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vova2212 [387]
3 years ago
7

Which equation represents the situation Please help me

Mathematics
2 answers:
stiv31 [10]3 years ago
6 0

A university charges a $100 application fee and $720 per month

x = the number of months

y = total cost


y = 720x + 100

[they charge $720 per month(x) plus a fee of $100]


Your answer is the last option.

Virty [35]3 years ago
4 0

100 dollars should be a one time fee so that has to be added to the equation, and since it is a one time fee it should not have a variable next to it.


It also says that you have to pay 720 dollars per month, so it has to have a variable next to it to show the number of months that she needs to pay that amount of money.


The answer should be D.

y= 720x +100

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110% of 300 (with work)
zepelin [54]

Answer: 36.67

Step-by-step explanation

110 of 300 can be written as:

110 /300

To find percentage, we need to find an equivalent fraction with denominator 100. Multiply both numerator & denominator by 100  

110 /300  ×  100 /100 = (  110 × 100 /300 ) ×  1 /100  =  36.67 /100

Therefore, the answer is 36.67%

If you are using a calculator, simply enter 110÷300×100 which will give you 36.67 as the answer.

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3 years ago
Find the Horizontal Tangent for x^3/3+3x^2-16x+9
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Answer:

The two horiz. tang. lines here are y = -3 and y = 192.

Step-by-step explanation:

Remember that the slope of a tangent line to the graph of a function is given by the derivative of that function.  Thus, we find f '(x):

f '(x) = x^2 + 6x - 16.  This is the formula for the slope.  We set this = to 0 and determine for which x values the tangent line is horizontal:

f '(x) = x^2 + 6x - 16 = 0.  Use the quadratic formula to determine the roots here:  a = 1; b = 6 and c = -16:  the discriminant is b^2-4ac, or 36-4(1)(-16), which has the value 100; thus, the roots are:

      -6 plus or minus √100

x = ----------------------------------- = 2 and -8.

                         2

Evaluating y = x^3/3+3x^2-16x+9 at x = 2 results in y = -3.  So one point of tangency is (2, -3).  Remembering that the tangent lines in this problem are horizontal, we need only the y-coefficient of (2, -3) to represent this first tangent line:  it is y = -3.

Similarly, find the y-coeff. of the other tangent line, which is tangent to the curve at x = -8.  The value of x^3/3+3x^2-16x+9 at x = -8 is  192, and so the equation of the 2nd tangent line is y=192 (the slope is zero).


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\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Let's solve ~

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That is :

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