Answer:
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Answer:
So, the times the ball will be 48 feet above the ground are t = 0 and t = 2.
Step-by-step explanation:
The height h of the ball is modeled by the following equation
The problem want you to find the times the ball will be 48 feet above the ground.
It is going to be when:
We can simplify by 16t. So
It means that
16t = 0
t = 0
or
t - 2 = 0
t = 2
So, the times the ball will be 48 feet above the ground are t = 0 and t = 2.
<h3>Answer is -9</h3>
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Work Shown:
(g°h)(x) is the same as g(h(x))
So, (g°h)(0) = g(h(0))
Effectively h(x) is the input to g(x). Let's first find h(0)
h(x) = x^2+3
h(0) = 0^2+3
h(0) = 3
So g(h(x)) becomes g(h(0)) after we replace x with 0, then it updates to g(3) when we replace h(0) with 3.
Now let's find g(3)
g(x) = -3x
g(3) = -3*3
g(3) = -9
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alternatively, you can plug h(x) algebraically into the g(x) function
g(x) = -3x
g( h(x) ) = -3*( h(x) ) ... replace all x terms with h(x)
g( h(x) ) = -3*(x^2 + 3) ... replace h(x) on right side with x^2+3
g( h(x) ) = -3x^2 - 9
Next we can plug in x = 0
g( h(0) ) = -3(0)^2 - 9
g( h(0) ) = -9
we get the same result.
Answer:9.9
Step-by-step explanation:
Add up the scores and dividing the total by the number of scores.
