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# Seven

Seven friends agree to send everyone a holiday card. How many postcards were sent?

Result

n =  42

#### Solution:

$n=7 \cdot \ (7-1)=42$

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text:
## anonymous 3 years ago Input in standard form the equation of the given line. The line with m = -3/8 and passing through (5, -4)

• This Question is Open
1. ash2326

Standard equation of a line passing through point (x1, y1) and slope m $\frac{y-y1}{x-x1}=m$ Just plugin the values and you'll get the standard form

2. phi

See http://en.wikipedia.org/wiki/Linear_equation#General_.28or_standard.29_form Ax + By = C is "standard form"

format: plain


text:
Spacing of fractions with prime denominator

Let $X\ge 2$ be large. Let $$A = \left\{\frac{a}{q}: 1\le a < q\le X,\ (a, q) = 1\right\},$$ and let $$B = \left\{\frac{a}{p}: 1\le a < p\le X,\ (a, p) = 1,\text{ and p is prime}\right\}.$$ Note that for distinct $\frac{a}{q}, \frac{a'}{q'}\in A$, we have $$\left\lVert \frac{a}{q} - \frac{a'}{q'}\right\rVert\ge\frac{1}{qq'}\ge\frac{1}{X^2},\quad\text{where}\quad\lVert\beta\rVert := \min_{n\in\mathbb{Z}} |\beta - n|.$$ It follows that in any interval of length $X^{-2}$ in $[0, 1]$, there are at most $O(1)$ elements of $A$, and this is clearly the best result possible, since $|A|\gg X^2$. Is it possible to get better results for $B$, since we have that $|B|\asymp X^2 / \log X$ by the prime number theorem. In particular, is it possible to get a bound of $o(\log X)$ for the number of elements of $B$ in an interval of length $$|B|^{-1}\ll X^{-2}\log X?$$ Note that the inclusion $B\subseteq A$ gives the trivial bound $O(\log X)$.

• With the current definition, we have $B = \{ A\}$, are you sure that is what you want? – Dirk Jul 10 '18 at 10:51
• @Dirk, I'm guessing OP has forgotten to tell us that $p$ is restricted to being a prime. – Gerry Myerson Jul 10 '18 at 12:35
• I've corrected these typos. The inequality was originally the wrong way around. – Mayank Pandey Jul 10 '18 at 19:50
• I wonder whether the paper Harman, Glyn, Numbers badly approximable by fractions with prime denominator, Math. Proc. Cambridge Philos. Soc. 118 (1995), no. 1, 1–5, MR1329453 (96c:11080) is relevant here. – Gerry Myerson Jul 11 '18 at 0:45
• This type of inequality on fractions is the starting point of the large sieve inequality. In case that you are actually interested in applications with the large sieve: Wolke observed that the sieve estimates with prime denominators can almost save the log factor: Wolke, D. On the large sieve with primes. Acta Math. Acad. Sci. Hungar. 22 (1971/72), 239–247. mathscinet.ams.org/mathscinet-getitem?mr=291121 – Christian Elsholtz Jul 17 '18 at 21:43

Wolke (On the large sieve with primes, Acta Math. Acad. Sci. Hungar. 22 (1971/72), 239–247, MathSciNet MR0291121 (45 #215)) has worked on this question. His motivation was Gallagher's approach to the large sieve. If I remember correctly, his result was almost saving a factor $\log x$, which is what one would expect.


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# Difference between revisions of "Algebra group is isomorphic to algebra subgroup of unitriangular matrix group of degree one more than logarithm of order to base of field size"

## Statement

### Version for algebra group over prime field

Suppose $p$ is a prime number, and $G$ is a finite p-group, i.e., a group of order a power of $p$. Suppose $q$ is a power of $p$. Then, if $G$ is an algebra group over $\mathbb{F}_q$, the following must be true:

1. The order of $G$ is a power of $q$.
2. $G$ must be isomorphic to a subgroup of the unitriangular matrix group $UT(m + 1,q)$ where $m = \log_q(|G|)$, and the image of this isomorphism must have the further property that subtracting 1 from all its elements gives a subring of the nilpotent associative ring $NT(m + 1,q)$ of strictly upper triangular matrices. Another way of saying this is that $G$ is isomorphic to an algebra subgroup of $UT(m+1,q)$.

### Version for algebra group over prime field

Suppose $p$ is a prime number, and $G$ is a finite p-group, i.e., a group of order $p^n$ for some nonnegative integer $n$. Then, if $G$ is an algebra group over $\mathbb{F}_p$, $G$ must be isomorphic to a subgroup of the unitriangular matrix group $UT(n + 1,p)$. Further, the image of this isomorphism must have the further property that subtracting 1 from all its elements gives a subring of the nilpotent associative ring $NT(m + 1,q)$ of strictly upper triangular matrices. Another way of saying this is that $G$ is isomorphic to an algebra subgroup of $UT(n+1,p)$.

Note that if $G$ is an algebra group over any field extension of $\mathbb{F}_p$, it is also an algebra group over $\mathbb{F}_p$, so if this condition is violated, $G$ cannot be an algebra group over any $\mathbb{F}_q$ for $q$ a power of $p$.

### Complete characterization of algebra groups

Note that it's obviously true that any algebra subgroup of a unitriangular matrix group is an algebra group. So the above gives a complete (necessary and sufficient) characterization of algebra groups.

## Facts used

1. Sylow's theorem, specifically Sylow implies order-dominating which says that any $p$-subgroup of a finite group is contained in a $p$-Sylow subgroup and hence has a conjugate contained in a predetermined $p$-Sylow subgroup.

## Proof

### Proof for finite field

Given: $G$ is a finite $p$-group that is an algebra group over $\mathbb{F}_q$, where $q$ is a power of $p$.

To prove: The order of $G$ is a power of $q$ and $G$ is isomorphic to a subgroup of $UT(m + 1,q)$ where $m = \log_q(|G|)$. Further, subtracting 1 from all the elements of this subgroup gives a subring of $NT(m+1,q)$.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Let $N$ be a nilpotent associative algebra over $\mathbb{F}_q$ for which $G$ is the corresponding algebra group. $G$ is an algebra group.
2 The order of $G$ equals the order of $N$, and $m = \log_q(|G|)$ is an integer that equals the dimension of $N$ as a vector space over $\mathbb{F}_q$. Step (1)
3 $N + F$ (the unitization of $N$) has dimension $m + 1$, and it acts faithfully on itself by left multiplication. Steps (1), (2)
4 $N + F$ is isomorphic to a subring of the matrix ring $M(n + 1,F)$. Under this identification, $G$, as a multiplicative subgroup, is isomorphic to a multiplicative subgroup of $GL(n+1,F)$ and $N$ is a nilpotent subring of $M(n+1,F)$. Step (3)
5 The image of $G$ under the isomorphism of Step (4) is conjugate to a subgroup of $UT(m + 1,F)$, which is the $p$-Sylow subgroup of $GL(m+1,F)$. Moreover, subtracting 1 from all the elements of this subgroup gives a nilpotent associative subring of $NT(m+1,F)$ (the strictly upper triangular matrices). Fact (1) $G$ is a $p$-group Step (4) By Step (4) and Fact (1), we know that the image of $G$ in $GL(m + 1,F)$ is conjugate to a subgroup of $UT(m+1,F)$, the upper triangular matrices. Under this conjugation operation, $N$ goes to a new ring. Since subtracting 1 commutes with conjugation, this conjugate of $N$ is the same as the ring obtained by subtracting 1 from the conjugate of $G$. Hence, it is a subring of $UT(m+1,F)$.
6 The image of $G$ under the isomorphism of Step (4) is conjugate to a subgroup of $UT(m + 1,F)$ with the property that subtracting 1 from all its element gives a subring of $NT(m + 1,F)$. Step (5) Follows from Step (5) and the observation that conjugation preserves the isomorphism type.

|}

### Proof for prime field

This follows from the general proof by setting $q = p$ and noting that $n = m$ in that case.


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# A medium power RF amplifier for lab use

Sometimes a small medium power broadband RF amplifier is handy for lab work if, for example, the output level available from a signal generator is not sufficient for a particular purpose. Applications include driving experimental circuits or testing the electromagnetic susceptibility of equipment or experimental setups. Of course, various connectorized amplifier modules are available from several manufacturers, and I already have a good collection of them, but in order to have some fun and to gain a bit of experience with modern LDMOS transistors I decided build a small broadband amplifier module myself.

The goal was to design a linear amplifier with a frequency range from 10 MHz to 500 MHz which is able to output about 1 W and has a gain of at least 35 dB, and which can run off a supply of 15 V. No effort was made to achieve a particular noise figure target or linearity requirement (e.g., a certain third order intercept point), but some care was taken to get a reasonably flat frequency response. Moreover, the amplifier was designed to be absolutely stable.

## Mechanical construction

The amplifier is built on a single FR4 board measuring 81 × 48 mm. The board is fitted in an RF tight die cast aluminium box. The box is Hammond model 1590B, which is also used as a heat sink for the amplifier. The circuit board is bolted head down into the aluminium box, and the input and output SMA sockets (Cinch model 142-0701-491) protrude through the top surface of the box. To ensure adequate thermal contact between the power devices on the board and the box, a 50 × 20 × 6 mm copper bar is glued to the back of the board with thermally conductive adhesive. This bar makes contact to the aluminium box, and thermal compound is applied to the contact surface. The power connection is routed through a feedthrough capacitor into the shielded box.

