Ask question

Ask question

All categories

emmainna [20.7K]

3 years ago

12

How do you do Pythagorean theorem

1 answer:

ser-zykov [4K]3 years ago

6 0

Pythagorean theorem is a^2 + b^2 = c^2
2. You are trying to find x which is c^2
so plug in the other numbers to the other letters. 9^2 + 17^2 = c^2
9^2= 81. 17^2= 289
81 + 289 = 370
370= c^2
Now all you have to do is square root the number
*square root sign* 370
19.24
c = 19.24
So, x = 19.24

You might be interested in

Mike hiked to big bear lake in 4.5 hours at an average rate of 3 1/5 miles per hour. Pedro hiked the same distance at a rate of

Goryan [66]

It took Pedro 4 hours.

3 0

3 years ago

3x+4=8y-3 5x+1=10y-4

dsp73

Answer:

x-3 and y=2

Step-by-step explanation:

4 0

2 years ago

If f(x)=3^x + 10^x and g(x)=4^x-2,find(f-g)(x)

tresset_1 [31]

H gc d rxr V unimoi oubyctvuimubyv

8 0

3 years ago

Suppose that surface σ is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩, 0≤u≤7 and 0≤v≤2π3 and f(x,y,z)=x2+y2+z2. Set up the sur

Bad White [126]

Looks like you have most of the details already, but you're missing one crucial piece.

$\sigma$ is parameterized by

$\vec r(u,v)=\langle u\cos3v,u\sin3v,v\rangle$

for $0\le u\le7$ and $0\le v\le\frac{2\pi}3$ , and a normal vector to this surface is

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left\langle\sin3v,-\cos3v,3u\right\rangle$

with norm

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{\sin^23v+(-\cos3v)^2+(3u)^2}=\sqrt{9u^2+1}$

So the integral of $f(x,y,z)=x^2+y^2+z^2$ is

$\displaystyle\iint_\sigma f(x,y,z)\,\mathrm dA=\boxed{\int_0^{2\pi/3}\int_0^7(u^2+v^2)\sqrt{9u^2+1}\,\mathrm du\,\mathrm dv}$

6 0

3 years ago

My kid asked me what was 5y-3+6 i couldnt answer it because this is all new to me we didnt do this when i was younger and i need

S_A_V [24]

Answer:

Add−3 and 6 .5y+3

4 0

2 years ago

Read 2 more answers