credit, and we would like them to be the same.

Since Mr. Cash is paying a lump sum of \$15,000, its future value is given by the lump sum formula:

$\ 15,000 \cdot\left(\frac{1+.09}{12}\right)^{60}$

Mr. Credit wishes to make a sequence of payments, or an annuity, of x dollars per month, and its future value is given by the annuity formula:

$\frac{x\left[\left(\frac{1+0.09}{12}\right)^{60}-1\right]}{\frac{0.09}{12}}$

We set the two future amounts equal and solve for the unknown:

\begin{aligned}&15,000 \cdot\left(\frac{1+.09}{12}\right)^{60}=\frac{\mathrm{x}\left[\left(\frac{1+0.09}{12}\right)^{60}-1\right]}{\frac{0.09}{12}} \\&15,000 \cdot 1.5657=\mathrm{x} \cdot 75.4241 \\&311.38=\mathrm{x}\end{aligned}


text:
# Parsing the lsblk output

I am a Python beginner learning Python 3. I have written two small functions that parse the lsblk output and return Linux physical and logical disks. Here is the first function:

from subprocess import run, PIPE

def physical_drives():
"""
Gets all physical drive names.

Gets all physical drive names on a Linux system,
parsing the lsblk utility output.

Parameters
----------

Returns
-------
list
A list of strings representing drive names.

"""

command = ['lsblk -d -o name -n']
output = run(command, shell=True, stdout=PIPE)

output_string = output.stdout.decode('utf-8')
output_string = output_string.strip()

results = output_string.split('\n')
return results

def main():
print(physical_drives())

if __name__ == '__main__':
main()

The second function:

from subprocess import run, PIPE

def partitions(disk):
"""
Gets all partitions for a given physical disk.

Gets all partitions present on a physical disk
on a Linux system.
The function parses the lsblk utility output.

Parameters
----------
disk : string
A string containing a disk name such as 'sda'

Returns
-------
list
A list of strings representing partitions.

"""

command = ['lsblk -o name -n -s -l']
output = run(command, shell=True, stdout=PIPE)

output_string = output.stdout.decode('utf-8')
output_string = output_string.strip()

results = list()
results.extend(output_string.split('\n'))
results = [x for x in results if x != disk and disk in x]

return results

def main():

from disks import physical_drives

for drive in physical_drives():

print(drive)
parts = partitions(drive)

for partition in parts:
print('\t' + partition)

if __name__ == '__main__':
main()

The functions are in two different files in the same directory. I would appreciate a quick review on anything that is not idiomatic/pythonic code. I also have one specific question. I have coded the functions in Ninja-Ide with lint and PEP8 suggestions turned on. In both files the IDE suggests that my print() statements:

print('\t' + partition)
print(physical_drives())

should be written with doubled parentheses:

print(('\t' + partition))
print((physical_drives()))

For Python 3 support, I have checked the print() documentation, but have been unable to find a reference for including function calls and string concatenation in double parentheses when calling print().

• The double-parentheses suggestion makes no sense. – 200_success Jan 13 '17 at 8:02

## lsblk

The -s option to lsblk was introduced to util-linux rather recently, in release 2.22. You may experience compatibility issues on slightly older GNU/Linux installations.

But I don't see why you would want the -s option at all — it just gives you an inverted device tree. For example, on my machine:

$lsblk -o name -n -s -l sda1 sda sda2 sda sr0 vg-root sda3 sda vg-var sda3 sda vg-data sda3 sda  In the output, sda appears multiple times. To understand the output, you need to drop the -l flag so that the list appears in tree form: $ lsblk -o name -n -s
sda1
└─sda
sda2
└─sda
sr0
vg-root
└─sda3
└─sda
vg-var
└─sda3
└─sda
vg-data
└─sda3
└─sda

Now, it's more apparent that the -s option isn't helpful. If you drop it, then the output makes more sense:

$lsblk -o name -n sda ├─sda1 ├─sda2 └─sda3 ├─vg-root ├─vg-var └─vg-data sr0$ lsblk -o name -n -l
sda
sda1
sda2
sda3
vg-root
vg-var
vg-data
sr0

To list the devices on sda, it would be better to run lsblk -o name -n -l /dev/sda — that would immediately drop sr0 from consideration, for example. Note that LVM volumes (such as vg-root above) would still appear in the output. I don't think that doing a substring search (if x != disk and disk in x in your code) is a reliable filter. It could be fooled if there are more than 26 physical disks: the 27th disk would be named sdaa. It might also be fooled by exceptionally tricky naming of LVM volumes.