The board is double sided and 0.6 mm thick. Its back side has a solid ground plane; the front side has a ground pour with via stitching to the back side. The main RF signal path is laid out in a straight line, and realized as a 50 Ohms grounded and fenced coplanar waveguide. The board thickness was chosen to obtain reasonable coplanar waveguide dimensions (conductor thickness: 1 mm, spacing: 0.5 mm) as well as to decrease the thermal resistance between the power devices and the copper bar on the back. To this end also a fair number of thermal vias have been placed.

The board also has four mounting holes to fit a standard quarter brick heat sink for DC/DC converters to the back of the board. When using such a heat sink, watch out for the thermal resistance as the board dissipates about 4.2 W in total.

## Circuit diagram and design data

The circuit diagram of the amplifier is available here as a pdf file. An image of the circuit diagram is shown below. The schematic and board design files are provided as a zip file here; both were designed with KiCad version 5.1.4.

## Circuit description and design

The amplifier is very simple and consists of two stages. The first stage U1 is a HMC311ST89E MMIC made by Analog Devices (this is actually a former Hittite part), which consists of an InGaP/GaAs heterojunction bipolar transistor together with the requisite bias and matching circuits. This device has about 16 dB of gain and typically outputs a level of 15.5 dBm up to 2 GHz at 1 dB compression, which is enough to drive the second stage. Its noise figure is quoted as 4.5 dB according to the datasheet; in view of its gain this noise figure will essentially be the overall noise figure of the amplifier.

The MMIC is biased by a series connection of two SMD inductors with different inductivities (47 nH and 10 µH, with the smaller one towards the device). This is to avoid series resonance of the inductor with the decoupling capacitors on the power rail within the useful frequency band of the amplifier, which requires a sufficiently large inductor in order to push the resonance to frequencies below the useful band. However, a large inductor with sufficient current handling capabilities may have a self resonance frequency well below the upper band limit of the amplifier (i.e., 500 MHz) and thus behave capacitively within that band. The first stage is supplied with 5 V from a standard TS78M05CP voltage regulator. This rather noisy regulator is well blocked from the active device by a ferrite bead and a fair amount of filtering capacitors. When choosing other regulators watch out for their maximum input voltage as the circuit is powered from 15 V; this can already be marginal for some regulators.

The second stage is built around Q2, a PD85004 N-channel enhancement-mode LDMOS transistor made by STMicroelectronics. This transistor is used in a common source configuration and biased for class A operation. Again, gate and drain biasing is fed through a series connection of two inductors. The drain inductor allows a maximum output voltage swing of twice the supply voltage. To avoid thermal drift of the quiescent drain current Q2 is actively controlled. In order to understand that this is necessary let us try to estimate the variability of the drain current $I_{\rm D}$ with temperature $T$ for constant gate-source voltage $U_{\rm GS}$. If the MOSFET is operating in the saturation region and $U_{\rm GS}>U_{\rm th}$, where $U_{\rm th}$ is the threshold voltage, we have (ignoring channel length modulation) $I_{\rm D}=\tfrac12K(U_{\rm GS}-U_{\rm th})^2$, where $K$ is the transconductance coefficient. Differentiating with respect to $T$ yields the relative temperature coefficient $\frac{1}{I_{\rm D}}\frac{\mathord{\rm d}I_{\rm D}}{\mathord{\rm d}T}=\frac1K\frac{\mathord{\rm d}K}{\mathord{\rm d}T}-\frac{2}{U_{\rm GS}-U_{\rm th}}\frac{\mathord{\rm d}U_{\rm th}}{\mathord{\rm d} T}.$ This equation explains the well-known effect that the temperature coefficient of the drain current can be positive or negative, depending on $U_{\rm GS}$ (in many cases, particularly in switching applications, the MOSFET is operated under conditions in which this temperature coefficient is negative). With $U_{\rm th}\approx3\,{\rm V}$ from the diagram in the datasheet of the PD85004, $U_{\rm GS}\approx4\,{\rm V}$ as measured in the circuit, $\mathord{\rm d}U_{\rm th}/\mathord{\rm d} T\approx-5\,{\rm mV/K}$ for a double diffused MOSFET, and $K^{-1}\mathord{\rm d}K/\mathord{\rm d}T\approx-5\cdot10^{-3}\,{\rm K}^{-1}$, we obtain at $I_{\rm D}=150\,{\rm mA}$ an absolute variability of $I_{\rm D}$ with temperature of $\frac{\mathord{\rm d}I_{\rm D}}{\mathord{\rm d}T}\approx0.75\,\frac{\rm mA}{\rm K}.$ Thus we expect a positive temperature coefficient of the quiescent current under the present conditions of class A operation, and a considerable variation over the expected temperature range. To stabilize the quiescent current and avoid thermal runaway due to the positive temperature coefficient, a standard circuit is employed: The quiescent current is proportional to the voltage drop across shunt R1, and this voltage drop controls the current through the voltage divider R2 and R5, which in turn sets the gate source voltage of Q2. The time constant of this control loop is determined by C13 as well as the resistors R2 and R5, and the quiescent current can be adjusted by RV1 (Bourns model 3314G—notice the reversed pins of the resistance element of this trimmer). Also note that there will be some temperature compensation between D1 and Q1.

For better gain flatness and to help bringing the input and output of Q2 closer to 50 Ohms, R4 and C6 apply some voltage feedback. The gate loading network R6 and C10 helps stabilizing Q2. This network could also be applied at the output, which is even favorable when low noise is desired. Matching of Q2 is achieved by a fourth order purely reactive low pass network at the input, and by a similar sixth order low pass network at the output. This closely follows the application circuit in the datasheet, however, here the inductors are discrete 0603 devices rather than transmission lines due to the lower frequencies involved. It is interesting to experiment with other matching networks, and I may come back to this problem.

## Matching the second stage

The input and output matching networks of the second stage were designed as follows. A prototype of the amplifier was built with only the second stage around Q2 populated, but without the matching networks and the stabilization network C10 and R6 (but with the feedback network C6 and R4 in place). The gaps due to the non-populated components (i.e., U1, L7, L8, L9, L10, and L11) in the coplanar waveguide from the input and output jacks to Q2 were bridged with zero ohms resistors.

Then the S-parameters of the second stage were measured with a VNA. To protect the VNA a 20 dB attenuator was connected between the amplifier output and the VNA port. Instead of de-embedding the attenuator, it was simply calibrated out by a standard UOSM calibration using my cheap SMA cal kit. To improve the signal to noise ratio during calibration, the VNA source level was increased to 10 dBm (notice that during reflection measurements the attenuator causes a 40 dB loss—still borderline acceptable for a high dynamic range VNA, but larger attenuators should be de-embedded, or an external high power directional coupler should be used when accurate S22 data is needed). For the actual measurement the source level was decreased to −10 dBm at the amplifier input (port 1), and remained at +10 dBm at the output (port 2). Finally, a port extension was performed to shift the reference planes of the measurement from the SMA jacks right to transistor Q2. The necessary offset lengths were determined by soldering shorts to ground in the appropriate places, then the phase delay to the short was measured and used as offset (modern VNAs can do this automatically).

The measured S-parameters of the bare second stage are plotted here; they were also exported as a Touchstone file and imported in QucsStudio in order to perform an S-parameter simulation of the second stage together with the matching networks. The project file, which includes the Touchstone file, can be downloaded here; a screenshot of the simulation of the final matching network is here. There are two S-parameter simulations included in the project file, one with the selected E12 component values and another one with the optimizer set up for maximizing S21 with priority while minimizing S11 and S22, and with proper initial values (see the picture).

One can come up with various solutions of this optimization problem by either optimizing the flatness of various gains (e.g., S21, or transducer gain for given source and load impedances) or input and output matching. The solution that I have adopted tries to maximize S21 flatness (i.e., transducer gain for source and load resistances of 50 ohms) up to 500 MHz and is mostly hand tweaked. You can try and experiment yourself with QucsStudio, or come up with another type of matching network. As can be seen, the output SWR of the second stage is not stellar, and it seems not possible to get a return loss better that 5 dB with a purely reactive matching network over a frequency band this broad. There are theoretical limits to matching accuracy and bandwith for lossless networks (e.g., the Bode–Fano theorem), and it may be that one needs to resort to a lossy network in order to achieve a better result. I have not done that since if output matching is critical, a 3 dB attenuator can always be attached to the output for a 6 dB improvement of return loss. Out of curiosity I also tried replacing the inductors by grounded coplanar waveguides, see here (this is a quick optimizer result for the parameters shown, and can still be tweaked). It does not yield a better return loss, but would consume a lot more board space (notice the lengths of the transmission lines).

## Some test results

The most basic measurement to be made are the S-parameters of the amplifier for a small signal input. They were measured for an input of −15 dBm, i.e., at an output power of about 20 dBm. Again, a 20 dB attenuator (calibrated out) at the amplifier output was used.

It is seen that the S21 gain is not as flat as anticipated by the simulation; the flatness is about 3.5 dB except for the small dip at 425 MHz. Among the reasons are the parasitics of the matching networks, which have not been included in the simulation. Moreover, the biasing chokes L1, L2, L3. L4, L5 and L6 have a critical influence on gain flatness, and some variability between different models and manufacturers has been found. In fact, I have changed L1 and L3 for 0805 inductors after the amplifier was assembled since L1 and L3 caused a resonance dip at about 320 MHz. Also the frequency response of the first stage U1 has not been included in the simulation.

Thus, in order to improve on flatness, one should redo the simulation with measured S-parameter of the first and the second stage, and include some parasitics of the matching networks. In this way, with some effort and optimization, a flatness of about 1 dB should be achievable.

Moreover, it is seen that the low pass character of the matching networks becomes already noticeable at about 475 MHz, and at 500 MHz the gain has fallen to 25 dB (this is not due to U1 or Q2 running out of steam, but entirely due to the matching networks). The lesson to be learned here is that when recalculating the matching networks for better flatness one should shoot for an upper band limit of, say, 550 MHz to achieve a useful band limit of at least 500 MHz.

