Chapter 8.5, Problem 14CYU

a.

To determine

Find the GCF of each term of the expression.

Answer to Problem 14CYU

The GCF of numbers and variables are $16\&t$ .

Explanation of Solution

Given:

It is given in the question that a rocket launched straight up with an initial velocity of $128$ feet per second. The height of the rocket in feet above sea level is modeled by the expression $128t-16t^2$ , where t is the time in seconds after the rocket is launched.

Concept Used:

In this, use the concept of separate the variable and number on each side and do the HCF separately.

Calculation: Here, the expression is $128t-16t^2$ .

The number is $128,16$ and the variable is $t,t^2$

GCF of $128,16$ = $16$

GCF of $t,t^2$ = $t$ .

Conclusion:

The GCF of numbers and variables are $16\&t$ .

b.

To determine

Find the factored form of the expression for the height of the rocket after t seconds.

Answer to Problem 14CYU

The factor for the expression is $t=0\&t=8$ .

Explanation of Solution

Given:

It is given in the question that a rocket launched straight up with an initial velocity of $128$ feet per second. The height of the rocket in feet above sea level is modeled by the expression $128t-16t^2$ , where t is the time in seconds after the rocket is launched.

Concept Used:

In this, use the concept of factorization of the expression.

Calculation: Here, the expression is $h=128t-16t^2$ .

  $\begin{array}{l}h=128t-16t^2\\ h=16t(8-t)\end{array}$

Put , $h=0$

  $\begin{array}{l}16t(8-t)=0\\ 16t=0,8-t=0\\ t=0,t=8\end{array}$

Conclusion:

The factor for the expression is $t=0\&t=8$ .

c.

To determine

Find the height of the rocket after $4$ seconds and $6$ seconds. Also, find the greater one.

Answer to Problem 14CYU

The greater one is $256$ in $4$ seconds.

Explanation of Solution

Given: It is given in the question that a rocket launched straight up with an initial velocity of $128$ feet per second. The height of the rocket in feet above sea level is modeled by the expression $128t-16t^2$ , where t is the time in seconds after the rocket is launched.

Concept Used:

In this, use the concept of putting the time value in the expression and get the answer.

Calculation: Here, the expression is $h=128t-16t^2$ .

Put $t=4\sec$ ,

  $\begin{array}{l}h=128t-16t^2\\ $ h=128(4)-16{(4)}^2$h=512-16\times 16\\ h=512-256\\ h=256\end{array}$

Put $t=6\sec$ ,

  $\begin{array}{l}h=128t-16t^2\\ $ h=128(6)-16{(6)}^2$h=768-16\times 36\\ h=768-576\\ h=192\end{array}$

The greater one is $256$ in $4$ seconds.

Conclusion:

The greater one is $256$ in $4$ seconds.