Answer:
512000
Step-by-step explanation:
8*10
80*80*80
We can make ratios equivalent to 18:4 by dividing and multiplying on it.
1. (18:4)/2=9:2
2. (18:4)*2=36:8
3. (18:4)*3=54:12
You're welcome :).
Answer: 2535
Step-by-step explanation:
use order of operations (parenthesis, exponents, multiplication/division, addition/subtraction)
first ill handle the two innermost parenthesis
(4 + 3(2 - 8))
(4 + 3(-6))
(4 - 18)
-14
now lets move on to the next parenthesis
( 3 - 5^2(-14))
(3 - 25( - 14)
( 3 + 350)
353
now onto the last part of the problem
8^2 + 7 (353)
64 + 2471
2535 :))
Given:
In the given circle, ∠ADB = 270° and length of ADB = 61.26 m
To find the radius of the circle.
Formula
The formula to determine the radius is given by

Substituting arc length = 61.26,
, we get;



Thus, the radius is 13 m
Finding the inverse function of

Remember that when you compose

with its inverse

you'll get the identity function:
![\mathsf{(f\circ f^{-1})(x)=x}\\\\ \mathsf{f\big[f^{-1}(x)\big]=x}\\\\](https://tex.z-dn.net/?f=%5Cmathsf%7B%28f%5Ccirc%20f%5E%7B-1%7D%29%28x%29%3Dx%7D%5C%5C%5C%5C%20%5Cmathsf%7Bf%5Cbig%5Bf%5E%7B-1%7D%28x%29%5Cbig%5D%3Dx%7D%5C%5C%5C%5C)
So if

then
![\mathsf{f\big[f^{-1}(x)\big]=10-[f^{-1}(x)]^2}\\\\ \mathsf{x=10-[f^{-1}(x)]^2}\\\\ \mathsf{[f^{-1}(x)]^2=10-x}\\\\ \mathsf{f^{-1}(x)=\pm \sqrt{10-x}}\\\\ \mathsf{f^{-1}(x)=-\sqrt{10-x}~~~or~~~f^{-1}(x)=\sqrt{10-x}}](https://tex.z-dn.net/?f=%5Cmathsf%7Bf%5Cbig%5Bf%5E%7B-1%7D%28x%29%5Cbig%5D%3D10-%5Bf%5E%7B-1%7D%28x%29%5D%5E2%7D%5C%5C%5C%5C%20%5Cmathsf%7Bx%3D10-%5Bf%5E%7B-1%7D%28x%29%5D%5E2%7D%5C%5C%5C%5C%20%5Cmathsf%7B%5Bf%5E%7B-1%7D%28x%29%5D%5E2%3D10-x%7D%5C%5C%5C%5C%20%5Cmathsf%7Bf%5E%7B-1%7D%28x%29%3D%5Cpm%20%5Csqrt%7B10-x%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bf%5E%7B-1%7D%28x%29%3D-%5Csqrt%7B10-x%7D~~~or~~~f%5E%7B-1%7D%28x%29%3D%5Csqrt%7B10-x%7D%7D)
The sign of the inverse depends on the domain of

itself, and where it's invertible.
If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2198197 I hope this helps.
Tags: <em>inverse function definition identity domain algebra</em>