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3 years ago

14

I'll give brainliest!!! I need help fast please I just need someone to check my answers!!! Don't respond if you don't know what

you're talking about!!!
LOOK AT THE SCREENSHOT!!

1 answer:

olga2289 [7]3 years ago

3 0

Answer:

your correct

Step-by-step explanation:

:D

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Find the area of the triangle.<br><br> A. 75 km²<br> B. 107.3 km²<br> C. 84.3 km²<br> D. 71.8 km²

suter [353]

Answer:

A. 75 km²

Step-by-step explanation:

The area of a triangle

A =1/2 ab sin C where a and b are the lengths of the sides and C is the included angle

A = 1/2 (14)(11) sin (77)

A = 75.0264949885

4 0

3 years ago

Some of these are wrong can someone help me please??!?!

stealth61 [152]

Product is the result of two numbers multiplied, and some of your answers are not multiplied

6 0

3 years ago

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G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

3 years ago

1. Find "V". Round to hundreds.<br>V = LWH<br>L= 29.4<br>W = 16.7<br>H = 21.8

postnew [5]

Volume = 10,700

V = LWH

L = 29.4

W = 16.7

H = 21.8

All we need to do is multiply them all together and then round the answer to get our final answer to the hundreds place.

V = LWH

V = 29.4(16.7)(21.8)

V = 490.98(21.8)

V = 10703.364

Then, we need to round it to the hundreds place.

V = 10,700

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3 years ago

How to do the math for the following questions

agasfer [191]

88.21 fram is the anserer

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3 years ago