Just like gain flatness, output matching is a little worse that predicted by the simulation. The input matching is entirely determined by the HMC311ST89E and its internal matching networks. The input and output SWR resulting from the S-parameters is plotted below.

The reverse isolation of the second stage is already quite appreciable, so it comes as no surprise to find that the amplifier is absolutely stable over the entire frequency band of 10 MHz to 500 MHz. In fact, its Rollet factor never falls below 1.5 in this band.

The next picture shows the harmonic content of the amplifier output at 100 MHz and +25 dBm output level. At this power level the first harmonic is at −15 dBc, the second harmonic is at −30 dBc, and the third at −39 dBc.

What remains to be measured are precise values of the first order compression point (estimated to be in excess of 25 dBm) over frequency, the third order intercept point, and noise figure.

These results should also be compared to Claudio Girardi's (IN3OTD), who has experimented with a single stage class A broadband amplifier using the PD85004, and published extensive test results.

## Conclusion

Not all the original design goals are fully met by the amplifier. However, by re-measuring the S-parameters of the first and second stage, re-running the simulation and tweaking the matching networks a flatter frequency response up to 500 MHz can be easily achieved. Moreover, when doing so special care has to be exercised when selecting the biasing chokes L1 to L6 to avoid resonances within the useful frequency band. Ideally the influence of the particular biasing choke should be included in the measurement data used for the simulation.

Also the output referred 1 dB compression point was found to be somewhat lower than expected. When higher output power is required, the supply voltage and quiescent current of Q2 can be increased. The PD85004 is rated up to 40 V drain-source voltage, 2 A drain current and 6 W total power dissipation, but the cooling concept used here would then have to be improved (the small die cast box would become insufficient as a heat sink). Moreover, R1 should then be decreased to avoid excessive power dissipation, and the associated circuitry be redesigned appropriately.

These experiences show the many difficulties in designing a good broadband amplifier, even at low power levels and over a moderate frequency band. Nevertheless, they can be overcome, and even in its present form the amplifier is a useful laboratory device and ready to be used in practice.


text:
# ecdf¶

Filename: ecdf.py

Authors: Thomas Sargent, John Stachurski

Implements the empirical cumulative distribution function given an array of observations.

class quantecon.ecdf.ECDF(observations)[source]

Bases: object

One-dimensional empirical distribution function given a vector of observations.

Parameters: observations : array_like An array of observations

Attributes

 observations (see Parameters)

Methods

 __call__(x) Evaluates the ecdf at x


text:
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# Observation of ‘third sound’ in superfluid 3He

## Abstract

Waves on the surface of a fluid provide a powerful tool for studying the fluid itself and the surrounding physical environment. For example, the wave speed is determined by the force per unit mass at the surface, and by the depth of the fluid1: the decreasing speed of ocean waves as they approach the shore reveals the changing depth of the sea and the strength of gravity. Other examples include propagating waves in neutron-star oceans2 and on the surface of levitating liquid drops3. Although gravity is a common restoring force, others exist, including the electrostatic force which causes a thin liquid film to adhere to a solid. Usually surface waves cannot occur on such thin films because viscosity inhibits their motion. However, in the special case of thin films of superfluid 4He, surface waves do exist and are called ‘third sound’. Here we report the detection of similar surface waves in thin films of superfluid 3He. We describe studies of the speed of these waves, the properties of the surface force, and the film's superfluid density.

This is a preview of subscription content

## Access options

from\$8.99

All prices are NET prices.

## References

1. 1

Landau, L. D. & Lifshitz, E. M. Fluid Mechanics 36–44 (Pergamon, Oxford, (1959)).

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Whitaker, D. L., Weilart, M. A., Vincente, C. L., Maris, H. J. & Seidel, G. M. Oscillations of charged helium II drops. J. Low Temp. Phys. 110, 173–178 (1998).

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Tilley, D. R. & Tilley, J. Superfluidity and Superconductivity 77–83 (Hilger, Bristol, (1986)).

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Atkins, K. R. Third and fourth sound in liquid helium II. Phys. Rev. 113, 962–965 (1959).

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Everitt, C. W. F., Atkins, K. R. & Denenstein, A. Third sound in liquid helium films. Phys. Rev. 136, A1494–A1499 (1964).

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Bergman, D. Hydrodynamics and third sound in thin HeII films. Phys. Rev. 188, 370–384 (1969).

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Rudnick, I. Critical surface density of the superfluid component in 4He films. Phys. Rev. Lett. 40, 1454–1455 (1978).

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Ellis, F. M., Hallock, R. B., Miller, M. D. & Guyer, R. A. Phase separation in films of 3He-4He mixtures. Phys. Rev. Lett. 46, 1461–1464 (1981).

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Ketola, K. S., Wang, S. & Hallock, R. B. Anomalous wetting of helium on cesium. Phys. Rev. Lett. 68, 201–204 (1992).

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Ellis, F. M. & Li, L. Quantum swirling of superfluid helium films. Phys. Rev. Lett. 71, 1577–1580 (1993).

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Sheldon, P. A. & Hallock, R. B. Third sound and energetics in 3He-4He mixture films. Phys. Rev. B 50, 16082–16085 (1994).

13. 13

Brisson, J. G., Mester, J. C. & Silvera, I. F. Third sound of helium on a hydrogen substrate. Phys. Rev. B 44, 12453–12462 (1991).

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Ekholm, D. T. & Hallock, R. B. Studies of the decay of persistent currents in unsaturated films of superfluid 4He. Phys. Rev. B 21, 3902–3912 (1980).

15. 15

Pickar, K. A. & Atkins, K. R. Critical velocity of a superflowing liquid-helium film using third sound. Phys. Rev. 178, 389–399 (1969).

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Cho, H. & Williams, G. A. Vortex core size in submonolayer superfluid 4He films. Phys. Rev. Lett. 75, 1562–1565 (1995).

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Eggenkamp, M. E. W. et al. On the possibility of observing third sound on 3He. J. Low Temp. Phys. 110, 299–304 (1998).

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Schechter, A. M. R., Simmonds, R. W. & Davis, J. C. Capacitive generation and detection of third sound resonances in saturated superfluid 4He films. J. Low Temp. Phys. 110, 603–608 (1998).

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Sachrajda, A., Harris-Lowe, R. F., Harrison, J. P., Turkington, R. R. & Daunt, J. G. 3He film flow: two-dimensional superfluidity. Phys. Rev. Lett. 55, 1602–1605 (1985).

20. 20

Daunt, J. G. et al. Critical temperature and critical current of thin-film superfluid 3He. J. Low Temp. Phys. 70, 547–568 (1988).

21. 21

Davis, J. C., Amar, A., Pekola, J. P. & Packard, R. E. Superfluidity of 3He films. Phys. Rev. Lett. 60, 302–304 (1988).

22. 22

Xu, J. & Crooker, B. C. Very thin films of 3He: a new phase? Phys. Rev. Lett. 65, 3005–3008 (1990).

23. 23

Freeman, M. R., Germain, R. S., Thuneberg, E. V. & Richardson, R. C. Size effects in thin films of superfluid 3He. Phys. Rev. Lett. 60, 596–599 (1988).

24. 24

Jacobsen, K. W. & Smith, H. Critical current for the flow of superfluid 3He in a confined geometry. J. Low Temp. Phys. 67, 83–89 (1987).

25. 25

Tilley, D. R. & Tilley, J. Superfluidity and Superconductivity 92–94 (Hilger, Bristol, (1986)).

26. 26

Porto, J. V. & Parpia, J. M. Superfluid 3He in aerogel. Phys. Rev. Lett. 74, 4667–4670 (1995).

27. 27

Freeman, M. R. Size Effects in 3He Films. Thesis, Cornell Univ.((1988)).

28. 28

Stein, D. L. & Cross, M. C. Phase transitions in two dimensional superfluid 3He. Phys. Rev. Lett. 42, 504–507 (1979).

29. 29

Volovik, G. E. An analog of the quantum Hall Effect in a superfluid 3He film. Sov. Phys. JETP 67, 1804–1811 (1988).

30. 30

Kurihara, S. Singlet superfluidity in 3He film on 4He film. J. Phys. Soc. Jpn 52, 1311–1316 (1983).

## Acknowledgements

We thank L. Bildsten, R. B. Hallock, D.-H. Lee, J. M. Goodkind, O. Ishikawa, J.Harrison and G. Williams for discussions. This work was supported by the NSF, the ONR and by the Packard Foundation.

## Author information

Authors

### Corresponding author

Correspondence to J. C. Davis.

