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coldgirl [10]
3 years ago
12

at a fourth of july celebration, a rocket is launched straight up with an initial velocity of 128 ft. The height of the rocket i

n feet above sea level is modeled by the expression 128t - 16t^2, where T is the time in seconds after the rocket is launched. (a) what is the gcf of each term of the expression? (b) write the factored form of the expression for the height of the rocket after T seconds. (c) well the height of the rocket be greater after four seconds or after six seconds?
Mathematics
1 answer:
DENIUS [597]3 years ago
7 0

Answer:

Step-by-step explanation:

Given

Height of the rocket is given by

h=128t-16t^2

h=8\times 16t-16t\times t

(a)Greatest common factor in each term is 16t

(b)

h=8\times 16t-16t\times t

h=16t(8-t)

(c)

After t=4\ s

h=16\times 4(8-4)

h=256\ m

after t=6\ s

h=16\times 6(8-6)

h=192\ m

Thus after 4\ s height of rocket is greater .

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3 years ago
Ryan flew from Wiley Post to Ponca City and back. Ryan maintained an average rate of 450 mph going to Ponca City and an average
Reil [10]
The answer is B, and here's why.  Set up a table for "there" and "back" and use the distance = rate * time formula, like this:
               d             r            t
there       d           450         t
back        d           400        1-t

Let me explain this table to you.  The distance is d, we don't know what it is, that's what we are actually looking for.  We only know that if we go somewhere from point A to point B, then back again to point A, the distance there is the same as the distance back.  Hence, the d in both spaces.  There he flew 450 mph, back he flew 400 mph.  If the total distance was 1 hour, he flew an unknown time there and one hour minus that unknown time back.  For example, if he flew for 20 minutes there, one hour minus 20 minutes means that he flew 60 minutes - 20 minutes = 40 minutes back.  See? Now, because the distance there = the distance back, we can set the rt in both equal to each other.  If d = rt there and d = rt back and the d's are the same, then we can set the rt's equal to each other.  450t = 400(1-t) and
450t = 400 - 400t and 850t = 400.  Solve for t to get t = .47058.  Now, t is time, not the distance and we are looking for distance. So multiply that t value by the rate (cuz d = r*t) to get that the distance one way is
d = 450(.470580 and d = 211. 76 or, rounded like you need, 212.
4 0
3 years ago
0.08 Divided by 3.84
Cloud [144]
Around 0.02 or more precise, 0.02083
8 0
4 years ago
Read 2 more answers
The difference between two integers is 8 and their product is 65. Assume that the larger of the two numbers is x. Write a quadra
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A quadratic equation in standard form that can be used to determine the value of x is x^{2} - 8x - 65 = 0.

The Factor of the equation from Part B is (x - 13)(x + 5)  and the possible solutions to the equation is x = 13 or -5.

The value for the second number is 5

<h3>Word Problem Leading To Quadratic Equation.</h3>

The general formula for quadratic equation in a in standard form is

ax^{2} + bx + c = 0

Given that the  difference between two integers is 8

Let the two integers = x and y,

and their product is 65. If the larger of the two numbers is x. Then,

x - y = 8 and xy = 65

Since we are looking for the value of x, make y the subject of formula in the first equation.

y = x - 8

Substitute y in the second equation.

x(x - 8) = 65

x^{2} - 8x - 65 = 0

x^{2} - 13x + 5x - 65 = 0

(x - 13)(x + 5) = 0

x = 13 or -5

We will ignore -5 since x is the larger number.

To get y Substitute x in the second equation.

xy = 65

13y = 65

y = 65/13

y = 5

Therefore, a quadratic equation in standard form that can be used to determine the value of x is x^{2} - 8x - 65 = 0. The Factor of the equation from Part B is (x - 13)(x + 5)  and the possible solutions to the equation is x = 13 or -5. The value for the second number is 5

Learn more about Quadratic Equation here: brainly.com/question/25841119

#SPJ1

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Zinaida [17]

Answer:

1/3!

Step-by-step explanation:

Look us Cymath on g0ogle, its a calculator that shows its work!

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