## Subprocess execution

Whenever practical, I recommend avoiding the shell when executing subprocesses. The shell introduces a set of potential security vulnerabilities — for example, shenanigans with the PATH environment variable. Best practice would be to run the command with a specific executable and pre-parsed command-line options:

run('/bin/lsblk -o name -n -s -l'.split(), stdout=PIPE)

## Alternative solution

I actually wouldn't bother with parsing the output of lsblk at all. After all, lsblk is just a way to report the contents of the sysfs filesystem. You would be better off inspecting /sys directly.

from glob import glob
from os.path import basename, dirname

def physical_drives():
drive_glob = '/sys/block/*/device'
return [basename(dirname(d)) for d in glob(drive_glob)]

def partitions(disk):
if disk.startswith('.') or '/' in disk:
raise ValueError('Invalid disk name {0}'.format(disk))
partition_glob = '/sys/block/{0}/*/start'.format(disk)
return [basename(dirname(p)) for p in glob(partition_glob)]

• I used the -s option because the output seemed easier to parse. I will definitelly check out your alternative solution. I didn't like using text parsing for my functions either, i was hoping for something like an API call but was unable to find one. Thank you. – Anonimista Jan 13 '17 at 16:17

This code looks okay, but I would simplify a few things:

This:

output_string = output.stdout.decode('utf-8')
output_string = output_string.strip()

results = output_string.split('\n')
return results

could be :

return output.stdout.decode('utf-8').strip().split('\n')

this:

def main():
print(physical_drives())

if __name__ == '__main__':
main()

could be:

if __name__ == '__main__':
print(physical_drives())

this:

output_string = output.stdout.decode('utf-8')
output_string = output_string.strip()

results = list()
results.extend(output_string.split('\n'))
results = [x for x in results if x != disk and disk in x]
return results

could be:

results = output.stdout.decode('utf-8').strip().split('\n'))
return [x for x in results if x != disk and disk in x]

this:

parts = partitions(drive)

for partition in parts:
print('\t' + partition)

could be:

for partition in partitions(drive):
print('\t' + partition)

And finally, I see no reason for the double paren to the print function.

• Thank you for your detailed suggestions. I am sorry I can mark only one post as the answer. – Anonimista Jan 13 '17 at 16:20
• @Anonimista No worries. The concept that a single code review is the answer is just odd. Hopefully code reviews are additive and they are all useful is some way. :-) – Stephen Rauch Jan 13 '17 at 16:22
• Especially because the answers complement one another. Your answer suggests how to write more pythonic code and the other answer is related to my use of lsblk. Oh, well. – Anonimista Jan 13 '17 at 16:35

The logic can be moved out to an external process, if you're able to install jq (command-line JSON parsing tool).

Command: apt install jq

Then the code can be reduced to just:

import os
resp = os.popen('lsblk -J  | jq -c .').readline().strip()

• Welcome to Code Review! You have presented an alternative solution, but haven't reviewed the code. Please edit to show what aspects of the question code prompted you to write this version, and in what ways it's an improvement over the original. It may be worth (re-)reading How to Answer. – Toby Speight Jul 11 '19 at 10:57
• @TobySpeight I disagree. I think that processing JSON is clearly simpler and better, and passes the "Oh, duh!" test. – 200_success Jul 11 '19 at 16:32
• @200 I've added an introductory sentence or two to make this a review. HTH. – Toby Speight Jul 11 '19 at 17:01


text:
QUESTION

# Between 1 and 31 are inserted m arithmetic means so that the ratio of the 7th and (m-1)th is 5:9. Find the value of m.

Hint: Take first term of an A.P series as a=1 and last term as b=31, find the common difference by using the formula $d = \dfrac{{b - a}}{{n + 1}}$. Later find the 7th and (m-1)th term and take the ratio and equate it to the given value to form a linear equation. In the linear equation substitute the value of common difference to find the value of ‘m’.

As we know that to insert n numbers between a and b the common difference $d = \dfrac{{b - a}}{{n + 1}}$

Here, we have to insert m numbers between 1 and 31.

So, b = 31, a = 1

And the number of terms to be inserted = n = m.

Therefore, $d = \dfrac{{31 - 1}}{{m + 1}}$

$d = \dfrac{{30}}{{m + 1}}$

Now, a = 1, $d = \dfrac{{30}}{{m + 1}}$, b = 31

We need to find 7th and (m - 1)th numbers inserted.

