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Paul [167]
3 years ago
7

The function h(t) = –16t2 + 96t + 6 represents an object projected into the air from a cannon. The maximum height reached by the

object is 150 feet.
Mathematics
1 answer:
jolli1 [7]3 years ago
6 0
If you've started pre-calculus, then you know that the derivative of  h(t)
is zero where h(t)  is maximum.

The derivative is            h'(t) = -32 t  +  96 .

At the maximum ...        h'(t) = 0

                                       32 t = 96 sec

                                           t  =  3 sec . 
___________________________________________

If you haven't had any calculus yet, then you don't know how to
take a derivative, and you don't know what it's good for anyway.

In that case, the question GIVES you the maximum height.
Just write it in place of  h(t), then solve the quadratic equation
and find out what  't'  must be at that height.

                                       150 ft = -16 t²  +  96  t  +  6 

Subtract 150ft from each side:    -16t²  +  96t  -  144  =  0 .

Before you attack that, you can divide each side by  -16,
making it a lot easier to handle:

                                                         t²  -  6t  +  9  =  0

I'm sure you can run with that equation now and solve it.    
The solution is the time after launch when the object reaches 150 ft.
It's 3 seconds.  
(Funny how the two widely different methods lead to the same answer.)

The answer is from AL2006

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The gallons of gas that each vehicle would consume for the different amounts of driving are as follows:

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<h3>Data and Calculations:</h3><h3>Rating (miles per gallon)</h3>

                                           City Driving     Highway Driving

Car A                                     32 mi/gal            29 mi/gal

Car B                                     27 mi/gal            34 mi/gal

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                   City Driving     Highway Driving

a)                       300 mi              100 mi

b)                       100 mi              300 mi

<h3>Car A's Consumption of Gas:</h3>

                   City Driving               Highway Driving              Total

a)              9.375 gal.(300/32)         3.45 gal. (100/29)        12.825 gal.

b)              3.125 gal.(100/32)         10.35 gal    (300/29)     13.475 gal.

<h3>Car B's Consumption of Gas:</h3>

                   City Driving               Highway Driving             Total

a)                11.11 gal. (300/27)         2.94 gal. (100/34)         14.05 gal.

b)              3.70 gal. (100/27)          8.82 gal. (300/34)        12.52 gal.

Thus, different driving ratings (miles per gallon) show the differences in gas consumption by vehicles in city driving and highway driving.

Learn more about the consumption of gas for driving at brainly.com/question/13244789

#SPJ1

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