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arlik [135]
3 years ago
15

3 equivalent ratios for : 4/3, 12/14, and 6/9

Mathematics
1 answer:
Mariana [72]3 years ago
5 0
<span>So the question is what are three equivalent ratios for 4/3, 12/14 and 6/9. The simplest way to get the equivalent ratio of some other ratio is to either multiply the nominator and the denominator by the same number or to divide the nominator and the denominator with the same number. I need to point out that it's not always possible to divide them and get a whole number. First: (4/3)*(2/2)=8/6, (4/3)*(3/3)=12/9 and (4/3)*(4/4)=16/12. Second: (12/14)/(2/2)=6/7, (12/14)*(2/2)=24/28 and (12/14)*(3/3)=36/42. Third: (6/9)/(3/3)=2/3, (6/9)*(2/2)=12/18 and finally (6/9)*(3/3)=18/27.</span>
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Step-by-step explanation:

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Sergio [31]

Answer:

The values of p in the equation are 0 and 6

Step-by-step explanation:

First, you have to make the denominators the same. to do that, first factor 2p^2-7p-4 = \left(2p+1\right)\left(p-4\right)2p

2

−7p−4=(2p+1)(p−4)

So then the equation looks like:

\frac{p}{2p+1}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{5}{p-4}

2p+1

p

−

(2p+1)(p−4)

2p

2

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=−

p−4

5

To make the denominators equal, multiply 2p+1 with p-4 and p-4 with 2p+1:

\frac{p^2-4p}{(2p+1)(p-4)}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{10p+5}{(p-4)(2p+1)}

(2p+1)(p−4)

p

2

−4p

−

(2p+1)(p−4)

2p

2

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(p−4)(2p+1)

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Since, this has an equal sign we 'get rid of' or 'forget' the denominator and only solve the numerator.

(p^2-4p)-(2p^2+5)=-(10p+5)(p

2

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2

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2

−4p)−(2p

2

+5) first:

(p^2-4p)-(2p^2+5)=-p^2-4p-5(p

2

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2

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2

−4p−5=−10p+5

Combine like terms:

-p^2-4p+0=-10p−p

2

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Factor:

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