Given:
The function for size of a square frame is

where, x is the side length of the picture.
The function for the price in dollars for the frame is

To find:
The single function for the price of a picture with an edge length of x.
Solution:
We know that, for a picture with an edge length of x.
Size of a square frame = f(x)
Price in dollars for the frame = p(x)
Single function for the price of a picture with an edge length of x is

![[\because f(x)=x+2]](https://tex.z-dn.net/?f=%5B%5Cbecause%20f%28x%29%3Dx%2B2%5D)
![[\because p(x)=3x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20p%28x%29%3D3x%5D)
Let the name of this function is c(x). So,

Therefore, the required function is
.
Answer:
7/36
Step-by-step explanation:
find the LCD write all the numerators above the least common denominator
LCD is 36
4 +3 /36 =7/36
Answer:
choice one
Step-by-step explanation:
fnjddjdjdjddjnddkkdd
Answer:
13 ft/s
Step-by-step explanation:
t seconds after the boy passes under the balloon the distance between them is ...
d = √((15t)² +(45+5t)²) = √(250t² +450t +2025)
The rate of change of d with respect to t is ...
dd/dt = (500t +450)/(2√(250t² +450t +2025)) = (50t +45)/√(10t² +18t +81)
At t=3, this derivative evaluates to ...
dd/dt = (50·3 +45)/√(90+54+81) = 195/15 = 13
The distance between the boy and the balloon is increasing at the rate of 13 ft per second.
_____
The boy is moving horizontally at 15 ft/s, so his position relative to the spot under the balloon is 15t feet after t seconds.
The balloon starts at 45 feet above the boy and is moving upward at 5 ft/s, so its vertical distance from the spot under the balloon is 45+5t feet after t seconds.
The straight-line distance between the boy and the balloon is found as the hypotenuse of a right triangle with legs 15t and (45+5t). Using the Pythagorean theorem, that distance is ...
d = √((15t)² + (45+5t)²)