Given:
To find:
Whether f(x) and g(x) are inverse of each other by using that f(g(x)) = x and g(f(x)) = x.
Solution:
We know that, two function are inverse of each other if:
and 
We have,
Now,
![[\because g(x)=\dfrac{8}{x}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20g%28x%29%3D%5Cdfrac%7B8%7D%7Bx%7D%5D)
![[\because f(x)=\dfrac{8}{x}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20f%28x%29%3D%5Cdfrac%7B8%7D%7Bx%7D%5D)
Similarly,
Since, and , therefore, f(x) and g(x) are inverse of each other.
Answer:
R = (-12, 22)
Step-by-step explanation:
Unfortunately it's not a straight line, so gonna need to put in some extra work. Basically think of it like breaking it into it's horizontal and vertical components then doubling them.
From Q to M you move 10 spaces to the left, so to go from M to R you will go another 10 to the left. Similarly, you start at -10 for the y value and go up 16 to 6, and you will need to go up another 16 to get to R.
So for x, 8 -10 = -2 then -2 - 10 = -12, so R's x value is -12
y we do the same thing. Gonna do it in one step though. -10 + 16 + 16 = 22
So R is (-12, 22)
Answer:
<em>The answer is Hence Proved</em>
Step-by-step explanation:
Given that CB║ED , CB ≅ ED
To prove Δ CBF ≅ Δ EDF
This means that the length of CB is equal to ED
As CB║ED The following conditions satisfies when a transversal cut
two parallel lines
- ∠ EDF = ∠ FBC ( Alternate interior points )
- ∠ DEF = ∠ FCB ( Alternate interior points )
∴ Δ CBF ≅ Δ EDF ( By ASA criterion)
The Δ CBF is congruent to Δ EDF By ASA criterion .
<em> Hence proved </em>
Hi! Sorry if this is wrong, but I believe it would be f(x)= |x+5|-4. This is because it shifts to the left, so it is adding 5 to the x in the absolute value bars. And it shifting 4 units down would be negative. Hope this helps!
