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pashok25 [27]
2 years ago
15

The height of a stack of DVD cases is proportional to the number of cases in the stack. The height of 6 DVD cases is 114 mm.

Mathematics
2 answers:
Zanzabum2 years ago
8 0
OK so my answer is from what you’ve just explained that means you multiply the six and the 114 is 684 so that means if you have to do that to 13 of them that means 684×13 is what your equation would be in the answer to 684×13 is 8892
Sholpan [36]2 years ago
5 0

88Answer:

Step-by-step explanation:

8892

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Use the given equation to find the number <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20x%20-%2019.7%20%3D%20
ziro4ka [17]

Answer:

x = 31.2

Step-by-step explanation:

\frac{1}{2}x - 19.7 = -4.1

\frac{1}{2}x = 15.6 --- add 19.7 to both sides to get rid of -19.7

x = 31.2 --- multiply both sides by 2 to get rid of \frac{1}{2}

7 0
2 years ago
Given that x = 3, y = -2 and z = 0, calculate the value of each of the following algebraic expressions: 2x + y. Answer quick ple
icang [17]
2 times 3+(-2)
6-2
4. Is the cord answer
4 0
3 years ago
Use synthetic division to find P (-10) for P(x)=2x^3+14x^2-58x
Savatey [412]
The polynomial remainder theorem states that the remainder upon dividing a polynomial p(x) by x-c is the same as the value of p(c), so to find p(-10) you need to find the remainder upon dividing

\dfrac{2x^3+14x^2-58x}{x+10}

You have

..... | 2 ...  14  ... -58
-10 |    ... -20  ... 60
--------------------------
..... | 2 ...  -6  ....  2

So the quotient and remainder upon dividing is

\dfrac{2x^3+14x^2-58x}{x+10}=2x-6+\dfrac2{x+10}

with a remainder of 2, which means p(-10)=2.
5 0
3 years ago
Which term describes the set of all possible output values for a function?
Firdavs [7]

The <em><u>correct answer</u></em> is:

C. Range

Explanation:

The definition of domain is the set of all possible input values, or x-values, of a function.

The definition of range is the set of all possible output values, or y-values, of a function.

3 0
3 years ago
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vredina [299]
I'm sorry no one can answer and where is the question?
Please put in the question and I might be able to help!
8 0
2 years ago
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