The answer is c. <span>a three-dimensional geometric shape with two parallel and circular bases and a curved side</span>
If you have given an equation, you see the x and y in it. Just taking second equation and think any common factor between x's variable. With common factor, multiply both and you will find the value of y and put it in any equation, you will find the value of X
Answer:
x² + y² = 16
Step-by-step explanation:
(x-h)² + (y-k)² = r² -- The standard form of equation of circle
(x-0)² + (y-0)² = 4²
Answer:
(a) The maximum height of the ball is 16 feet at 1 seconds.
(b) The ball hits the ground at 2 seconds.
PART (A) :
Given function: h(t) = -16t² + 32t
Comparing to quadratic function: ax² + bx + c
In this function: a = -16, b = 32, c = 0
<u>To find the maximum height</u>, use the vertex formula:

<u>Insert values</u>

Then find h(t) = -16(1)² + 32(1) = 16 feet
Conclusion: The maximum height of the ball is 16 feet at 1 seconds.
PART (B) :
When the ball hits the ground, the height [h(t)] will be 0 ft
-16t² + 32t = 0
-16t(t - 2) = 0
-16t = 0, t - 2 = 0
t = 0, t = 2
The ball hits the ground at 2 seconds. Note: the other 0 seconds is for when the ball was launched at the beginning.
Answer: 33
Step-by-step explanation: