Ask question

Ask question

All categories

3 years ago

11

The order of transformations in a sequence matters; reversing the order will not necessarily give the same image. Is this statem

ent True of False?

2 answers:

qaws [65]3 years ago

6 0

Answer:

False

If the transformations include a reflection, then the orientation will change. A translation or rotation will preserve the original orientation

Kamila [148]3 years ago

4 0

Answer:

I think true I'm not sure

You might be interested in

Factorise the following expressions.<br> a) 6x + 5x + 1<br> b) 6x2 13x + 6

fredd [130]

Answer:

a) 1 1 + 1

b)1 3 + 1 8

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Work out the size of angle x

marin [14]

Answer:

129°

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Distributions and Comparing Data Project

LenKa [72]

Answer:

1. The shape of the histogram created with MS Excel is approximately bell shaped and approximately evenly spread about the central (largest counts) values

The shape of the box plot created with MS Excel is approximately evenly distributed

2. The appropriate measure of central tendency for a bell shaped histogram is the mean and median, due to the approximately equal distribution about the highest frequency class

3. Part A

The 15 numbers between 1 and 20 generated by a random generator are;

1, 8, 8, 15, 4, 18, 11, 6, 17, 1, 18, 15, 10, 12, 11

The measure of center is the mean = (1+8+8+15+4+18+11+6+17+1+18+15+10+12+11)/15 = 155/15 = 31/3

The measure of spread used is the variance, s² = 32.38

$Where, \,s^2 =\dfrac{\sum \left (x_i-\bar x \right )^{2} }{n - 1}$

${\sum \left (x_i-\bar x \right )^{2} }$ = 453.333

n = 15

s² = 453. $\overline 3$ /(15 - 1) = 32.38

Part B

The measure of central tendency are the mean and the median because the size of the data (n = 15), is 75% of the population (N = 20) nd therefore the data is approximately normal and can be represented by the mean and the standard deviation

Step-by-step explanation:

4 0

3 years ago

Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find

Viktor [21]

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk $x^2+y^2\le3$ , I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere $x^2+y^2+z^2=3$ .

a. Let $C$ denote the hemispherical <u>c</u>ap $z=\sqrt{3-x^2-y^2}$ , parameterized by

$\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $C$ to be

$\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k$

Then the upward flux of $\vec F=(2z+2)\,\vec k$ through $C$ is

$\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du$

$\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du$

$=\boxed{2(3+2\sqrt3)\pi}$

b. Let $D$ be the disk that closes off the hemisphere $C$ , parameterized by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le\sqrt3$ and $0\le v\le2\pi$ . Take the normal to $D$ to be

$\vec s_v\times\vec s_u=-u\,\vec k$

Then the downward flux of $\vec F$ through $D$ is

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv$

$=\boxed{-6\pi}$

c. The net flux is then $\boxed{4\sqrt3\pi}$ .

d. By the divergence theorem, the flux of $\vec F$ across the closed hemisphere $H$ with boundary $C\cup D$ is equal to the integral of $\mathrm{div}\vec F$ over its interior:

$\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV$

We have

$\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2$

so the volume integral is

$2\displaystyle\iiint_H\mathrm dV$

which is 2 times the volume of the hemisphere $H$ , so that the net flux is $\boxed{4\sqrt3\pi}$ . Just to confirm, we could compute the integral in spherical coordinates:

$\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi$

4 0

4 years ago

True or False: (x + 3)2 = x2 + 9

malfutka [58]

Answer:

false

Step-by-step explanation:

3 0

3 years ago