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Rzqust [24]
3 years ago
11

A boy throws a ball into the air. The equation h=−16t^2+23t+4 models the path of the ball, where h is the height (in feet) of th

e ball t seconds after it is thrown. How long is the ball in the air?
Mathematics
1 answer:
lina2011 [118]3 years ago
8 0

Answer:

0.72secs

Step-by-step explanation:

Given the height of the ball in air modeled by the equation:

h=−16t²+23t+4

Required

Total time it spent in the air

To get this we need to calculate its time at the maximum height

At the maximum height,

v = dh/dt = 0

-32t + 23 = 0

-32t = -23

t = 23/32

t = 0.72secs

<em>Hence the total time it spend in the air will be 0.72secs</em>

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4 0
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Line JK passes through points J(–4, –5) and K(–6, 3). If the equation of the line is written in slope-intercept form, y = mx + b
andrey2020 [161]
(-4,-5)(-6,3)
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\bf \begin{array}{lccclll}&#10;&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\&#10;&------&------&------\\&#10;Upstream&10&40-4&u\\&#10;Downstream&10&40+4&d&#10;\end{array}&#10;\\\\\\&#10;10=(40-4)u\implies \cfrac{10}{40-4}=u\implies \cfrac{10}{36}=u\implies \cfrac{5}{18}=u&#10;\\\\\\&#10;10=(40+4)d\implies \cfrac{10}{40+4}=d\implies \cfrac{10}{44}=d\implies \cfrac{5}{22}=d

so, "u" is 16 minutes and 40 seconds.

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4 0
3 years ago
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