All categories

4 years ago

1. What is the area of the shape? 6cm 4cm 2cm 3cm

1 answer:

3 0

Hello,
I’m not sure how to answer your question because looks like you forgot to put an image of the shape your talking about.

You might be interested in

I need help please

devlian [24]

I am pretty sure its B (minimum at -1, -4) because that would be were the first point would be at. The Graphing of this equation would be a "upside down u" shape.

4 0

3 years ago

Evaluate the expression when x=20, y=4, and z=7. 3z+8

Finger [1]

Answer:

3z+8

3(7)+8

21+8

29 is your answer ☺️☺️☺️

3 0

3 years ago

Read 2 more answers

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

3 years ago

Julie is photocopying a letter. The original letter was 8 in wide and 11 in. long. The new copy is 12 in. wide. How long is the

lions [1.4K]

It would be 16.5 inches long.
First you have to divide 11 by 8 (1.375) to get how long would the letter be if it was 1 inch wide. Then multiply 1.375 by 12 which is 16.5.

8 0

3 years ago

A can of orange juice has a circular base with an area of 156 square centimeters. The height of the can is 20 centimeters.

docker41 [41]

A Because if you do 156 x 20 it equals 3120. It’s A because the volume formula is Length x Width x height, And 156 and 20 were the only numbers given.

4 0

4 years ago