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>
Answer:
Options A, D and F
Step-by-step explanation:
From the picture attached,
lines 'l' ad 'm' are the parallel lines and line 't' is a transversal intersecting each line at two different points.
From the angles formed at the points of intersection,
∠5 ≅ ∠8 [Vertical angles]
∠5 ≅ ∠4 [Alternate interior angles]
∠5 ≅ ∠1 [Corresponding angles]
Therefore, Options A, D and F will be the correct options.
Answer:
6 mm
Step-by-step explanation:
So, the area he needs to fill isn't really important, you don't need to calculate it. What is necessary is the individual dimensions of that area (width/length). The width and the length must be divisible by the lengths of the tiles. This is so you don't need to cut anything. For example let's say the area needed to be filled was 15 mm (width) by 10 mm (length). If I chose 3mm square tiles, it could perfectly fill up the width, but when it came to the length there would be a small 1 mm gap. And you would need to cut one of the square tiles, so it would have a length of 1 mm. In this case the 15mm was divisible by 3, but the 10 mm wasn't. So using this logic, it can't be the 4 mm square tile since 150mm / 4 mm = 37.5. It can be the 6m since 150mm / 6mm = 25 and 180 mm / 6mm = 30. It can't be the 8 mm since 150 mm / 8mm = 18.75 mm
The formula for this is an easy, straightforward one:
where r is the radius and l is the slant height. You have your area and your radius, so solve it for slant height.
and
. Multiply that pi in to the 32 and then divide that into 587 and you'll get 5.8
