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valentinak56 [21]
3 years ago
8

The function H(t) = −16t2 + 60t + 95 shows the height H(t), in feet, of a projectile after t seconds. A second object moves in t

he air along a path represented by g(t) = 20 + 38.7t, where g(t) is the height, in feet, of the object from the ground at time t seconds.
Part A: Create a table using integers 1 through 4 for the 2 functions. Between what 2 seconds is the solution to H(t) = g(t) located? How do you know? (6 points)

Part B: Explain what the solution from Part A means in the context of the problem. (4 points)

I need help with part B
Mathematics
1 answer:
EleoNora [17]3 years ago
5 0
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>
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Step-by-step explanation:

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(8, 6)→(8\times1.5,6\times1.5)=(12,9)

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(4, −4)→(4\times1.5,-4\times1.5)=(6,-6)

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Three equivalent ratios for 6: 54
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A cone has a diameter of 3 inches the cone holds 12 cubic inches of water to the nearest in what is the height
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5 inches

Step-by-step explanation:

A cone's volume can be found using the formula V = \frac{1}{3} \pi r^2 h. If the cone has a diameter of 3 inches then its radius is half r= 1.5 inches. Substitute r = 1.5 and V = 12 cubic inches.

V = \frac{1}{3} \pi r^2 h \\12 = \frac{1}{3} \pi (1.5)^2 h \\12 = \frac{1}{3} \pi (2.25) h\\12 = 2.355h\\5.10 = h

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