Answer:
B) translated 6 units to the right then translated 9 units up
Step-by-step explanation:
Just use the graph to follow each translation until you get thee right answer, or you can count how many spaces are from the pre-image to the image. Just make sure to always use the same point.
Use distributive method
5r - 50 = -51
Add 50
5r = -1
Divide by 5
r = -1/5
Answer:
x=4, MN= 37, LM= 37, y=7.
Step-by-step explanation:
If MP is a perpendicular bisector to LN, then NP and LP are equivalent.
(Solve for y)
2y+2= 16
(Move the +2 to the right side of the equation)
2y= 14
(Divide both sides by 2 to isolate the variable)
y=7
To find x and the measure of MN and LM, solve for x in the following equation:
7x+9 = 11x-7
(Move 7x to the right side of the equation)
9 = 4x-7
(Move -7 to the right side of the equation.)
16= 4x
(Divide both sides by 4 to isolate the variable.)
4= x
Plug x back into both equations to get the measure of MN and ML
MN=7(4)+9
MN= 28+9
MN= 37
LM= 11(4)-7
LM= 44-7
LM= 37
I hope this helps!
Given:

From the function above, there was a horizonal stretch.
using the transformation rule:

Therefore, we can say that there is a horizontal shift of the graph to the right.
ANSWER:
B. The graph of the function g(s) is a horizontal shift of the graph of the function to the right.
Answer:

Step-by-step explanation:
First, notice that:



We proceed to use the chain rule to find
using the fact that
to find their derivatives:

Because we know
then:
We substitute in what we had:

Now we put in the values
in the formula:

Because of what we supposed:

And we operate to discover that:


and this will be our answer