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3 years ago

15

The base of the pyramid is a square. what is the surface area if the pyramid's height is 4 feet and the area of the base is 36ft

? show your work.

1 answer:

leonid [27]3 years ago

6 0

the answer to you're question is 2623.61440185 for the measurements for you're pyramid

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Help pls <img src="https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D%20-%20%5Cfrac%7B1%7D%7B3%7D" id="TexFormula1" title="\frac{2}{3

ivann1987 [24]

Step-by-step explanation:

$\frac{2}{3} - \frac{1}{3} = \frac{2 - 1}{3} = \frac{1}{3} \\$

6 0

3 years ago

A room is 3.4 m wide and 2.6 m long. What is the perimeter of the room

polet [3.4K]

Answer:

The perimeter of the room is 12m

7 0

3 years ago

Read 2 more answers

Speeeed brainlest<br> eeee

egoroff_w [7]

The answer is 3 3/8. You’re welcome :)

6 0

3 years ago

Read 2 more answers

Isnt this suppose to be 2/5?

olasank [31]

Answer:

Yes and no. 2/5 is the simplified form of 40/100. they are the same amount, so i would choose j

Step-by-step explanation:

4 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

Strike441 [17]

Parameterize $S$ by

$\vec r(u,v)=u\,\vec\imath+v\,\vec\jmath+(7-u^2-v^2)\,\vec k$

with $0\le u\le 1$ and $0\le v\le1$ . Take the normal vector to be

$\vec r_u\times\vec r_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k$

Then the flux across $S$ is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S$

$\displaystyle=\int_0^1\int_0^1(uv\,\vec\imath+(7-u^2-v^2)v\,\vec\jmath+(7-u^2-v^2)u\,\vec k)\cdot(\vec r_u\times\vec r_v)\,\mathrm du\,\mathrm dv$

$\displaystyle\int_0^1\int_0^1(2u^2v+(u+2v^2)(7-u^2-v^2))\,\mathrm du\,\mathrm dv=\frac{1343}{180}$

8 0

3 years ago