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svlad2 [7]
3 years ago
5

Are the fractions 1/5 , 5/5 , and 5/1 equivalent?Explain.

Mathematics
1 answer:
NeX [460]3 years ago
5 0
<span><u><em>Answer:</em></u>
They are not equal

<u><em>Explanation:</em></u>
<u><em>Let's take a look at each one of them separately:</em></u>
</span>\frac{1}{5}<span> : This fraction is in the simplest form. It is equivalent to 0.2
The fraction is formed by dividing one part over five

</span>\frac{5}{5}<span> : This fraction is not in the simplest form. We can divide 5 by 5 and the answer would be 1. Therefore, </span>\frac{5}{5}<span> is equivalent to 1

</span>\frac{5}{1}<span> : This fraction is not in the simplest form. We can divide 5 by 1 and the answer would be 5. Therefore, </span>\frac{5}{1}<span> is equivalent to 5.

Based on the above, we can cnfirm that 0.2 , 1 and 5 are not equivalent.
Therefore, the given fractions are not equivalent

Hope this helps :)</span>
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Solve the system by elimination.(show your work)
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Answer:

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Step-by-step explanation by elimination:

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-2 x - y + z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

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{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x - 3 y - 2 z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y+0 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1 , y = 1 , z = 0

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Answer:

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3) LHS  = Cosec² A + Cot² A

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