Answer:
Vertical stretching by a factor of 12, followed by upward translation of 2 units.
Step-by-step explanation:
Let's assume you're starting with f(x), the parent function.
Multiplying f(x) by 12 will stretch the graph vertically by a factor of 12. A point (1,1) on the graph of f(x) will re-appear as (1,12) after this vertical stretching. Once you've done that, translate the entire graph upward by 2 units.
Answer:

Step-by-step explanation:
we know that
the formula to calculate the distance between two points is equal to

we have

step 1
Find the distance AB
we have

substitute in the formula



step 2
Find the distance BC
we have

substitute in the formula



step 3
Find the distance DE
we have

substitute in the formula




step 4
Find the distance AE
we have

substitute in the formula




step 5
Find out the area of the four walls
To determine the area sum the length sides and multiply by the height of the walls
so

substitute the given values
The height of the walls is 10 ft



step 6
Determine the number of gallons needed to paint the four walls
we know that
one gallon of paint is enough to cover 400 square feet
using proportion

Round up

Answer:
x=9
Step-by-step explanation:
Equation: 5.60x=50.40
Solve: 5.60x=50.40
Divide both sides by 5.60 ( 5.60x/5.60 ) and ( 50.40/5.60 )
x=9
|DF| = |DE| + |EF| |DF| = 9x -36 |DE| = 47 |EF| = 3x+10 Substitute: 9x - 39 = 47 + 3x + 10 9x - 39 = 3x + 57 |+39 9x = 3x + 96 |-3x 6x = 96 |:6 x = 16 Put the value of x to the equation |EF| = 3x + 10 |EF| = (3)(16) + 10 = 48 + 10 = 58 Answer: |EF| = 58
Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/4046336-if-df-9x-39-find-ef.html
Answer:

Step-by-step explanation:





For the Interval 
Length of the Curve 
Using T1-Calculator
