H(t) = -16t² + 60t + 95
g(t) = 20 + 38.7t
h(1) = -16(1²) + 60(1) + 95 = -16 + 60 + 95 = -16 + 155 = 139
h(2) = -16(2²) + 60(2) + 95 = -16(4) + 120 + 95 = -64 + 215 = 151
h(3) = -16(3²) + 60(3) + 95 = -16(9) + 180 + 95 = -144 + 275 = 131
h(4) = -16(4²) + 60(4) + 95 = -16(16) + 240 + 95 = -256 + 335 = 79
g(1) = 20 + 38.7(1) = 20 + 38.7 = 58.7
g(2) = 20 + 38.7(2) = 20 + 77.4 = 97.4
g(3) = 20 + 38.7(3) = 20 + 116.1 = 136.1
g(4) = 20 + 38.7(4) = 20 + 154.8 = 174.8
Between 2 and 3 seconds.
The range of the 1st object is 151 to 131.
The range of the 2nd object is 97.4 to 136.1
h(t) = g(t) ⇒ 131 = 131
<span>It means that the point where the 2 objects are equal is the point where the 1st object is falling down while the 2nd object is still going up. </span>
Answer:
f(x) = (-1/2)(x^2 + 8x - 15)
Step-by-step explanation:
This function has two roots: -3 and 5. Most likely it is a quadratic (all of which have two roots).
Then f(x) = a(x + 3)(x - 5)
The graph goes through (1. 8): Therefore, y = 8 when x = 1:
f(1) = a(1 + 3)(1 - 5) = 8, or
a(4)(-4) = 8, or
-16a = 8, which leads to a = -1/2.
Thus the quadratic in question is f(x) = (-1/2)(x + 3)(x - 5), or
f(x) = (-1/2)(x^2 + 8x - 15)
Answer:
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System c, plan out a variable system as the first buy, is 5 peaches and two watermelons, totaling to 7.75$, then the second buy is 3 watermelons and 4 peaches, totaling to 10.95$, system c is the only system that matches up
First add 1/2 and 3/4 together by making 1/2, 2/4 then add 3/4 to get 5/4 or 1 1/4 then 10 - 1 1/4 is 8 3/4 which is the answer
