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3 years ago

If y = x + 5 and y = 11, then x = what?

Mathematics

2 answers:

Anestetic [448]3 years ago

5 0

Answer:

Step-by-step explanation:

Solnce55 [7]3 years ago

4 0

Answer:

x=6

Step-by-step explanation:

y(11)= x + 5

-5 -5

6= x or x = 6

Hope this helps!!

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Solve for p: 3/8 +p/8= 3 1/8

Ivenika [448]

Solve for p:
p/8 + 3/8 = 25/8
p/8 + 3/8 = (p + 3)/8:
(p + 3)/8 = 25/8

Multiply both sides of (p + 3)/8 = 25/8 by 8:
(8 (p + 3))/8 = (8×25)/8
(8×25)/8 = (8×25)/8:
(8 (p + 3))/8 = (8×25)/8
(8 (p + 3))/8 = 8/8×(p + 3) = p + 3:p + 3 = (8×25)/8
(8×25)/8 = 8/8×25 = 25:
p + 3 = 25

Subtract 3 from both sides:
p + (3 - 3) = 25 - 3
3 - 3 = 0:
p = 25 - 3
25 - 3 = 22:
Answer: p = 22

4 0

3 years ago

Please help as soon as possible.

aniked [119]

Answer:

C = 37.6991

Step-by-step explanation:

The equation for the circumference of a circle is given: C = π · d

where, d is the diameter of the circle.

Plug in the value of d = 12 and use the π button on the calculator.

C = π · 12

C = 37.6991

5 0

3 years ago

Read 2 more answers

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:

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