So,7th term is $a + 7d = 1 + 7d$ and (m - 1)th term is $a+(m - 1)d = 1 +(m-1)d$

Now it is given that ratio of

$\dfrac{{7th{\text{ number}}}}{{\left( {m - 1} \right){\text{th number}}}} = \dfrac{5}{9}$

$\dfrac{{1 + 7d}}{{1 + \left( {m - 1} \right)d}} = \dfrac{5}{9}$

So when we do cross multiplication we will get:

$\left( {1 + 7d} \right)9 = 5\left[ {1 + \left( {m - 1} \right)d} \right]$

By further solving we get,

$\Rightarrow 9 + 63d = 5 + 5d(m -1)$

$\Rightarrow 9 + 63d = 5 + 5dm - 5d$

$\Rightarrow 4 + 68d = 5dm$

Now when we put $d{\text{ = }}\dfrac{{30}}{{m + 1}}$ in the above equation we will get:

$\Rightarrow 4 + 68\left( {\dfrac{{30}}{{m + 1}}} \right) = 5\left( {\dfrac{{30}}{{m + 1}}} \right)m$

Now making the fractions common,

$\Rightarrow \dfrac{{4\left( {m + 1} \right) + 68 \times 30}}{{m + 1}} = \dfrac{{5 \times 30 \times m}}{{m + 1}}$

$\Rightarrow 4\left( {m + 1} \right) + 2040 = 150m{\text{ , }}m \ne - 1$

$\Rightarrow 4m + 4 + 2040 = 150m$

$\Rightarrow 2044 = 150m - 4m$

$\Rightarrow 2044 = 146m$

$\Rightarrow m = \dfrac{{2044}}{{146}}$

$\Rightarrow m = 14$

Hence the value of m is 14.

NOTE: We must remember the equation to find common differences when inserting n elements in A.P. It also should be noted that when we tried to find the 7th and the (m - 1)th terms of inserted elements, we were actually finding the 8th and the mth elements of the A.P.


text:
# tennessine

The Simple English Wikipedia has an article on:

livermorium${\displaystyle \longleftarrow }$ 117 ${\displaystyle \longrightarrow }$oganesson

## Noun

 Pluralnone
1. (uncountable) Tennessine is a synthetic element with an atomic number of 117 and symbol Ts.


text:
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Can we live on renewables? | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: Sustainable Energy – Without the Hot Air Go to the latest version.

# 2.18: Can we live on renewables?

Created by: CK-12

The red stack in figure 18.1 adds up to 195 kWh per day per person. The green stack adds up to about 180 kWh/d/p. A close race! But please remember: in calculating our production stack we threw all economic, social, and environmental constraints to the wind. Also, some of our green contributors are probably incompatible with each other: our photovoltaic panels and hot-water panels would clash with each other on roofs; and our solar photovoltaic farms using 5% of the country might compete with the energy crops with which we covered 75% of the country. If we were to lose just one of our bigger green contributors – for example, if we decided that deep offshore wind is not an option, or that panelling 5% of the country with photovoltaics at a cost of £200000 per person is not on – then the production stack would no longer match the consumption stack.

Furthermore, even if our red consumption stack were lower than our green production stack, it would not necessarily mean our energy sums are adding up. You can’t power a TV with cat food, nor can you feed a cat from a wind turbine. Energy exists in different forms – chemical, electrical, kinetic, and heat, for example. For a sustainable energy plan to add up, we need both the forms and amounts of energy consumption and production to match up. Converting energy from one form to another – from chemical to electrical, as at a fossil-fuel power station, or from electrical to chemical, as in a factory making hydrogen from water – usually involves substantial losses of useful energy. We will come back to this important detail in Chapter Five energy plans for Britain, which will describe some energy plans that do add up.

Here we’ll reflect on our estimates of consumption and production, compare them with official averages and with other people’s estimates, and discuss how much power renewables could plausibly deliver in a country like Britain.

The questions we’ll address in this chapter are:

1. Is the size of the red stack roughly correct? What is the average consumption of Britain? We’ll look at the official energy-consumption numbers for Britain and a few other countries.
2. Have I been unfair to renewables, underestimating their potential? We’ll compare the estimates in the green stack with estimates published by organizations such as the Sustainable Development Commission, the Institution of Electrical Engineers, and the Centre for Alternative Technology.
3. What happens to the green stack when we take into account social and economic constraints?

Figure 18.1: The state of play after we added up all the traditional renewables.

Red reflections

Our estimate of a typical affluent person’s consumption (figure 18.1) has reached 195 kWh per day. It is indeed true that many people use this much energy, and that many more aspire to such levels of consumption. The average American consumes about 250 kWh per day. If we all raised our standard of consumption to an average American level, the green production stack would definitely be dwarfed by the red consumption stack.

What about the average European and the average Brit? Average European consumption of “primary energy” (which means the energy contained in raw fuels, plus wind and hydroelectricity) is about 125 kWh per day per person. The UK average is also 125 kWh per day per person.

