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Sergio039 [100]
3 years ago
14

A ball is kicked into the air and follows the path described by h(t)= – 4.9t^2 + 6t + 0.6, where t is the time in seconds and h

is the height in meters above the ground. Find the maximum height of the ball. What value would you have to change in the equation if the maximum height of the ball is more than 2.4 meters?
Mathematics
1 answer:
Natali [406]3 years ago
7 0
I'm going to do the calculus way because it's easier
max height,
take derititive of the equation

h'(t)=-9.8t+6
find what value of t makes it equal to 0
t=6/9.8
if we inputed it for t we get
h(6/9.8)=2.43673...

2.43673...>2.4 so we don't need to change anything

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(a) The number of shoes that Marquise has is 40 pairs of shoes

(b) The number of pairs of shoes that Bron'Tavious has is 20 pairs of shoes

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Step-by-step explanation:

The given information are;

The number of shoes that Marquise has = 2 × The number of pairs of shoes that Bron'Tavious

The number of pairs of shoes that JayDarius has = The number of pairs of shoes that Bron'Tavious + 15

The total number of pairs of shoes they have all together = 95 pairs of shoes

Let the number of pairs of shoes that Bron'Tavious has = X. Therefore, we have;

The number of shoes that Marquise has = 2 × X

The number of pairs of shoes that JayDarius has = X + 15

The total number of pairs of shoes they have all together = X + X + 15 + 2 × X = 95

X + X + 15 + 2 × X = 95

4 × X + 15 = 95

4 × X = 95 - 15 = 80

X = 80/4 = 20

∴ X = 20

(a) The number of shoes that Marquise has = 2 × X = 2 × 20 = 40

The number of shoes that Marquise has = 40 pairs of shoes

(b) The number of pairs of shoes that Bron'Tavious has = X = 20

The number of pairs of shoes that Bron'Tavious has = 20 pairs of shoes

(c) The number of pairs of shoes that JayDarius has = X + 15 = 20 + 15 = 35

The number of pairs of shoes that JayDarius has = 35 pairs of shoes.

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