I HAVE 5 MINUTES PLEASE HELP ! ( slope )
2 answers:
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The number of calories per ounce of soda is 10
<h3>Part A: Represent the relationship between the number of calories and the number of ounces</h3>The given parameters are:
Calories = 50
Ounces = 5
Let the number of calories be y and the ounces be x.
So, we have:
y = kx
Substitute y = 50 and x = 5
50 = 5k
Divide by 5
k = 10
Substitute k = 10 in y = kx
y = 10x
See attachment for the graph of the relationship between the number of calories and the number of ounces
<h3>Part B: What is the number of calories per ounce of soda?</h3>In (a), we have:
k = 10
This means that the number of calories per ounce of soda is 10
<h3>Part C: How does the unit rate relate to the slope of the line in the graph above? </h3>The unit rate and the slope represent the same and they have the same value
Read more about linear graphs at:
#SPJ1
Answer:
a) starting height: 5.5 ft
b) hang time: 5.562 seconds
c) maximum height: 126.5 ft
d) time to maximum height: 2.75 seconds
Step-by-step explanation:
a) The starting height is the height at t=0.
h(0) = -16·0 +88·0 +5.5
h(0) = 5.5
The starting height is 5.5 feet.
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b) The ball is in the air between t=0 and the non-zero time when h(t) = 0. We can find the latter by solving ...
-16t^2 + 8t +5.5 = 0
t^2 -(11/2)t = 5.5/16 . . . . . subtract 5.5, then divide by -16
t^2 -(11/2)t +(11/4)^2 = (5.5/16) +(11/4)^2 . . . . complete the square
(t -11/4)^2 = 126.5/16 . . . . . . . . . . . . . . . . . . . . call this [eq1] for later use
t -11/4 = √7.90625
t = 2.75 +√7.90625 ≈ 5.562
The ball will be in the air about 5.562 seconds.
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c) If we multiply [eq1] above by -16 and add the constant on the right, we get the vertex form of the height equation:
h(t) = -16(t -11/4) +126.5
The vertex at (2.75, 126.5) tells us ...
The maximum height of the ball is 126.5 feet.
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d) That same vertex point tells us ...
The maximum height will be reached at t = 2.75 seconds.
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If you really need answers fast, a graphing calculator can give them to you in very short order (less than a minute).