These official averages do not include two energy flows. First, the “embedded energy” in imported stuff (the energy expended in making the stuff) is not included at all. We estimated in Chapter Stuff that the embedded energy in imported stuff is at least 40 kWh/d per person. Second, the official estimates of “primary energy consumption” include only industrial energy flows – things like fossil fuels and hydroelectricity – and don’t keep track of the natural embedded energy in food: energy that was originally harnessed by photosynthesis.

Another difference between the red stack we slapped together and the national total is that in most of the consumption chapters so far we tended to ignore the energy lost in converting energy from one form to another, and in transporting energy around. For example, the “car” estimate in Part I covered only the energy in the petrol, not the energy used at the oil refinery that makes the petrol, nor the energy used in trundling the oil and petrol from A to B. The national total accounts for all the energy, before any conversion losses. Conversion losses in fact account for about 22% of total national energy consumption. Most of these conversion losses happen at power stations. Losses in the electricity transmission network chuck away 1% of total national energy consumption.

Figure 18.2: Energy consumption, broken down by end use, according to the Department for Trade and Industry.

When building our red stack, we tried to imagine how much energy a typical affluent person uses. Has this approach biased our perception of the importance of different activities? Let’s look at some official numbers. Figure 18.2 shows the breakdown of energy consumption by end use. The top two categories are transport and heating (hot air and hot water). Those two categories also dominated the red stack in Part I. Good.

Railways Petroleum 0.4
Water transport Petroleum 1.0
Aviation Petroleum 7.4
All modes Electricity 0.4
All energy used by transport 31.6

2006 breakdown of energy consumption by transport mode, in kWh/d per person. Source: Dept. for Transport (2007).

Figure 18.4: Power consumption per capita, versus GDP per capita, in purchasing-power-parity US dollars. Squares show countries having “high human development;” circles, “medium” or “low.” Figure 30.1 shows the same data on logarithmic scales.

Let’s look more closely at transport. In our red stack, we found that the energy footprints of driving a car 50 km per day and of flying to Cape Town once per year are roughly equal. Table shows the relative importances of the different transport modes in the national balance-sheet. In the national averages, aviation is smaller than road transport.

How do Britain’s official consumption figures compare with those of other countries? Figure 18.4 shows the power consumptions of lots of countries or regions, versus their gross domestic products (GDPs). There’s an evident correlation between power consumption and GDP: the higher a country’s GDP (per capita), the more power it consumes per capita. The UK is a fairly typical high-GDP country, surrounded by Germany, France, Japan, Austria, Ireland, Switzerland, and Denmark. The only notable exception to the rule “big GDP implies big power consumption” is Hong Kong. Hong Kong’s GDP per capita is about the same as Britain’s, but Hong Kong’s power consumption is about 80 kWh/d/p.

Figure 18.5: Hong Kong. Photo by Samuel Louie and Carol Spears.

The message I take from these country comparisons is that the UK is a fairly typical European country, and therefore provides a good case study for asking the question “How can a country with a high quality of life get its energy sustainably?”

Green reflections

People often say that Britain has plenty of renewables. Have I been mean to green? Are my numbers a load of rubbish? Have I underestimated sustainable production? Let’s compare my green numbers first with several estimates found in the Sustainable Development Commission’s publication The role of nuclear power in a low carbon economy. Reducing $CO_2$ emissions – nuclear and the alternatives. Remarkably, even though the Sustainable Development Commission’s take on sustainable resources is very positive (“We have huge tidal, wave, biomass and solar resources”), all the estimates in the Sustainable Development Commission’s document are smaller than mine! (To be precise, all the estimates of the renewables total are smaller than my total.) The Sustainable Development Commission’s publication gives estimates from four sources detailed below (IEE, Tyndall, IAG, and PIU). Figure 18.6 shows my estimates alongside numbers from these four sources and numbers from the Centre for Alternative Technology (CAT). Here’s a description of each source.

IEE The Institute of Electrical Engineers published a report on renewable energy in 2002 – a summary of possible contributions from renewables in the UK. The second column of figure 18.6 shows the “technical potential” of a variety of renewable technologies for UK electricity generation – “an upper limit that is unlikely ever to be exceeded even with quite dramatic changes in the structure of our society and economy.” According to the IEE, the total of all renewables’ technical potential is about 27 kWh/d per person.

Figure 18.6: Estimates of theoretical or practical renewable resources in the UK, by the Institute of Electrical Engineers, the Tyndall Centre, the Interdepartmental Analysts Group, and the Performance and Innovation Unit; and the proposals from the Centre for Alternative Technology’s “Island Britain” plan for 2027.

Tyndall The Tyndall Centre’s estimate of the total practicable renewable-energy resource is 15 kWh per day per person.