## Rights and permissions

Reprints and Permissions

Schechter, A., Simmonds, R., Packard, R. et al. Observation of ‘third sound’ in superfluid 3He. Nature 396, 554–557 (1998). https://doi.org/10.1038/25090

• Accepted:

• Issue Date:

• ### Extreme quantum nonlinearity in superfluid thin-film surface waves

• Y. L. Sfendla
• C. G. Baker
• W. P. Bowen

npj Quantum Information (2021)

• ### The A-B transition in superfluid helium-3 under confinement in a thin slab geometry

• N. Zhelev
• T. S. Abhilash
• J. M. Parpia

Nature Communications (2017)

• ### Anomalous Resonance Frequency Shift of a Microelectromechanical Oscillator in Superfluid $$^3$$He-B

• P. Zheng
• W. G. Jiang
• H. B. Chan

Journal of Low Temperature Physics (2017)

• ### Study of Superfluid $$^3$$ 3 He Under Nanoscale Confinement

• L. V. Levitin
• R. G. Bennett
• N. Zhelev

Journal of Low Temperature Physics (2014)

• ### Superfluid Phases of 3He in a Periodic Confined Geometry

• J. J. Wiman
• J. A. Sauls

Journal of Low Temperature Physics (2014)


text:
# \input gets slower after each \input

Phelype Oleinik phe.h.o1 at gmail.com
Tue Apr 27 20:04:54 CEST 2021

```On 27/04/2021 14:51, Hartmut Henkel wrote:
> On Tue, 27 Apr 2021, Hartmut Henkel wrote:
>>
>> ...not sure if this has to do with a problem tackled here, long long
>> ago: http://hhenkel.de/pdftex/patch3/avl-patch3.html
>
> ooops, this one is already part of pdftex, so it's likely not related to
> the slow-down problem. Sorry for the noise.

Yes, that seems to be already implemented for hyperlinks.  However, if I
read that correctly, the AVL Tree search you mention is for searching
strings.  If so, that may as well be implemented for this exact case, as
it doesn't seem to require a change in the underlying data structure...

I will investigate when I have the time.


text:
Accelerating the pace of engineering and science

• Trials

# ridge

Ridge regression

## Syntax

b = ridge(y,X,k)
b = ridge(y,X,k,scaled)

## Description

b = ridge(y,X,k) returns a vector b of coefficient estimates for a multilinear ridge regression of the responses in y on the predictors in X. X is an n-by-p matrix of p predictors at each of n observations. y is an n-by-1 vector of observed responses. k is a vector of ridge parameters. If k has m elements, b is p-by-m. By default, b is computed after centering and scaling the predictors to have mean 0 and standard deviation 1. The model does not include a constant term, and X should not contain a column of 1s.

b = ridge(y,X,k,scaled) uses the {0,1}-valued flag scaled to determine if the coefficient estimates in b are restored to the scale of the original data. ridge(y,X,k,0) performs this additional transformation. In this case, b contains p+1 coefficients for each value of k, with the first row corresponding to a constant term in the model. ridge(y,X,k,1) is the same as ridge(y,X,k). In this case, b contains p coefficients, without a coefficient for a constant term.

The relationship between b0 = ridge(y,X,k,0) and b1 = ridge(y,X,k,1) is given by

       m = mean(X);
s = std(X,0,1)';
b1_scaled = b1./s;
b0 = [mean(y)-m*b1_scaled; b1_scaled]

This can be seen by replacing the xi (i = 1, ..., n) in the multilinear model y = b00 + b10x1 + ... + bn0xn with the z-scores zi = (xiμi)/σi , and replacing y with yμy.

In general, b1 is more useful for producing plots in which the coefficients are to be displayed on the same scale, such as a ridge trace (a plot of the regression coefficients as a function of the ridge parameter). b0 is more useful for making predictions.

Coefficient estimates for multiple linear regression models rely on the independence of the model terms. When terms are correlated and the columns of the design matrix X have an approximate linear dependence, the matrix (XTX)–1 becomes close to singular. As a result, the least-squares estimate

becomes highly sensitive to random errors in the observed response y, producing a large variance. This situation of multicollinearity can arise, for example, when data are collected without an experimental design.

Ridge regression addresses the problem by estimating regression coefficients using

where k is the ridge parameter and I is the identity matrix. Small positive values of k improve the conditioning of the problem and reduce the variance of the estimates. While biased, the reduced variance of ridge estimates often result in a smaller mean square error when compared to least-squares estimates.

## Examples

expand all

### Ridge Regression

load acetylene

acetylene has observations for the predictor variables x1 , x2 , x3 , and the response variable y .

Plot the predictor variables against each other.

subplot(1,3,1)
plot(x1,x2,'.')
xlabel('x1'); ylabel('x2'); grid on; axis square

subplot(1,3,2)
plot(x1,x3,'.')
xlabel('x1'); ylabel('x3'); grid on; axis square

subplot(1,3,3)
plot(x2,x3,'.')
xlabel('x2'); ylabel('x3'); grid on; axis square

Note the correlation between x1 and the other two predictor variables.

Compute coefficient estimates for a multilinear model with interaction terms, for a range of ridge parameters using ridge and x2fx .

X = [x1 x2 x3];
D = x2fx(X,'interaction');
D(:,1) = []; % No constant term
k = 0:1e-5:5e-3;
b = ridge(y,D,k);

Plot the ridge trace.

figure
plot(k,b,'LineWidth',2)
ylim([-100 100])
grid on
xlabel('Ridge Parameter')
ylabel('Standardized Coefficient')
title('{\bf Ridge Trace}')
legend('x1','x2','x3','x1x2','x1x3','x2x3')

The estimates stabilize to the right of the plot. Note that the coefficient of the x2x3 interaction term changes sign at a value of the ridge parameter .

## References

[1] Hoerl, A. E., and R. W. Kennard. "Ridge Regression: Biased Estimation for Nonorthogonal Problems." Technometrics. Vol. 12, No. 1, 1970, pp. 55–67.

[2] Hoerl, A. E., and R. W. Kennard. "Ridge Regression: Applications to Nonorthogonal Problems." Technometrics. Vol. 12, No. 1, 1970, pp. 69–82.

[3] Marquardt, D.W. "Generalized Inverses, Ridge Regression, Biased Linear Estimation, and Nonlinear Estimation." Technometrics. Vol. 12, No. 3, 1970, pp. 591–612.

[4] Marquardt, D. W., and R.D. Snee. "Ridge Regression in Practice." The American Statistician. Vol. 29, No. 1, 1975, pp. 3–20.


text:
My Math Forum Grading on a Curve

 August 2nd, 2017, 08:08 PM #11 Global Moderator   Joined: Dec 2006 Posts: 18,061 Thanks: 1396 There's too little data for complicated mathematics to be worthwhile. The nature of the quiz (untimed, open book, etc.) suggests that a student should have to do really well to get an A. From what you stated, I would guess that the number of points attained reflected not so much the ability of each student, but mainly the amount of careful thought that was applied in reading and answering the questions. I chose grade bands of equal "width", making that width quite generous. Showing the results (for this particular quiz) on a histogram suggests that the width shouldn't be larger. Being any more generous would allow students to "pass" even if they've answered most of the questions incorrectly. The grade width could be smaller, but that would split up the students who score 18-21 points, which could attract criticism unless you feel strongly that it's right to do so. As it is, more than 75% of your students "passed". If you made the grades tougher, the significant effect would probably be that 14-17 points would no longer be considered satisfactory. Letter grades are not intended to be converted back to points. Thanks from AplanisTophet
 August 3rd, 2017, 04:02 AM #12 Senior Member   Joined: Jun 2014 From: USA Posts: 308 Thanks: 21 Thank you everyone
August 3rd, 2017, 05:13 AM   #13
Senior Member

Joined: May 2016
From: USA

Posts: 805
Thanks: 320

Quote:
 Originally Posted by AplanisTophet Here are my thoughts: I can't really use the 'old-fashioned method' because the student having aced the quiz is a curve breaker. No curve would result. I could say that I want the average score to be 85% and then use the following formula to change the maximum number of points from 25 to something else (denoted $x$): $$\frac{\text{current class avg.}}{x} = \frac{\text{85}}{100}$$ The problem with the above is that my ace student would end up with 'extra credit' in contradiction to the terms of the syllabus, as would any other student with an actual score exceeding $x$. Let's say I wanted a cutoff for those getting an A and I've decided that the top 8% of grades should earn an A. I believe then that I'm left with an integral I cannot solve (that of the probability density of the normal distribution), but could otherwise approximate: $$\Pr( \,x \geq a) \, = \int_{a}^{\infty} \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{( \,x-\mu) \,^2}{2\sigma^2}} \, dx = 0.08$$ I'm not willing to go through some complex approximation process though. Further, after getting my approximation, I would have to convert the grades back to points because the overall grade is based on a point system. That brings me back to your suggestion. How would I convert those grades you conjured above back into points as opposed to letter grades? I have no idea how you came up with that. Ok, maybe I'm thinking about this too much, but also trying to nail down an approach I could possibly use going forward and technically even include in a syllabus (no, I won't be using Latex to create my syllabus ). I was curious if anyone here had a good method.
You are thinking too much.

Decide where you want failure to be and fail those below that bright line. Decide where you want an A to be, and give an A to all those above that bright line. Decide the number of gradations you want for the remainder, for example B, C, and D would be 3, calculate the reciprocal (1/3rd in my example) and assign the top 1/3rd a B, the middle 1/3rd a C, and the bottom 1/3rd a D.

Grading on a curve is an exemption, not a science.

 August 3rd, 2017, 09:03 AM #14 Global Moderator   Joined: Dec 2006 Posts: 18,061 Thanks: 1396 Regarding my previous post, the histograms below show that using slightly broader grades affects the appearance a lot. Hist4.PNG Hist5.PNG Thanks from EvanJ and AplanisTophet

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text:
