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o-na [289]
3 years ago
6

Simplify this problem. |3r−15| if r<5

Mathematics
1 answer:
enot [183]3 years ago
8 0

Answer:

We have the problem:

|3r−15| if r<5

First we see the equality, if r = 5 we have:

I3r - 15I = I3*5 - 15I = I0I = 0.

Then the only restriction that we have is:

I3r - 15I > 0.

now, we could simplify it a bit further:

if r < 5, then the thing inside the absolute value will always be negative:

Then we can write:

I3*r - 15I = -(3*r -15) > 0

multiplying by -1 in both sides

(3r - 15) < 0.

if we keep simplifying this, we will get our initial restriction:

3r - 15 < 0

3r < 15

r < 15/3 = 5

r < 5

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Answer:

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Step-by-step explanation:

Given

s = 10|x - 4| + 50

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30= 10|x - 4|

Divide through by 10

\frac{30}{10}= \frac{10|x - 4| }{10}

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Reorder

|x - 4| = 3

This can be split into

x - 4 = 3 or x - 4 = -3

Solve for x

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Hence:

<em>The numbers of measure of music played are 1 and 7</em>

4 0
3 years ago
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Answer:

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Step-by-step explanation:

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It should go like this:

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(4z +3) / (z + 2)
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