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3 years ago

a friend uses the distributive property to simplify 4(2b-5) and gets 8b - 5 as the result. describe and correct the error.

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

3 0

well it would be 8b - 20 because you have to multuply by 4 on 5 too not just 2b

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Mandy divided a 12-pound bag of oranges into 5 equal piles. Which rational number represents the weight of the oranges in each p

Gnom [1K]

12÷5= 2.4 pounds

2.4 pounds =2 4/10 pounds

=2 2/5

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3 years ago

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Please help!!!! (don’t comment for free points, this isn’t the place)

Allushta [10]

tan 44 = h/ x

x = 15 ft * sin 26

h = x * tan 44

= (15 ft * sin 26) (tan 44 )

so the anser is D

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3 years ago

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PLZZZZZZZZZZZZZZZZ HELP

Slav-nsk [51]

Answer:

The first box in the girl row is 18, the second one in the boy row is 8, and the bottom one is also 18.

Step-by-step explanation:

To get 81 to 9, you have to divide by 9, which means you have to do the same to 72. You get the ration 9 to 8, 9 being girls and 8 being guys. Then to get from 8 guys to 16, you have to multiply by 2, which means you do the same to 9 and get the answer of 18 girls. Hope this helps :)

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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:

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

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3 years ago

of all the students in class, 1/5 of them walk to school. What percent of students in the class walk to school

bagirrra123 [75]

Answer:

20%

Step-by-step explanation:

1/5 is the same as 2/10

2/10 is the same as 0.2

0.20 is the same as 20%

=) u got it!

7 0

3 years ago

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