All categories

3 years ago

What is the surface area of this right rectangular prism?

Mathematics

2 answers:

hammer [34]3 years ago

5 0

Answer: Here's what I got:

Step-by-step explanation:

Is this what you wanted? My answer is A=22in²

Here's how I got it:

A=2(wl+hl+hw)=2·(1·3+2·3+2·1)=22in²

This was right? Make sure to like and rate! Appericate it.

mojhsa [17]3 years ago

3 0

Answer:

A=22in²

Step-by-step explanation:

A=2(wl+hl+hw)=2·(1·3+2·3+2·1)=22in²

You might be interested in

Solve.<br> X = -7<br> 4x - 4y = 12<br> (x,y)

Fynjy0 [20]

Answer:

(-7, -10)

$x=-7\\y=-10$

Step-by-step explanation:

$4x-4y=12$

Substitute x for -7.

$4(-7)-4y=12$

Multiply 4 by -7.

$-28-4y=12$

Add 28 on both sides.

$-4y=40$

Divide -4 on both sides.

$y=-10$

4 0

3 years ago

There are 42 lorries, 54 cars and 284 motorbikes on a ferry.

fiasKO [112]

Answer:

x = 27

Step-by-step explanation:

5 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

A cereal box has a height of 32 centimeters. it has a base with an area of 160-square centimeters. What is the volume,in cubic c

lutik1710 [3]

Volume=legnth times widht times height
legnth times width=area of base
so
volume=area of base times height for a prism

area of base=160
height=32
volume=160 times 32
volume=5120

volume is 5120 cubic centimeters or
5120 cm^3

3 0

3 years ago

HELP ME (How many cubes with side length 1/2 cm does it take to fill the prism

Natalija [7]

answer:

step-by-step explanation:

for this you would find the volume and then divide by 1/2

v = l x w x h

v = (1)(2)(3/2)

v = 3

now, divide

3 ÷ 1/2 = 6

therefore, it will take 6 half cubes to fill up the container

8 0

3 years ago