The maximum height of an object is the greatest height the object can reach. The maximum height the rocket reaches is 40.2 feet
<h3>Vertex form of an equation</h3>
The maximum height of an object is the greatest height the object can reach. Given the equation that represent the height of the rocket.
f(t) = -15t² + 48t
The maximum height of the rocket occur at the point where:
t = -b/2a
t = -48/2(-15)
t =48/30
t = 1.6secs
Substitute t = 1.6 into the function as shown:
f(1.6) = -15(1.6)² + 48(1.6)
f(1.6) = -38.4 + 76.8
f(1.6) = 40.2
Hence the maximum height the rocket reaches is 40.2 feet
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Answer:
Step-by-step explanation:
p = 2*3 = 6 pieces
0.2(x + 20) - 3 > -7 - 6.2x.....distribute trhr the parenthesis
0.2x + 4 - 3 > -7 - 6.2x...simplify
0.2x + 1 > - 7 - 6.2x...addition property...add 6.2x to both sides
0.2x + 6.2x + 1 > -7 ...simplify
6.4x + 1 > - 7.....subtraction property....subtract 1 from both sides
6.4x > -7 - 1....simplify
6.4x > -8....division property....divide both sides by 6.4
x > -8 / 6.4...simplify
x > - 1.25 <==
Answer:
Hello your question is incomplete below is the complete question
What is wrong with the equation? integral^2 _3 x^-3 dx = x^-2/-2]^2 _3 = -5/72 f(x) = x^-3 is continuous on the interval [-3, 2] so FTC2 cannot be applied. f(x) = x^-3 is not continuous on the interval [-3, 2] so FTC2 cannot be applied. f(x) = x^-3 is not continuous at x = -3, so FTC2 cannot be applied The lower limit is less than 0, so FTC2 cannot be applied. There is nothing wrong with the equation. If f(2) = 14, f' is continuous, and f'(x) dx = 15, what is the value of f(7)? F(7) =
answer : The value of f(7) = 29
Step-by-step explanation:
Attached below is the detailed solution
Hence : F(7) - 14 = 15
F(7) = 15 + 14 = 29
