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olganol [36]
2 years ago
11

Under her cell phone plan, Sarah pays a flat cost of $69 per month and $4 per gigabyte. She wants to keep her bill at $83.80 per

month. Write and solve an equation which can be used to determine gg, the number of gigabytes of data Sarah can use while staying within her budget.
Mathematics
1 answer:
zalisa [80]2 years ago
7 0

Using a linear function, it is found that Sarah can use 3.7 gigabytes while staying within her budget.

<h3>What is a linear function?</h3>

A linear function is modeled by:

y = mx + b

In which:

  • m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.
  • b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.

Considering the flat cost as the y-intercept and the cost per gigabyte as the slope, the cost of using g gigabytes is:

C(g) = 4g + 69.

She wants to keep her bill at $83.80 per month, hence:

C(g) = 83.80

4g + 69 = 83.80

4g = 14.80

g = 14.80/4

g = 3.7.

Sarah can use 3.7 gigabytes while staying within her budget.

More can be learned about linear functions at brainly.com/question/24808124

#SPJ1

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If you've started pre-calculus, then you know that the derivative of  h(t)
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The derivative is            h'(t) = -32 t  +  96 .

At the maximum ...        h'(t) = 0

                                       32 t = 96 sec

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___________________________________________

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

6 0
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