Answer:
![f^{-1}(x) = \sqrt[3]{x+5}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx%2B5%7D)
Step-by-step explanation:
A function 
is the inverse of f if whenever y =f(x) and 
To find the inverse of f(x) = 
- 5, we use the following steps:
- Step 1 : Put y for f(x) and solve for x
y = - 5
<=> = y + 5
<=> x = ![\sqrt[3]{y+ 5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%2B%205%7D)
- Step 2: Put
for x, we have:
![f^{-1}(y) =\sqrt[3]{y+5}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28y%29%20%3D%5Csqrt%5B3%5D%7By%2B5%7D)
- Step 3: Interchange y =x, we have
Let Calculate 3 points on f(x)
x = 0 => y = -5
x = 1 => y = -4
x = 2 => y = 3
Let Calculate 3 points on (x)
x = -5 => y = 0
x = -4 => y = 1
x = 3 => y = 2
Yes, the inverse is correct because:
- the domain of the inverse function is the range of the original function
- the range of the inverse function is the domain of the original function
Answer:
No
Step-by-step explanation:
The given point does not satisfy the first equation.
y = -8(3) -3 = -27
4 ≠ -27 . . . . . the first equation is not satisfied.
There is more than 1 way to do this but the way i use is to go back to zero and move that amount of spaces to left if the original integer is positive or to the right if the original integer is a negative
Answer:
The correct option is;
The points on the perpendicular bisector of a side of a triangle are equidistant from the vertices of the side it bisects
Step-by-step explanation:
With the assumption that that the question meant that ΔACD and ΔBCD are congruent, we have;
Segment 
is congruent to segment 
by Congruent Parts of Congruent Triangles are Congruent, CPCTC
Similarly, segment 
is congruent to segment 
by Congruent Parts of Congruent Triangles are Congruent, CPCTC
Therefore, the points on the perpendicular bisector of a side of a triangle are equidistant from the vertices of the side it bisects
The first one is Corresponding Angles
The second one is Alternate interior angles
The third one is Co-interior angles
The fourth one is Vertical angles
The fifth one is Alternate exterior angles
Hope it helps