IAG The Interdepartmental Analysts Group’s estimates of renewables, take into account economic constraints. Their total practical and economical resource (at a retail price of 7p/kWh) is 12 kWh per day per person.

PIU The “PIU” column shows the “indicative resource potential for renewable electricity generation options” from the DTI’s contribution to the PIU review in 2001. For each technology I show their “practical maximum,” or, if no practical maximum was given, their “theoretical maximum.”

CAT The final column shows the numbers from the Centre for Alternative Technology’s “Island Britain” plan Helweg-Larsen and Bull (2007).

Bio-powered Europe

Sometimes people ask me “surely we used to live on renewables just fine, before the Industrial Revolution?” Yes, but don’t forget that two things were different then: lifestyles, and population densities.

Turning the clock back more than 400 years, Europe lived almost entirely on sustainable sources: mainly wood and crops, augmented by a little wind power, tidal power, and water power. It’s been estimated that the average person’s lifestyle consumed a power of 20 kWh per day. The wood used per person was 4 kg per day, which required 1 hectare $(10000 \ m^2)$ of forest per person. The area of land per person in Europe in the 1700s was $52000 \ m^2$. In the regions with highest population density, the area per person was $17500 \ m^2$ of arable land, pastures, and woods. Today the area of Britain per person is just $4000 \ m^2$, so even if we reverted to the lifestyle of the Middle Ages and completely forested the country, we could no longer live sustainably here. Our population density is far too high.

Green ambitions meet social reality

Figure 18.1 is bleak news. Yes, technically, Britain has “huge” renewables. But realistically, I don’t think Britain can live on its own renewables – at least not the way we currently live. I am partly driven to this conclusion by the chorus of opposition that greets any major renewable energy proposal. People love renewable energy, unless it is bigger than a figleaf. If the British are good at one thing, it’s saying “no.”

Wind farms? “No, they’re ugly noisy things.”

Solar panels on roofs? “No, they would spoil the visual amenity of the street.”

Figure 18.7: The state of play after we add up all the traditional renewables, and then have a public consultation.

After the public consultation. I fear the maximum Britain would ever get from renewables is in the ballpark of 18 kWh/d per person. (The left-hand consumption number, 125 kWh/d per person, by the way, is the average British consumption, excluding imports, and ignoring solar energy acquired through food production.)

Figure 18.8: Where the wild things are. One of the grounds for objecting to wind farms is the noise they produce. I’ve chopped out of this map of the British mainland a 2-km-radius exclusion zone surrounding every hamlet, village, and town. These white areas would presumably be excluded from wind-farm development. The remaining black areas would perhaps also be largely excluded because of the need to protect tranquil places from industrialization. Settlement data from www.openstreetmap.org.

More forestry? “No, it ruins the countryside.”

Waste incineration? “No, I’m worried about health risks, traffic congestion, dust and noise.”

Hydroelectricity? “Yes, but not big hydro – that harms the environment.”

Offshore wind? “No, I’m more worried about the ugly powerlines coming ashore than I was about a Nazi invasion.”

Wave or geothermal power? “No, far too expensive.”

After all these objections, I fear that the maximum Britain would ever get from renewables would be something like what’s shown in the bottom right of figure 18.7.

Figure 18.8 offers guidance to anyone trying to erect wind farms in Britain. On a map of the British mainland I’ve shown in white a 2-km radius exclusion zone surrounding every hamlet, village, and town. These white areas would presumably be excluded from wind-farm development because they are too close to the humans. I’ve coloured in black all regions that are more than 2 km from any human settlement. These areas are largely excluded from wind-farm development because they are tranquil, and it’s essential to protect tranquil places from industrialization. If you want to avoid objections to your wind farm, pick any piece of land that is not coloured black or white.

Figure 18.9: Production of renewables and nuclear energy in the UK in 2006. All powers are expressed per-person, as usual. The breakdown of the renewables on the right hand side is scaled up 100-fold vertically.

Some of these environmentalists who have good hearSts but confused minds are almost a barrier to tackling climate change.

Malcolm Wicks, Minister of State for Energy

We are drawing to the close of Part I. The assumption was that we want to get off fossil fuels, for one or more of the reasons listed in Chapter Motivations – climate change, security of supply, and so forth. Figure 18.9 shows how much power we currently get from renewables and nuclear. They amount to just 4% of our total power consumption.

The two conclusions we can draw from Part I are:

1. To make a difference, renewable facilities have to be country-sized.

For any renewable facility to make a contribution comparable to our current consumption, it has to be country-sized. To get a big contribution from wind, we used wind farms with the area of Wales. To get a big contribution from solar photovoltaics, we required half the area of Wales. To get a big contribution from waves, we imagined wave farms covering 500 km of coastline. To make energy crops with a big contribution, we took 75% of the whole country.

