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oksian1 [2.3K]
3 years ago
12

An object is launched straight into the air. The projectile motion of the object can be modeled using h(t) = 96t – 16t2, where t

is the time since launch and h(t) is the height in feet of the projectile after time t in seconds.
Mathematics
1 answer:
ratelena [41]3 years ago
6 0

Answer with Step-by-step explanation:

We are given that height of projectile after t seconds is given by

h(t)=96t-16t^2

a.h(t)=144 ft

144=96t-16t^2

16t^2-96t+144=0

t^2-6t+9=0

t^2-3t-3t+9=0

t(t-3)-3(t-3)=0

(t-3)(t-3)=0

t-3=0

t=3

After 3 s, the height of the project will be 144 feet in the air.

b.h(t)=0

96t-16t^2=0

16t(6-t)=0

16t=0\implies t=0

6-t=0\implies t=6

At t=0, the initial position of projectile

At t=6 s , the projectile will hit the ground.

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Coach Rivas can spend up to $750 on 30 swimsuits for the swim team. The inequality shown can be used to find the maximum amount
777dan777 [17]

Answer:

$0 < p ≤ $25

Step-by-step explanation:

We know that coach Rivas can spend up to $750 on 30 swimsuits.

This means that the maximum cost that the coach can afford to pay is $750, then if the cost for the 30 swimsuits is C, we have the inequality:

C ≤ $750

Now, if each swimsuit costs p, then 30 of them costs 30 times p, then the cost of the swimsuits is:

C = 30*p

Then we have the inequality:

30*p ≤ $750.

To find the possible values of p, we just need to isolate p in one side of the inequality.

So we can divide both sides by 30 to get:

(30*p)/30 ≤ $750/30

p ≤ $25

And we also should add the restriction:

$0 < p ≤ $25

Because a swimsuit can not cost 0 dollars or less than that.

Then the inequality that represents the possible values of p is:

$0 < p ≤ $25

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2 years ago
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2 years ago
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Step-by-step explanation:

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Step-by-step explanation:

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now, the sum of both salines, must add up to the 77% mixture, let's say is "y"
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anyway   thus

\bf \begin{array}{lccclll}&#10;&amount&concentration&&#10;\begin{array}{llll}&#10;concentrated\\&#10;amount&#10;\end{array}\\&#10;&-----&-------&-------\\&#10;\textit{first sol'n}&11&x&11x\\&#10;\textit{second sol'n}&4&0.55&2.20\\&#10;------&-----&-------&-------\\&#10;mixture&y&0.77&0.77y&#10;\end{array}\\\\&#10;-------------------------------\\\\&#10;\begin{cases}&#10;11+4=y\implies 15=\boxed{y}\\&#10;11x+2.2=0.77y\\&#10;----------\\&#10;11x+2.2=0.77\cdot \boxed{15}&#10;\end{cases}

solve for "x"
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3 years ago
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