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:
Step-by-step explanation:
LCM of 5(y - 4)(y + 4) and (y + 4) is 5(y - 4)(y + 4)
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Answer: 5(y - 4)(y + 4) (Answer D)-----------------------------------------------------------------------
Answer:
4 is a positive number
Step-by-step explanation:
Cos θ = adjacent / hypothenus = 5/13
opposite = sqrt(hypothenus^2 - adjacent^2) = sqrt(13^2 - 5^2) = sqrt(169 - 25) = sqrt(144) = 12
sec θ = 1/cos θ = 1/(adjacent / hypothenus) = hypothenus / adjacent = 13/5
cot θ = 1/tan θ = 1/(opposite/adjacent) = adjacent / opposite = 5/12
sec θ = 13/5
cot θ = 5/12
