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3 years ago

the area of a rectangle is 42 square inches . the length is 7 inches . what is the perimeter of the rectangle ?

Mathematics

1 answer:

Sphinxa [80]3 years ago

6 0

Area = 42 sqaure inches

Length =7

Area = l x width

if the width is X then 7x=42 and x= 42/7 = 6

there X (width) is 6

perimeter thus is 7+7+6+6= 26

The answer is 26

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

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

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

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

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

4 years ago

The 60-foot rope was cut into two pieces. One of the pieces was 10 feet longer than 4 times the length of the other piece. How l

koban [17]

Answer:

50 and 10

Step-by-step explanation:

x+4x+10=y

y is 60

x+4x+10=60

5x+10=60

5x=50

x=10

10*4=40

40+10=50

50+10=60

3 0

3 years ago

HELP ASAP <br> What is the slope of the line on the graph?

ollegr [7]

The slope of this line is -1/6.

2 Point: (6,2) (-6,6)
x₁ y₁ x₂ y₂

Remember:

rise change in y y₂ - y₁ 6 - 2 4
---------- = ------------------- = ------------- = ----------- = ----------
run change in x x₂ - x₁ -6 - 6 -12

Simplify 4/-12

4 ÷ 4 1
------- = ------------- = -1/3
-12 ÷ 4 -3

The slope of the line is -1/3.

3 0

4 years ago

Help much needed ✨ <br> Thank you

Radda [10]

1 = 64
2 = 44
3 = 5 over 1

3 0

3 years ago

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Sergio039 [100]

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