Renewable facilities have to be country-sized because all renewables are so diffuse. Table summarizes most of the powers-per-unit-area that we encountered in Part I.

To sustain Britain’s lifestyle on its renewables alone would be very difficult. A renewable-based energy solution will necessarily be large and intrusive.

2. It’s not going to be easy to make a plan that adds up using renewables alone. If we are serious about getting off fossil fuels, Brits are going to have to learn to start saying “yes” to something. Indeed to several somethings.

In Part II I’ll ask, “assuming that we can’t get production from renewables to add up to our current consumption, what are the other options?”

Power per unit land or water area
Wind $2 \ W/m^2$
Offshore wind $3 \ W/m^2$
Tidal pools $3 \ W/m^2$
Tidal stream $6 \ W/m^2$
Solar PV panels $5-20 \ W/m^2$
Plants $0.5 \ W/m^2$
Rain-water (highlands) $0.24 \ W/m2$
Hydroelectric facility $11 \ W/m^2$
Geothermal $0.017 \ W/m^2$

Renewable facilities have to be country-sized because all renewables are so diffuse.

UK average energy consumption is 125 kWh per day per person. I took this number from the UNDP Human Development Report, 2007.

The DTI (now known as DBERR) publishes a Digest of United Kingdom Energy Statistics every year. [uzek2]. In 2006, according to DUKES, total primary energy demand was 244 million tons of oil equivalent, which corresponds to 130 kWh per day per person.

I don’t know the reason for the small difference between the UNDP number and the DUKES number, but I can explain why I chose the slightly lower number. As I mentioned, DUKES uses the same energy-summing convention as me, declaring one kWh of chemical energy to be equal to one kWh of electricity. But there’s one minor exception: DUKES defines the “primary energy” produced in nuclear power stations to be the thermal energy, which in 2006 was 9 kWh/d/p; this was converted (with 38% efficiency) to 3.4 kWh/d/p of supplied electricity; in my accounts, I’ve focused on the electricity produced by hydroelectricity, other renewables, and nuclear power; this small switch in convention reduces the nuclear contribution by about 5 kWh/d/p.

Losses in the electricity transmission network chuck away 1% of total national energy consumption. To put it another way, the losses are 8% of the electricity generated. This 8% loss can be broken down: roughly 1.5% is lost in the long-distance high-voltage system, and 6% in the local public supply system. Source: MacLeay et al. (2007).

Figure 18.4. Data from UNDP Human Development Report, 2007. [3av4s9]

In the Middle Ages, the average person’s lifestyle consumed a power of 20 kWh per day. Source: Malanima (2006).

“I’m more worried about the ugly powerlines coming ashore than I was about a Nazi invasion.” Source: [6frj55].

Feb 23, 2012

Dec 02, 2014


text:
# zbMATH — the first resource for mathematics

Periodic products of groups. (English. Russian original) Zbl 1377.20026
J. Contemp. Math. Anal., Armen. Acad. Sci. 52, No. 3, 111-117 (2017); translation from Izv. Nats. Akad. Nauk Armen., Mat. 52, No. 3, 3-15 (2017).
Summary: In this paper, we provide an overview of the results relating to the $$n$$-periodic products of groups that have been obtained in recent years by the authors of the present paper, as well as some results obtained by other authors in this direction. The periodic products were introduced by the first author [Proc. Steklov Inst. Math. 142, 1–19 (1976; Zbl 0424.20020); translation from Tr. Mat. Inst. Steklov 142, 3–21 (1976)] to solve the Maltsev’s well-known problem. It was shown that the periodic products are exact, associative and hereditary for subgroups. They also possess some other important properties such as the Hopf property, the $$C*$$-simplicity, the uniform non-amenability, the $$SQ$$-universality, etc. It was proved that the $$n$$-periodic products of groups can uniquely be characterized by means of certain quite specific and simply formulated properties. These properties allow to extend to $$n$$-periodic products of various families of groups a number of results previously obtained for free periodic groups $$B(m,n)$$. In particular, we describe the finite subgroups of $$n$$-periodic products, Also, we analyze and extend the simplicity criterion of $$n$$-periodic products obtained previously by the first author [Math. Notes 88, No. 6, 771–775 (2010; Zbl 1230.20025); translation from Mat. Zametki 88, No. 6, 803–810 (2010)].