# Monitoring adverse events during peritoneal dialysis

## 2007-11-15

One of the doctors I was working with had an interesting data set examining adverse events in patients with peritoneal dialysis. These patients start treatment with peritoneal dialysis on a specific day and are followed until they stop this treatment. Reasons for stopping peritoneal dialysis might be that the patient got better and no longer needed any treatment, the patient got worse and needed to switch to hemodialysis, or the patient died. Patients who moved out of town presumably continued their dialysis, but they were lost to follow-up in this particular study. There were two adverse events examined: exit site infections, and peritonitis. Although I ran several complex analyses on this data set, I thought it might be useful to look at a simpler approach to monitoring the frequency of adverse events using control charts.

Here’s the data on when patients began treatment and when they ended their treatment.

id t0 t1 0 58 680 1 95 1416 2 189 1247 3 207 532 4 402 1136 5 412 501 6 431 1851 7 550 1414 8 581 1339 9 770 2325 10 809 1498 11 841 1339 12 880 1563 13 903 2664 14 939 1451 15 980 2920 16 1106 2103 17 1107 1291 18 1155 1792 19 1159 1654 20 1200 1968 21 1247 1574 22 1422 2247 23 1449 1544 24 1499 2310 25 1582 1755 26 1677 2142 27 1786 2920 28 1799 2108 29 1803 2729 30 2097 2723 31 2101 2639 32 2225 2419 33 2254 2920 34 2496 2920 35 2500 2920 36 2507 2920 37 2562 2730 38 2590 2870 39 2663 2920 40 2701 2838 41 2711 2920

The first column is the patient id, the second column is when the patient started dialysis (number of days since the start of the review period), and the third column is when the patient ended dialysis (again in number of days).

There were 42 patients in this study. The average patient stayed in the study for 640 days (1.8 years). We will divide the number of days by 365 in most graphical presentations of the data to show time in years rather than days.

This graph shows the cumulative number of patients at risk of an adverse event at any time point in the review period.

The vertical line segments at the bottom of the graph represent the times that patients began dialysis and the segments at the top of the graph represent the times that patients ended dialysis. Each time a patent begins dialysis, the graph jumps up one unit. Every time a patient ends dialysis, the graph drops one unit. At the busiest time, there were 17 patients on dialysis. The graph falls to zero at 8 years, not because all patients were removed from dialysis, but because that represented the end of the observation period.

Notice that this graph generally climbs for the first three years, drops of a bit then is roughly level for years 4 through 8. This is not surprising because only patients who began dialysis during the eight year observation window were included in the study.

Here are the times of exit site infections:

id tx 0 643 0 657 1 143 1 203 1 756 1 900 1 1122 1 1331 2 596 3 231 4 905 4 953 4 986 6 680 6 1273 6 1410 6 1485 6 1584 7 967 9 1292 12 1442 13 1331 13 1351 13 2528 13 2658 14 949 14 1261 14 1447 16 1250 21 1267 27 1947 33 2520 33 2767 35 2769

Notice that some patients experience more than one exit site infection and some patients experience no infections. You might already notice something interesting with the data. About half of the events have three digit days (days in the hundreds). Since the full time range is from zero to a bit less than 3 thousand, you would expect only about a third of these events to have three digit days. This is possible evidence that these events occurred more frequently early in the review period rather than later. A careful analysis, of course, would have to control for the number of patients at risk during these time intervals.

Here is the data on times of peritonitis.

id tx 0 218 1 652 1 1265 1 1328 3 237 7 641 7 1004 7 1080 9 978 9 1036 9 1236 9 1974 9 2116 10 904 10 1305 13 1815 13 1983 13 2082 14 949 15 2859 20 1959 24 1977 24 2089 25 1620 25 1641 27 1803 27 2354 27 2520 28 1809 28 2054 33 2351 41 2740 41 2755

The graph below shows the cumulative number of patient years over the eight year study (you can get this by integrating the figure shown above). This graph also notes the occurrence of exit site infections using a red plus sign.

Here’s a second graph with occurrence of peritonitis marked with a red plus sign.

The horizontal distance between successive plus signs represents the waiting time in years between the successive patients. The vertical distance between successive plus signs represents the waiting time in patient-years. It’s a subtle but important difference. By calculating the number of patient-years between successive events, you can It adjusts for the fact that during the first two or three years of observation, there were fewer patient-years of data compared to successive years.

If you plot the patient-years between successive exit site infections, you will see the following control chart.

The gray region represents the first four years of the study. After the fourth year, an intervention was implemented to reduce adverse events. You can see that it had a dramatic impact at first, with the hospital twice waiting over 10 patient year between events shortly after the intervention started. More recently, it looks as if the waiting time has slipped back at least partway to the norm.

Here’s the plot for peritonitis events.

Notice that there is no evidence that peritonitis is becoming a rarer event after the intervention.


text:
### Home > INT3 > Chapter 10 > Lesson 10.3.1 > Problem10-145

10-145.

What is the sum of each of the infinite geometric series? If it is not possible to calculate the sum, explain why not. Homework Help ✎

1. $8 + 4 + 2 + \ldots$

Find the common ratio.

$\frac{4}{8}=\frac{1}{2}$

$\frac{2}{4}=\frac{1}{2}$

Use the equation for an infinite geometric series:

$\displaystyle\sum_{n=1}^{\infty}{ar^{n-1}}=\frac{a}{1-r}$

$16$

1. $3 + 9 + 27 + 81 + \ldots$

See part (a).

Not possible. $r>1$, and the terms keep increasing.


text:
# Show transcribed image text a. Use the given Taylor polynomial P2 to approximate the given quantity.

Show transcribed image text a. Use the given Taylor polynomial P2 to approximate the given quantity. b. Compute the absolute error in the approximation assuming the exact value is given by a calculator. Approximate e^-0.15 using f(x) = e^-x and p2(x) = 1 - x + x^2/2. a. Using the Taylor polynomial p2, e^-0.15 (Do not round until the final answer. Then round to four decimal places as needed.) b. absolute error (Use scientific notation. Use the multiplication symbol in the math palette as needed. Round to two decimal places as needed.)


text: By using this website, you agree to our Cookie Policy. Online calculators and formulas for an annulus and other geometry problems. See our full terms of service. Thank you for your questionnaire.Sending completion, Area of a parallelogram given base and height, Area of a parallelogram given sides and angle. 1. It does not matter which side you take as base, as long as the height you use it perpendicular to it. A = (b * h) … Some functions are limited now because setting of JAVASCRIPT of the browser is OFF. We start by moving this marked triangle over to the right hand side. The formula for the area of a parallelogram is base x height. An enormous range of area of parallelograms worksheets for grade 5 through grade 8 have been included here. The height is not the side length like you might use in a rectangle, but instead it is the altitude. Enter the base a and height h and when you click on the button "Calculate the area of parallelogram", the area of parallelogram is calculated from the base and height. Area R = ab sin (A) = 53.35 * 12 * sin (112.52°) = 640.2 * 0.923. The formula for the height of the parallelogram is as follows: Each parallelogram can be rearranged to form a rectangle. The result is always the square unit: for example, square feet, square inches, square yards or square miles. A KS3 Bitesize 11-14 Maths guide showing how to prove the area of a parallelogram is base multiplied by height. A rectangle is a quadrilateral with four right angles. Otherwise known as a quadrangle, a parallelogram is a 2D shape that has two pairs of parallel sides. Step 2: Enter length of horizontal sides into Length 1 and Length 2. A circle is a perfect round shape. Free Parallelogram Area & Perimeter Calculator - calculate area & perimeter of a parallelogram step by step This website uses cookies to ensure you get the best experience. = 591.39. Calculate the area of a parallelogram whose base is 24 in and a height of 13 in. Your feedback and comments may be posted as customer voice. We are not to be held responsible for any resulting damages from proper or improper use of the service. A parallelogram is a quadrilateral with two pairs of parallel sides. The diagonals of a parallelogram bisect each other. Area of a parallelogram formula. The area of a polygon is the number of square units inside the polygon. This can be easily understood visually - if you use the height to cut a triangle from the shape, then spin it 180° and move it to other side, it will perfectly match, and the shape will be that of a rectangle. It’s that easy! An easy to use, free area calculator you can use to calculate the area of shapes like square, rectangle, triangle, circle, parallelogram, trapezoid, ellipse, octagon, and sector of a circle. Each tool is carefully developed and rigorously tested, and our content is well-sourced, but despite our best effort it is possible they contain errors. Our online calculators, converters, randomizers, and content are provided "as is", free of charge, and without any warranty or guarantee. 