Ask question

Ask question

All categories

3 years ago

15

Two parallel lines are cut by a transversal. Angle 1 measures (4x + 28)°, and the angle adjacent to the alternate exterior angle

with angle 1 measures (14x + 8)°. What is the value of x?
A.1/2
B.2
C.8
D.12

1 answer:

PilotLPTM [1.2K]3 years ago

5 0

C,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

You might be interested in

For what values of m does the graph of y = mx?- 5x-2 have no<br>5x - 2 have no x-intercepts?

Rasek [7]

Answer:

no

Step-by-step explanation:

It does.

5x-2=0

5x=2

x=2/5

(2/5, 0)

is an x-intercept

6 0

3 years ago

Read 2 more answers

Select the correct answer.

Marrrta [24]

Answer:

D makes an annuity. for annuity u have to make equal deposit regularly. so i think its D

8 0

3 years ago

If f(x) = 16x2 - 20x, what is the value of 13)?

aksik [14]

I don’t get this question right here

6 0

4 years ago

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

3 years ago

What is the surface area of a cylinder with a diameter of 8 and a height of 18?Express the answer in terms of pi

viktelen [127]

Step-by-step explanation:

diameter: 8

radius: 4

height: 18

$A=2\pi rh+2\pi r^2\\A=2\pi(4)(18)+2\pi(4)^2\\A=2\pi(72)+2\pi(16)\\A=144\pi+32\pi\\A=176\pi$

4 0

4 years ago