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german
3 years ago
13

Evaluate 3(c + d) if c = 8 and d = -5

Mathematics
1 answer:
Sloan [31]3 years ago
4 0
Substitution time :D

c = 8

d = -5


All you have to do is replace the variables(letters) with the numbers!


3(8 + (-5))

Do parentheses first! And remember that adding a negative is the same as subtracting :)

3(3) = 9
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Write the function in standard form<br><br> f(x) = -4 (x - 6)^2 + 15
amid [387]

9514 1404 393

Answer:

  f(x) = -4x^2 +48x -129

Step-by-step explanation:

It usually works well to compute the square first. That is, simplify according to the order of operations.

  f(x) = -4(x^2 -12x +36) +15

  f(x) = -4x^2 +48x -144 +15

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7 0
2 years ago
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The first- and second-year enrollment values for a technical school are shown in the table below: Enrollment at a Technical Scho
lyudmila [28]

Answer:

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Explanation:

<u>Rewrite the table and the choices for better understanding:</u>

<em>Enrollment at a Technical School </em>

Year (x)       First Year f(x)      Second Year s(x)

2009                  785                        756

2010                   740                        785

2011                    690                        710

2012                   732                         732

2013                   781                          755

Which of the following statements is true based on the data in the table?

  • The solution to f(x) = s(x) is x = 2012.
  • The solution to f(x) = s(x) is x = 732.
  • The solution to f(x) = s(x) is x = 2011.
  • The solution to f(x) = s(x) is x = 710.

<h2>Solution</h2>

The question requires to find which of the options represents the solution to f(x) = s(x).

That means that you must find the year (value of x) for which the two functions, the enrollment the first year, f(x), and the enrollment the second year s(x), are equal.

The table shows that the values of f(x) and s(x) are equal to 732 (students enrolled) in the year 2012,<em> x = 2012. </em>

Thus, the correct choice is the third one:

  • The solution to f(x) = s(x) is x = 2012.
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Answer:

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If you need help, don't hesitate to ask.

I'm here and ready to respond to the hardest questions.

May I please have brainliest?

Have a great day!!!!

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