2. The area is the size of the parallelogram. Area of a Parallelogram Formula If you know the length of base b , and you know the height or width h , you can now multiply those two numbers to get area using this formula: Opposite sides are equal in length and opposite angles are equal in measure. Area of a parallelogram given sides and angle Calculator, $$\normalsize Parallelogram\ (a,b,\theta\rightarrow S)\\. The formula is actually the same as that for a rectangle, since it the area of a parallelogram is basically the area of a rectangle which has for sides the parallelogram's base and height. He decides to use the area formula for parallelograms in order to find the missing dimension. The formula for finding the area of a parallelogram is base times the height, but there is a slight twist. In Euclidean geometry, a parallelogram is a simple quadrilateral with two pairs of parallel sides. The calculator will now display the area of the parallelogram, which is 600cm². Use this calculator to easily calculate the area of a parallelogram by its base and height. Multiply the base of the parallelogram by the height to find the area. [3] 2019/04/18 11:24 Female / Under 20 years old / High-school/ University/ Grad student / A little / Purpose of use And Width of the vertical sides into Width 1 and Width 2. Calculator online for an parallelogram. Calculating The Area Of A Parallelogram A parallelogram is a 4-sided shape formed by two pairs of parallel lines. Now, you will be able to easily solve problems on the area of parallelogram vectors, area of parallelogram proofs, and area of a parallelogram without height, and use the area of parallelogram calculator. The area of the parallelogram is the magnitude of the cross product of \overrightarrow{AB}  and  \overrightarrow{AD} = AB\cdot AD\cdot sin(\theta)  so you would directly get the area if that's what you want, but you could then just divide by the base to get the height. Navigate through these 40+ worksheets on calculating the area using the formula with the base and height measures expressed as integers, fractions or decimals, area problems involving unit conversion, finding the missing dimensions and much more. The area of a parallelogram is the \(base \times perpendicular~height~(b \times h)$$.. You can see that this is true by rearranging the parallelogram to make a rectangle. [1]  2020/03/24 03:52   Female / Under 20 years old / High-school/ University/ Grad student / A little /, [2]  2019/07/14 23:58   Female / 40 years old level / Self-employed people / Useful /, [3]  2019/04/18 11:24   Female / Under 20 years old / High-school/ University/ Grad student / A little /, [4]  2018/12/06 01:45   Male / Under 20 years old / Others / A little /, [5]  2018/11/22 12:22   Male / Under 20 years old / Elementary school/ Junior high-school student / Very /, [6]  2018/04/12 13:28   Male / Under 20 years old / High-school/ University/ Grad student / Very /, [7]  2018/01/18 21:09   Male / Under 20 years old / High-school/ University/ Grad student / Not at All /, [8]  2017/06/13 10:13   Male / 60 years old level or over / Self-employed people / Very /, [9]  2017/03/26 19:00   Female / Under 20 years old / High-school/ University/ Grad student / Useful /, [10]  2015/10/10 11:45   Female / 30 years old level / A teacher / A researcher / Very /. Find the length of the height of a parallelogram without area if given side and angle : The acute angle of intersection of the heights is equal to the acute angle of the parallelogram. Area of a triangle calculation using all different rules, side and height, SSS, ASA, SAS, SSA, etc. Parallelogram Area Without Height If the height of the parallelogram is unknown to us, then we can use trigonometry concept here to find its area. Area is 2-dimensional like a carpet or an area rug. The formula is: The formula is actually the same as that for a rectangle, since it the area of a parallelogram is basically the area of a rectangle which has for sides the parallelogram's base and height. Learn how to find the area and perimeter of circles. Calculate area of parallelogram needed to build a fractile pattern for wooden art project. Calculate the area of a parallelogram given base and height using the formula. As with other similar calculations, it is important to multiply in identical units. Base and height as in the figure below: You need two measurements to calculate the area using our area of parallelogram calculator. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. These online calculators use the formula and properties of the parallelogram listed below. Parallelogram Calculator Directions Just tell us what you know by selecting the image below, then you can enter your information and we will calculate everything. (See image above). Learn how to find the area and perimeter of a parallelogram. Step 3: Press calculate button. To find the area of a pallelogram-shaped surface requires information about its base and height. This becomes the height and as it was previously mentioned area of the parallelogram is base times height (bh) we now know both the values of base (b) and height (h) and calculating this, the area of the parallelogram will be 15.8 which will be equal to 120. The area of a rectangle is found by multiplying the base by its height. In this case, we’ll say the base and height are 30cm and 20cm respectively. We need to find the width (or height) h of the parallelogram; that is, the distance of a perpendicular line drawn from base C D to A B. Easy to use online calculators to calculate the area Ap, sides, diagonals, height and angles of a parallelogram. 3. Each of the 4 parallelograms must have a minimal area of 720 square inches and a set base of 36 inches, which means Michael has to determine the minimal height of each parallelogram to ensure the area requirements. Area = ab sin (x) https://www.gigacalculator.com/calculators/area-of-parallelogram-calculator.php. Then you can apply the simple multiplication formula above, making sure both lengths are of the same unit. If you'd like to cite this online calculator resource and information as provided on the page, you can use the following citation: Georgiev G.Z., "Area of a Parallelogram Calculator", [online] Available at: https://www.gigacalculator.com/calculators/area-of-parallelogram-calculator.php URL [Accessed Date: 20 Jan, 2021]. Step 1: Measure all sides of the area in one unit (Feet, Meter, Inches or any other). A parallelogram is a 4-sided shape formed by two pairs of parallel lines. Real-world examples related to the area of a parallelogram; Once you get these concepts, you will then learn about the area of parallelogram using vectors, as well as learn about the area of a parallelogram without height. For example, if the base of a parallelogram is 8 inches and the height to it is 4 inches, then its area is 8 x 4 = 32 square inches. The height (altitude) is found by drawing a perpendicular line from the base to the highest point on the shape. The formula for the area of a parallelogram is base x height. Perimeter P = 2a + 2b. Rectangle. There are lots of options below. This is the way the area of parallelogram … How to use irregular area calculator? Calculate the unknown defining areas, lengths and angles of a paralellogram. For example, if the base is 5, and the height 3, then your area is. Parallelogram Area Calculator. To find the area of a parallelogram, multiply the base by the height. You need to take two measurements - the base is easily measured, while for the height you would need to build a right angle at one of the parallelogram's angles. The base and height of the parallelogram are perpendicular. Solution. To use the parallelogram area calculator simply enter the base and height into the form fields and our area calculator will perform the calculations for you. = 2 (53.35 + 12) = 2 (65.35) = 130.7. Opposite sides are equal in length and opposite angles are equal in measure. Base and height as in the figure below: You need two measurements to calculate the area using our area of parallelogram calculator. If your problem gives you a measurement of the base and height of a parallelogram, simply multiply them to get your area. It is one of the simplest shapes, and … For this example, we’ll go with the calculator that solves the parallelogram’s area when given the base and height. Area Ar of a parallelogram may be calculated using different … To find the area of a parallelogram, multiply the base by the height. The Parallelogram Area Calculator can calculate the area of a parallelogram instantly if you enter in the height and width of the parallelogram.The height of a parallelogram is the distance vertically from the bottom edge to the top edge moving in a straight line. Parallelogram calculator The area, perimeter, height, diagonals and angles of a parallelogram The parallelogram is a quadrangle with opposite sides parallel to each other, and consequently, the opposite sides have the same length, the opposite angles are equal, and the two diagonals intersect each other at … 15 i … Enter the right values in the right sections. It doesn't matter which one, as long as one of the angle's arms is the base. The height is the distance between the base and the top of the parallelogram. Through our simulation, you will also understand how the area of parallelogram calculator works. Parallelograms - area. In the metric systems you can get square centimeters, square meters, square kilometers, and others. ∴ Perimeter of the parallelogram is 130.7 cm, area is 591.39 cm², height is 49.2 cm, diagonals are 59 cm, 50 cm, side length is 53.35 cm, angles are 112.5°, 67.47°. The output our calculator provides is independent of the unit of measurement. We hope you enjoyed learning about the area of a parallelogram with the simulations and practice questions. Formulas, explanations, and graphs for each calculation. The area of a rectangle is base times height and so, the area of a parallelogram is also base times height. Area Ap, sides, diagonals, height and angles of a triangle calculation all... Held responsible for any resulting damages from proper or improper use of the parallelogram listed.... The service both lengths are of equal measure limited now because setting of JAVASCRIPT the.: Enter length of horizontal sides into Width 1 and length 2 for any resulting from! Arms is the base to the right hand side base, as long as the.... Cookie Policy perpendicular line from the base to the right hand side s ).... Number of square units inside the polygon parallelogram area calculator without height for wooden art project or facing sides a! 