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vekshin1
3 years ago
8

If an object is propelled upward from a height of 48 feet at an initial velocity of 96 feet per second, then its height after t

seconds is given by the equation h(t)= -16t2+96t +48, where height is in feet. After how many seconds will the object reach a height of 192 feet?
Mathematics
1 answer:
ivolga24 [154]3 years ago
5 0

Answer:

the number of seconds to reach the height is 3 seconds

Step-by-step explanation:

The computation of the seconds that reach the height is as follows;

Given that

h = -16t^2 + 96t + 48

here

H = 192 feet

So,

192 = -16t^2 + 96t + 48

-16t^2 + 96t - 144 = 0

Divide by -16

t^2 - 6t + 9 = 0

t^2 - 3t - 3t + 9

t(t - 3) - 3(t - 3)

t = 3 seconds

Hence, the number of seconds to reach the height is 3 seconds

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Galina-37 [17]

Given rational expression is:


\frac{(4x^2-32x+48)}{(3x^2-17x-6)}

Now we can simplify this by factoring as shown below:


=\frac{4(x^2-8x+12)}{(3x^2-17x-6)}


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8 0
3 years ago
Recall the scenario about Eric's weekly wages in the lesson practice section. Eric's boss have been very impressed with his work
Solnce55 [7]

Answer:  

1)\quad f(x)=\bigg\{\begin{array}{ll}12x&0\leq x

2) D: x = [0, 24]

3) R: y = [0, 384]

4) see graph

<u>Step-by-step explanation:</u>

Eric's regular wage is $12 per hour for all hours less than 9 hours.

The minimum number of hours Eric can work each day is 0.

f(x) = 12x    for   0 ≤ x < 9

Eric's overtime wage is $18 per hour for 9 hours and greater.

The maximum number of hours Eric can work each day is 24 (because there are only 24 hours in a day).

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f(x)=\bigg\{\begin{array}{ll}12x&0\leq x

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The minimum hours he can work in one day is 0 and the maximum he can work in one day is 24.

D:  0 ≤ x ≤ 24        →        D: x = [0, 24]

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