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3 years ago

What is the surface area of the box?

Mathematics

2 answers:

rusak2 [61]3 years ago

6 0

Use the formula length times width times height

jonny [76]3 years ago

3 0

1700 because I multiply 30 x 20 then 20 x 5 then 30 x 5 then add them all up and multiply your answer by 2

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What’s the new value of x ??! Will mark Brianliest !!!!!!!! PLEASE help !!!!!!!!!!!!!

gayaneshka [121]

Answer:

Step-by-step explanation:

add 67 plus 64 and subtract that sum from 180

3 0

2 years ago

The big desk in your boss's office is 75 inches long. How long is that in feet?

masya89 [10]

Answer:

The big desk in your boss's office is 6.25 feet long.

Step-by-step explanation:

As given

The big desk in your boss's office is 75 inches long.

As the relationship between inches and feet.

As 1 feet = 12 inches

$1\ inches = \frac{1}{12}\ feet$

Now convert 75 inches into feet.

$75\ inches = \frac{75}{12}\ feet$

$75\ inches = 6.25\ feet$

Therefore the big desk in your boss's office is 6.25 feet long.

4 0

3 years ago

Read 2 more answers

Which of the following give the highest future value is 6000000 is invested at 6% for 3 years

Blababa [14]

Answer:

$5084745.76271

Step-by-step explanation:

Given data

Final amount= $6000000

Rate=6%

Time= 3 years

Now let us find the initial amount which is the principal

using the simple interest formula we have

6000000 = P(1+0.06*3)

6000000 =P(1+0.18)

6000000 =P*1.18

P= 6000000 /1.18

P=$5084745.76271

Hence the initial deposite is $5084745.76271

3 0

2 years ago

Add. 4 7/12+2 3/4+7 5/12

mihalych1998 [28]

Answer:

C.14¾

3 9⁄10

Step-by-step explanation:

$\frac{4}{5} + 3 \frac{1}{10} = \frac{8}{10} + 3 \frac{1}{10} = 3 \frac{9}{10}$

$4 \frac{7}{12} + 2 + \frac{3}{4} + 7 \frac{5}{12} = 13 \frac{12}{12} + \frac{3}{4} = 14 + \frac{3}{4} = 14 \frac{3}{4}$

I am joyous to assist you anytime.

3 0

3 years ago

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