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Class-11Maharashtra Board - Algebra of Complex Numbers - LearnNext offers animated video lessons with neatly explained examples, Study Material, FREE NCERT Solutions, Exercises and Tests. Ch 5 Maths Class 11 complex numbers form a part of Unit 2- Algebra of CBSE Class 11 Maths syllabus, which has a weightage of 30 marks (out of total 80 marks). Complex Numbers. If z1=3,z2=-7i, and z3=5+4i, show that z1(z2+z3)=z1z2+z1z3, If z1=2+5i, z2=-3-4i, and z3=1+i, find the additive and multiplicate inverse of z1,z2 and z3. Class-11Maharashtra Board - Introduction to Complex Numbers - LearnNext offers animated video lessons with neatly explained examples, Study Material, FREE NCERT Solutions, Exercises and Tests. =12 + 5i – (4 − 2i) =8 +7i. Question 1. } Solution) From complex number identities, we know how to subtract two complex numbers. function gtag(){dataLayer.push(arguments);} This will clear students doubts about any question and improve application skills while preparing for board exams. a) Find b and c b) Write down the second root and check it. For a complex number z = p + iq, p is known as the real part, represented by Re z and q is known as the imaginary part, it is represented by Im z of complex number z. Chapter 5 Class 11 Complex Numbers. = denticity × number of ligandCoordination number = 2 × 2 + 2 × 1 = 6Charge on ligand + O.S. Register online for Maths tuition on Vedantu.com to … DONE! Practice to excel and get familiar with the paper pattern and the type of questions. The Inverse Trigonometric Functions Class 12 PDF complete solutions have been provided for Chapter 2 of Class 12 NCERT Maths. "name": "QB365", Argand plane and polar representation of complex numbers. Dr. K. Thirumurugan. It is denoted by z i.e. Download all subjects previous year question papers with answers from our website. Complex Number Part 1 (Definition Algerba of complex number) Complex Number Part 2 (Identifier Modulus Conjugate) Complex Number … Uses of Complex numbers - Here are the uses of Complex numbers, Complex numbers are … NCERT Solutions For Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are prepared by the expert teachers at BYJU’S. "name": "QB365", Question Papers 1851. Here you can get Class 11 Important Questions Maths based on NCERT Text book for Class XI. Since these complex numbers have imaginary parts, it is not possible to find out the greater complex number between them. Complex Numbers Problems with Solutions and Answers - Grade 12. Also, this Intermediate II stage Maths syllabus put emphasis on discovery and understanding of Basic Mathematical concepts by students. These solutions for Complex Numbers are extremely popular among Class 12 Science students for Math Complex Numbers Solutions come handy for quickly completing your homework and preparing for exams. e.g. Expert Teachers at SamacheerKalvi.Guru has created Tamilnadu State Board 12th Maths Solutions Book Pdf Free Download New Syllabus of Volume 1 and Volume 2 in English Medium and Tamil Medium are part of Samacheer Kalvi 12th Books Solutions.Here we have given TN Board Samacheer Kalvi 12th Std Maths Guide Pdf Free Download of Text Book Back Questions and Answers, Notes, … (ii) Advertisement Remove all ads. This chapter provides detailed information about the complex numbers and how to represent complex numbers in the Arg and plane. NCERT Exemplar Class 12 Chemistry is very important resource for students preparing for XII Board Examination. We often use z for a complex number. If $$\frac { (a+i)^{ 2 } }{ 2a-i }$$ =p+iq, show that p2+q2 =$$\frac { ({ a }^{ 2 }+i)^{ 2 } }{ 4a^{ 2 }+1 }$$. } Concept ... Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Arts; Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Commerce; Free Practice for SAT, ACT and Compass Math tests. Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Exercise 5.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. "@context": "https://schema.org", Important Solutions 4565. CBSE Class 12 Maths formulas available for Chapterwise on Vedantu.com. Maharashtra state board books are prepared by the international standards that facilitate the overall development of students. $$\left[ Re\left( iz \right) \right] ^{ 2 }=3$$, If (cosθ + i sinθ)2 = x + iy, then show that x2+y2 =1. Here we have provided NCERT Exemplar Problems Solutions along with NCERT Exemplar Problems Class 12.. Click here to download Free PDFs of NCERT Solutions prepared by … The syllabus is divided into three sections: A, B and C. Section A is compulsory for all candidates. Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board chapter 1 (Complex Numbers) include all questions with solution and detail explanation. 01:00:00 Hrs, $$\sum _{ i=1 }^{ 13 }{ \left( { i }^{ n }+i^{ n-1 } \right) }$$, $$\left( \frac { 1 }{ \sqrt { 2 } } \right)$$, $$\frac { 1 }{ 1-cos\theta -isin\theta }$$, $$\left[ Re\left( iz \right) \right] ^{ 2 }=3$$, $$\frac { ({ a }^{ 2 }+i)^{ 2 } }{ 4a^{ 2 }+1 }$$, $$\left( \cfrac { z-i }{ z+2 } \right) =\cfrac { \pi }{ 4 }$$, $$\left( \cfrac { \sqrt { 3 } }{ 2 } +\cfrac { i }{ 2 } \right) ^{ 5 }+\left( \cfrac { \sqrt { 3 } }{ 2 } -\cfrac { i }{ 2 } \right) ^{ 5 }=-\sqrt { 3 }$$, Seventh Standard Social Science CIV - Market and Consumer Protection English Medium Free Online Test 1 Mark Questions with Answer Key 2020 - 2021, 12th Standard Maths English Medium Free Online Test One Mark Questions with Answer Key 2020, 12th Standard Maths English Medium Free Online Test 1 Mark Questions 2020, 12th Standard Maths English Medium Free Online Test 1 Mark Questions 2020 - Part Two, 12th Standard Maths English Medium Free Online Test Book Back One Mark Questions, 12th Standard Maths English Medium Free Online Test Book Back 1 Mark Questions with Answer Key, 12th Standard Maths English Medium Free Online Test Book Back One Mark Questions - Part Two, 12th Standard Maths English Medium Free Online Test Book Back 1 Mark Questions with Answer Key - Part Two, 12th Standard Maths English Medium Free Online Test Book Back One Mark Questions - Part Three, 12th Standard Maths English Medium Free Online Test Book Back 1 Mark Questions with Answer Key - Part Three, 12th Standard Maths English Medium Free Online Test Creative 1 Mark Questions, 12th Standard Maths English Medium Free Online Test Creative One Mark Questions with Answer Key, 12th Standard Maths English Medium Free Online Test Creative 1 Mark Questions - Part Two, 12th Standard Maths English Medium Free Online Test Creative One Mark Questions with Answer Key - Part Two, 12th Standard Maths English Medium Free Online Test Creative 1 Mark Questions - Part Three, 12th Maths - Complex Numbers Model Question Paper 1, QB365 Mobile App For Practice Question Papers, Introductory Microeconomics and Macroeconomics, à®à®£à®¿à®©à®¿ à®ªà®¯à®©à¯à®ªà®¾à®à¯à®à®³à¯, à®à®£à®¿à®©à®¿ à®¤à¯à®´à®¿à®²à¯à®¨à¯à®à¯à®ªà®®à¯, à®à®£à®à¯à®à¯à®ªà¯à®ªà®¤à®¿à®µà®¿à®¯à®²à¯, à®à®£à®¿à®©à®¿ à®ªà®¯à®©à¯à®ªà®¾à®à¯à®à®³à¯. ... Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. z > 0, 4 + 2i < 2 + 4 i are meaningless. Complex Numbers Class 11 – A number that can be represented in form p + iq is defined as a complex number. Write the conjugates of the following complex numbers: -5-7 i . Find the locus of Z if |3z - 5| = 3 |z + 1| where z=x+iy. Need for complex numbers, especially √1, to be motivated by inability to solve some of the quardratic equations. window.dataLayer = window.dataLayer || []; Register online for Maths tuition on Vedantu.com to … If z=x+iy and arg $$\left( \cfrac { z-i }{ z+2 } \right) =\cfrac { \pi }{ 4 }$$, then show that x2+y2+3x-3y+2=0, Show that $$\left( \cfrac { \sqrt { 3 } }{ 2 } +\cfrac { i }{ 2 } \right) ^{ 5 }+\left( \cfrac { \sqrt { 3 } }{ 2 } -\cfrac { i }{ 2 } \right) ^{ 5 }=-\sqrt { 3 }$$, Verify that arg(1+i) + arg(1-i) = arg[(1+i) (1-i)]. Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Complex numbers are useful abstract quantities that can be used in calculations and result in physically meaningful solutions. } { Find the modulus and argument of the following complex numbers and convert them in polar form. Question 2) Subtract the complex numbers 12 + 5i and 4 − 2i. Chapter 6 Class 11 Linear Inequalities. Here we have covered Important Questions on Complex Numbers and Quadratic Equations for Class 11 Maths subject. All the examples listed here are in Cartesian form. Students are advised to refer to the attached assignments and practise them regularly. Class. Download free printable worksheets for CBSE Class 11 Trigonometry with important topic wise questions, students must practice the NCERT Class 11 Trigonometry worksheets, question banks, workbooks and exercises with solutions which will help them in revision of important concepts Class 11 Trigonometry. { Class 11 Maths Complex Numbers . "@type": "Organization", Contents1 Important Questions for CBSE Class 12 Chemistry – The d- and f- Block Elements1.1 PREVIOUS YEARS QUESTIONS1.2 20151.3 Very Short Answer Type Questions [1 Mark]1.4 Short Answer Type Questions [I] [2 Marks]1.5 Short Answer Type Questions [II] [3 Marks]1.6 20141.7 Very Short Answer Type Questions [1 Mark]1.8 Short Answer Type Questions [I] [2 Marks]1.9 […] 1 PM. And Re() for the real part and Im() for the imaginary part, like this: Which looks like this on the complex plane: The Mandelbrot Set. If z=$$\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 107 }+\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 107 }$$ , then show that Im (z) =0. Find the values of the real numbers x and y, if the complex numbers (3−i)x−(2−i)y+2i +5 and 2x+(−1+2i)y+3+ 2i are equal. Download 12th Standard Stateboard Maths Complex Numbers Model question paper 2. Complex Numbers DEFINITION: Complex numbers are definited as expressions of the form a + ib where a, b ∈ R & i = $$\sqrt { -1 }$$ . The textbook is approved by the Tamilnadu State Board on the basis of the Tamilnadu Board Class 12 Maths syllabus. Next Tamilnadu State Board – Class XII Maths – 6 Marks Important Questions. A number of the form a + ib, where a and b are real numbers, is called a complex number, a is called the real part and b is called the imaginary part of the complex number. The Maths textbook use very simple steps and examples to explain topics. of metal ion = charge on complex– 2 + x = 0⇒ x = +2 Subject. Get Important questions for class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations here.Students can get different types of questions covered in this chapter. (i) Q 2. If $$\frac { (a+i)^{ 2 } }{ 2a-i }$$ =p+iq, show that p2+q2 =$$\frac { ({ a }^{ 2 }+i)^{ 2 } }{ 4a^{ 2 }+1 }$$. 5. Write the following in the rectangular form: (i) $$\overline{(5+9 i)+(2-4 i)}$$ Samacheer Kalvi 12th Books Solutions Menu Toggle. " /> --> If z1=3,z2=-7i, and z3=5+4i, show that z1(z2+z3)=z1z2+z1z3. PGT in Mathematics, GHSS Vazhudhavur, Villupuram Dist, Tamilnadu, India, He has been teaching Math for 15 years. Here we have provided NCERT Exemplar Problems Solutions along with NCERT Exemplar Problems Class 12.. Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.1. (iv) Q 3. " /> --> We hope the given Tamilnadu State Board Class 12th Maths Solutions Book Volume 1 and Volume 2 Pdf Free Download New Syllabus in English Medium and Tamil Medium will help you. Find the values of the real number x and y if 3x + (2x - 3y) i = 6 + 3i9. Also, BYJU’S provides step by step solutions for all NCERT problems, thereby ensuring students understand them and clear their exams with flying colours. gtag('js', new Date()); Maharashtra State Board of Education has prescribed Mathematics and Statistics Syllabus for Higher Secondary Level i.e. TN Stateboard Maths Class 12 Sample Question Papers for students appearing in TN Stateboard Hr Sec Public Exams. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ … 2. If you have any queries regarding TN Board 12th Standard Samacheer Kalvi Maths Guide Pdf Free Download of Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with … It also lays the foundation for many of the concepts that are commonly asked for the boards as well as in many competitive exams. "sameAs": "https://www.qb365.in/materials/" Textbook Solutions 13411. "@type": "Organization", The chapter Complex Number and Quadratic Equation Class 11 consists of several mathematical concepts, theorems, and formulas that are fundamental for solving sums of algebra. "description": "12th Maths - Complex Numbers Important Questions, the complete list of 12th Standard STATEBOARD question papers, syllabus, exam tips, study material, previous year exam question papers, centum tips, formula, answer keys, etc.."?>", Find the radius and centre of the circle $$z\bar { z }$$-(2+3i)z-(2-3i)$$\bar { z }$$+9 =0 where z is a complex number. Download the important maths formulas and equations to solve the problems easily and score more marks in your Class 12 CBSE Board … Study materials, centum tips,formula, Syllabus, Previous Year Question Papers, online practice tests and more, portal for Students & Parents 2019 - 2020 Comparison of complex numbers Consider two complex numbers z 1 = 2 + 3i, z 2 = 4 + 2i. Algebraic Operations: Inequalities in complex numbers are not defined. ‘a’ is called as real part of z (Re z) and ‘b’ is called as imaginary part of z (Im z). "description": "12th Maths - Complex Numbers Model Question Paper 1, the complete list of 12th Standard STATEBOARD question papers, syllabus, exam tips, study material, previous year exam question papers, centum tips, formula, answer keys, etc.."?>", Find the modulus of the following complex numbers NCERT Exemplar Class 12 Maths is very important resource for students preparing for XII Board Examination. Doing Ph.D. in Bharathiar University, on a problem of Stability analysis in Non – Newtonian fluids. The Telangana Class 12 Maths Syllabus is designed by the Telangana State Board of Intermediate Education (TSBIE). 2 + i3, -5 + 6i, 23i, (2-3i), (12-i1), 3i are some of the examples of complex numbers. Any complex number is : (i) Purely real, if b = 0 (ii) Purely imaginary, if a = 0 (iii) Imaginary, if b not equal to 0 . function gtag(){dataLayer.push(arguments);} Question from very important topics are covered by NCERT Exemplar Class 12.You also get idea about the type of questions and method to answer in your class 12th examination. Class-12Tamilnadu Samacheer Kalvi Board - Introduction to Complex Numbers - LearnNext offers animated video lessons with neatly explained examples, … HSSLIVE Plus One(XI) Maths Previous Year Chapter Wise Question Paper of Kerala Board Class 11 is a very important resource for students who are going to appear for the upcoming board exams. Check important questions and answers for Class 12 Chemistry Board Exam 2020 from 9 - Coordination Compounds. "@type": "Course", Contents1 Important Questions for CBSE Class 12 Chemistry – Coordination Compounds1.1 PREVIOUS YEARS QUESTIONS1.2 20151.3 Very Short Answer Type Question [1 Mark]1.4 Short Answer Type Questions [I] [2 Marks]1.5 Short Answer Type Questions [II] [3 Marks]1.6 20141.7 Very Short Answer Type Questions [1 Mark]1.8 Short Answer Type Questions [l] [2 Marks]1.9 20131.10 Short Answer Type […] Find all complex numbers z such that z 2 = -1 + 2 sqrt(6) i. z = a + ib. For any suggestion or improvement regarding plus one previous year question paper do let us know in the comments below. These questions are based on the latest CBSE Class 12 Chemistry Syllabus. Every Complex Number Can Be Regarded As Therefore, the combination of both the real number and imaginary number is a complex number.. Hello friends Complex Numbers Class 11th Maharashtra Board Part - 1 This video is based on complex numbers class 11th Maharashtra Board new syllabus. Maths Class 11 Important Questions are very helpful to score high marks in board exams. Samacheer Kalvi 12th Books Solutions Menu Toggle. Show that $$\left( \cfrac { \sqrt { 3 } }{ 2 } +\cfrac { i }{ 2 } \right) ^{ 5 }+\left( \cfrac { \sqrt { 3 } }{ 2 } -\cfrac { i }{ 2 } \right) ^{ 5 }=-\sqrt { 3 }$$. Statement of Fundamental Theorem of Algebra, solution of quadratic equations in the complex number system. The complex number 2 + 4i is one of the root to the quadratic equation x 2 + bx + c = 0, where b and c are real numbers. Share this Video Lesson with your friends Support US to Provide FREE Education Subscribe to Us on YouTube Next > Try Further learning steps . The Telangana Class 12 Maths Syllabus is designed by the Telangana State Board of Intermediate Education (TSBIE). What type of isomerism is exhibited by the complex [Co(en)3]3+? Session starts at 8 PM and will end at approx. NCERT Solution Complex Number . Maths Important Questions Class 11 are given below. Complex Numbers Class 11 – A number that can be represented in form p + iq is defined as a complex number. Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.5 Additional Problems. Thus, the ordering relation (greater than or less than) of complex numbers, that is greater than or less than, is meaningless. Vedantu has a team of passionate and devoted educators who are ready to depart their knowledge and expertise in the most effective way possible. All the topics will be discussed in detail and would be helpful for the aspirants preparing for IIT JEE exam. yes its there in class 12..and it is interesting chapter. Pre Boards. Algebraic properties of complex numbers. Let z_1= a + ib \text{ and } z_2 = c + id . Free PDF download of Class 11 Maths revision notes & short key-notes for Chapter-5 Complex Numbers and Quadratic Equations to score high marks in exams, prepared by expert mathematics teachers from latest edition of CBSE books. gtag('config', 'UA-97495421-1'); The value of $$\sum _{ i=1 }^{ 13 }{ \left( { i }^{ n }+i^{ n-1 } \right) }$$ is, If |z-2+i|≤2, then the greatest value of |z| is, If |z1|=1,|z2|=2|z3|=3 and |9z1z2+4z1z3+z2z3|=12, then the value of |z1+z2+z3| is, If z=cos$$\frac { \pi }{ 4 }$$+i sin$$\frac { \pi }{ 6 }$$, then, |z|=$$\frac { \sqrt { 3 } }{ 2 }$$, arg(z)=$$\frac { 5\pi }{ 24 }$$, |z| =$$\frac { \sqrt { 3 } }{ 2 }$$, arg (z) =tan-1$$\left( \frac { 1 }{ \sqrt { 2 } } \right)$$, If z=$$\frac { 1 }{ 1-cos\theta -isin\theta }$$, the Re(z) =, $$\frac{1}{2}$$cot$$\frac { \theta }{ 2 }$$, Obtain the Cartesian form of the locus of z=x+iy in Chapter 11 Class 11 Conic Sections. Also, this Intermediate II stage Maths syllabus put emphasis on discovery and understanding of Basic Mathematical concepts by students. The board has revised this syllabus by keeping current trends in view. Textbook Solutions 7155 Question Bank Solutions 4607. The course includes 18 sessions of 90 minutes each. gtag('js', new Date()); 12th Standard Stateboard Maths Complex Numbers Model question paper 2 Covered with important questions in Complex Numbers, Best website for Tamilnadu School Students, parents & teachers, download free question papers and answer keys for 12th Standard. A complex number is usually denoted by the letter ‘z’. 2 + 3i 4 − 5i × 4 + 5i 4 + 5i = 8 + 10i + 12i + 15i 2 16 + 25 = −7 + 22i41. A learner can find anything and everything related to Class 12 Maths Chapter 2. Here, p and q are real numbers and $$i=\sqrt{-1}$$. Free PDF download of Important Questions with solutions for CBSE Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations prepared by expert Maths teachers from latest edition of CBSE(NCERT) books. Class 11 Maths; Class 12 Maths; Other Courses; PYQ Log In; Select Page. Areas and will help them to identify their weak areas and will help to... Any question and improve application skills while preparing for Board exams is no validity if we say that number! Here, p and q are real numbers and Quadratic Equations in the complex numbers Model question paper 2,!: free JEE Exam Preparation Group complex numbers class 12 state board Eckovation two complex numbers Ex 2.4 where students can interact teachers/experts/students. > 0, 4 + 2i √1, to be motivated by inability to solve some of the quardratic.. Stage of their preparations will benefit from the course = denticity × of... Non – Newtonian fluids { -1 } \ ) are real numbers and \ ( i=\sqrt { }... This Chapter provides detailed information about the complex numbers Ex 2.4 this will students! For practice numbers z 1 = 2 × 1 = 6Charge on ligand + O.S Chapter 5 numbers... Maths help the students in solving the Problems quickly, accurately and efficiently and. Tutor Recommended for you download 12th standard Stateboard Maths complex numbers: -5-7 i is essential Class... 6 marks Important questions are based on NCERT Text book for Class 12.. and it is not possible find. To a lot of Problems is a crucial stage for the complex number system q 3 Online Maths... Cbse Class 12 Maths Chapter 5 complex numbers will be provided in English the. Ways of representation for the aspirants preparing for IIT JEE Exam × 1 = 6Charge on ligand O.S. Are based on NCERT Text book for Class 12 Maths Chapter 2 complex numbers have imaginary,. > 0, 4 + 2i Equations for Class XI check it = 6 + 3i9 of 12. 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## Chemistry and Chemical Reactivity (9th Edition)

a) The number of moles of $ClF_3$ is: $n=PV/RT$ $n=(250/760×1\ atm)×2.5\ L/((0.082)(293\ K))$ $n=0.034\ mol$ From stoichiometry, we find: $0.034×6/4=0.051\ mol\ NiO$ $0.051\ mol×74.69\ g/mol=3.83\ g$ b) Since T and V are constant, it follows: $P=250\ mmHg×(2+3)/4=312.5\ mmHg$ $P_{Cl_2}=250\ mmHg×2/4=125\ mmHg$ $P_{O_2}=250\ mmHg×3/4=187.5\ mmHg$