##### MSC:
 20F50 Periodic groups; locally finite groups 20F05 Generators, relations, and presentations of groups 20E06 Free products of groups, free products with amalgamation, Higman-Neumann-Neumann extensions, and generalizations 20F28 Automorphism groups of groups 22D25 $$C^*$$-algebras and $$W^*$$-algebras in relation to group representations
Full Text:
##### References:
 [1] Adian, S. I., Periodic products of groups, Trudy MIAN SSSR, 142, 3-21, (1976) [2] Adian, S. I., Once more on periodic products of groups and on a problem of A. I.maltsev, Mat. Zametki, 88, 803-810, (2010) [3] S. I. Adian, The Burnside problem and identities in groups (Nauka, Moscow, 1975). [4] Adian, S. I., New estimates of odd exponents of infinite Burnside groups, Trudy MIAN, 289, 41-82, (2015) · Zbl 1343.20040 [5] Adian, S. I., On the simplicity of periodic products of groups, Dokl. Akad. Nauk SSSR, 241, 745-748, (1978) · Zbl 0416.54023 [6] Adian, S. I.; Atabekyan, V. S., The Hopfian property of n-periodic products of groups, Mat. Zametki, 95, 483-491, (2014) · Zbl 1326.20040 [7] Adian, S. I., Random walks on free periodic groups, Izv. Akad. Nauk SSSR Ser. Mat., 46, 1139-1149, (1982) · Zbl 0512.60012 [8] Adian, S. I.; Atabekyan, V. S., Characteristic properties and uniform non-amenability of n-periodic products of groups, Izv.RAN. Ser. Mat., 79, 3-18, (2015) · Zbl 1360.20018 [9] S. I. Adian, V. S. Atabekyan, “On free groups of infinitely based varieties of S.I. Adian”, Izv. RAN.Ser. Mat., 81 (5), 2017. · Zbl 1436.20044 [10] Atabekyan, V. S., Uniform nonamenability of subgroups of free Burnside groups of odd period, Mat. Zametki, 85, 516-523, (2009) · Zbl 1213.20036 [11] Atabekyan, V. S., Monomorphisms of free Burnside groups, Mat. Zametki, 86, 483-490, (2009) · Zbl 1200.20028 [12] Atabekyan, V. S., On normal subgroups in the periodic products of S.I. Adian, TrudyMIAN, 274, 15-31, (2011) [13] Ivanov, S. V., On subgroups of free Burnside groups of large odd exponent, Illinois J. Math., 47, 299-304, (2003) · Zbl 1036.20029 [14] A. Yu. Olshanskii, The Geometry of Defning Relations in Groups (Kluwer, Amsterdam, 1991). [15] Harpe, P., On simplicity of reduced $$C$$*-algebras of groups, Bull. Lond. Math. Soc., 39, 1-26, (2007) · Zbl 1123.22004 [16] Breuillard, E.; Kalantar, M.; Kennedy, M.; Ozawa, N., $$C$$*-simplicity and the unique trace property for discrete groups, ArXiv:1410., 2518, 1-20, (2014) [17] Adian, S. I.; Atabekyan, V. S., $$C$$*-simplicity of n-periodic products, Mat. Zametki, 99, 643-648, (2016) [18] Olshanskii, A. Yu.; Osin, D. V., $$C$$*-simple groups without free subgroups, Groups Geom.Dyn., 8, 933-983, (2014) · Zbl 1354.22009 [19] V. S. Atabekyan, A.L. Gevorgyan, Sh.A. Stepanyan, “The uniqueness trace property of n-periodic products of groups”, Journal of ContemporaryMathematical Analysis, 52 (5), 2017. · Zbl 1378.20041 [20] Neshadim, M. V., Free products of groups have no outer normal automorphisms, Algebra i Logika, 35, 562-566, (1996) [21] Atabekyan, V. S., Normal automorphisms of free Burnside groups, Izv.RAN. Ser.Mat., 75, 3-18, (2011) · Zbl 1227.20030 [22] Atabekyan, V. S., Non-f-admissible normal subgroups of free Burnside groups, Journal of Contemporary Mathematical Analysis, 45, 112-22, (2010) · Zbl 1299.20046 [23] Cherepanov, E. A., Normal automorphisms of free Burnside groups of large odd exponents, Internat. J. Algebra Comput, 16, 839-847, (2006) · Zbl 1115.20024 [24] Atabekyan, V. S.; Gevorgyan, A.L., On outer normal automorphisms of periodic products of groups, Journal of ContemporaryMathematical Analysis, 46, 289-92, (2011) · Zbl 1299.20047 [25] Gevorgyan, A. L., On automorphisms of periodic products of groups, Proceedings of the YSU, Physics and Mathematics, 2, 3-9, (2012) · Zbl 1309.20028 [26] Gevorgyan, A. L.; Stepanyan, Sh. A., On automorphisms of some periodic products of groups, Proceedings of the YSU, Physics and Mathematics, 2, 7-10, (2015) · Zbl 1322.20023 [27] Atabekyan, V. S., Splitting automorphisms of free Burnside groups, Sb. Math., 204, 182-189, (2013) · Zbl 1282.20025 [28] Atabekyan, V. S., Splitting automorphisms of order pk of free Burnside groups are inner, Math. Notes, 95, 586-589, (2014) · Zbl 1315.20030
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# How do you subtract \frac { x ^ { 2} - 10x } { x ^ { 2} - 11x - 12} - \frac { 24} { x ^ { 2} - 11x - 12}?

