All categories

2 years ago

∠A and ∠B are adjacent. The sum of their measures is 92∘. ∠A measures (2x+5)∘. ∠B is three times the size of ∠A.

Mathematics

1 answer:

Ilia_Sergeevich [38]2 years ago

5 0

Answer:

Equation: 2x + 5 + 6x + 15 = 92

Solution: x = 9

m∠A =23° m∠B = 69°

Step-by-step explanation:

measure of ∠B = 3(2x + 5) = 6x + 15

The sum of the angles = 92 = 2x + 5 + 6x + 15

92 = 8x + 20

72 = 8x

x = 9

m∠A = 2(9) + 5 = 18 + 5 = 23

m∠B = 6(9) + 15 = 54 + 15 = 69

Check: 23 + 69 = 92 and 23(3) = 69

You might be interested in

Can anyone give me the answer and tell me how you did it?

pav-90 [236]

Answer:

the answer is 150

Step-by-step explanation:

8 0

2 years ago

In this graph what is the slope of the line?

sashaice [31]

Answer:

The slope of the line is 10

Step-by-step explanation:

To find the slope, simply identify 2 points on the line. I chose: (1,10) and (2,20)

20-10=10

2-1=1

The slope of the line is 10

To check this answer, just multiply X with 10 and you will always get Y!

7 0

3 years ago

Identify the four step used to solve this equation 4(x-7)=2x-6

irinina [24]

1. Solve the bracket 4×x and 4 × -7
4x-28=2x-6
2. Get the x on one side and the numbers on the other so: first +28
4x=2x+22 now -2x
so 2x=22 and now you only need to divide into 2
x=11

3 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

3 years ago

Please help pic below

kvasek [131]

Πr^2=area 9=r 9^2=81π=254.47

7 0

3 years ago