By using distance formula (which is attached below), you can solve this problem:
√(x2-x1)^2+(y2-y1)^2
(5, -1) and (5, 3)
√(5-5)^2+(3-(-1))^2
√0^2+4^2
√0+16
√16
4
The distance between the two points is 4.
Answer:
I think it is 123 for number 9 i can't see, the other problems
Step-by-step explanation:
he would need to rent it for a year or 12 months
The function is

, and according to the description of the function in the problem statement, we have the following:
at t=0 after being thrown (that is, at initial time), the height of the ball is calculated by h(0) as follows:

(ft), which is the initial height, as expected.
At t=1 (sec), the height would be

.
etc.
The path is parabolic, as we know by seeing that the function is a quadratic polynomial function. This function has been given in factored form as well. From that we can see that the zeros of the function are t=7 and t=-2.
This means that at t=7 sec, the height h is 0, which means that the ball has hit the ground. t=-2 has no significance in the context of our problem so we just neglect it.
Answer: B) 7 sec
Answer:
f(g(5)) = 64
g(f(5)) = 28
Step-by-step explanation:
Given that f(x) = x^2 and g(x) = x+3
f(g(x) = f(x+3)
f(x+3) = (x+3)^2
f(g(x)) = (x+3)^2
f(g(5)) = (5+3)^2
f(g(5)) = 8^2
f(g(5)) = 64
b) g(f(x)) = g(x^2)
g(f(x)) = x^2 + 3
g(f(5)) = 5^2 +3
g(f(5)) = 25 + 3
g(f(5)) = 28
Hence the value of g(f(5)) is 28