Jun 7, 2017

It's $\frac{x + 2}{x + 1}$.

#### Explanation:

$\frac{{x}^{2} - 10 x}{{x}^{2} - 11 x - 12} - \frac{24}{{x}^{2} - 11 x - 12}$

Noting that the two fractions have common denominators, we can simplify the expression as so:

$\frac{{x}^{2} - 10 x - 24}{{x}^{2} - 11 x - 12}$

Then, we can factor both the numerator and the denominator into a $\left(x + a\right) \left(x + b\right)$ format:

$\frac{\left(x + 2\right) \left(x - 12\right)}{\left(x + 1\right) \left(x - 12\right)}$

Then, we cancel the $x - 12$s and finish the problem with the final answer:

$\frac{x + 2}{x + 1}$


text:
# Mathematician:Edward J. Goodwin

## Mathematician

Indiana physician and amateur mathematician who believed he had squared the circle, trisected the angle and doubled the cube.

He proposed a bill to allow for the charging of royalties for the use of the value of $\pi$ (pi) that he had calculated.

It was rejected before the second reading through the efforts of Clarence Abiathar Waldo.

American

## History

• Born: c. 1825
• Died: 1902

## Critical View

... his solutions of the trisection of the angle, doubling the cube and quadrature of the circle having been already accepted as contributions to science by the American Mathematical Monthly ... And be it remembered that these noted problems had been long since given up by scientific bodies as unsolvable mysteries and above man's ability to comprehend.

It is necessary to point out that those contributions to the American Mathematical Monthly were published "By the request of the author" and therefore have no intellectual standing, not having been subject to the necessary level of peer review.

## Also known as

Some sources cite him (probably erroneously) as Edwin J. Goodwin.


text:
# Rate of Return Calculator

## What is Rate of Return Calculator?

A 'Rate of Return Calculator' is an online tool which finds the rate of return in percentage for the given investment, returns, and the time period. Cuemath's 'Rate of Return Calculator' helps you to find the rate of return within a few seconds.

## How to Use Rate of Return Calculator?

Follow the below steps to find the rate of return.

• Step 1: Enter the invested money, return money, in the input boxes.
• Step 2: Click on "Calculate" to find the rate of return.
• Step 3: Click on "Reset" to enter the new set of values.

## How to find the Rate of Return?

The following formula helps to find the rate of return.

Rate of return = (Return Money - Invested Money)/Invested Money ×100

The rate of return obtain is a percentage value and need to have a % symbol after the answer.

Let us look at the following example to understand more about the rate of return.

### Solved Example:

Joseph invests $5000 in the share of a company and receives$8000 by selling the shares.  Find his rate of return.

### Solution:

The invested amount is $5000 and the returns is$8000.  This is for a time period of 3 years.

Let us use the below formula to find the rate of return.

Rate of return = (Return Money - Invested Money)/Invested Money ×100

= (8000 - 5000)/5000 × 100

= 3000/5000× 100

= 3/5 × 100

= 60%

Therefore the rate of return is 20%.

Now, try the calculator to find the rate of return for the following values.

a)  $870 has been invested and the return amount is$1280

b)  $1250 Yen has been invested, and the return amount is$1480


text:
## anonymous 4 years ago can someone explain to me how to solve 5x^2-21x-20/5x^2-16x-16

1. anonymous

factor: ((x-5)(5x+4))/((x-4)(5x+4)) cancel out 4x+4 (x-5)/(x-4)

2. anonymous

*5x+4

3. anonymous

I think you mean how to symplify or do the division.

4. anonymous

both?

5. anonymous

Did you understand how to solve the problem? Can you do a similiar problem by yourself?

6. anonymous

i can do problems such as : 4x+3/5 but when it becomesmore than that it becomes confusing

7. anonymous

that was just an example

8. anonymous

So if you have something like $\frac{4x+3}{x+5}$ you have problems

9. anonymous

i've never done a problem like that in my lessons, i've had things like x^2-15x+56/ x-7 or with more terms

10. anonymous

practice factoring, if you are good at factoring those problems become easy