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SashulF [63]
3 years ago
7

Q bisects PR, PQ=3y, and PR=42. Find y and QR.

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
3 0
|PR| = |PQ| + |QR|; |PQ| = |QR| conclusion |PR| = 2|PQ|

|PQ| = 3y; |PR| = 42; |QR|=?

subtitute

42 = 2(3y)
6y = 42    |divide both sides by 6
y = 7

|QR| = 3(7) = 21
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I GIVE BRAINLIEST PLS HELP
ioda
The answer is 8
Here's why:
{ ( \frac{( {6}^{7}) \times ( {3}^{3})  }{(  {6}^{6}) \times ( {3}^{4}  ) } )}^{3}  =   \\ ( \frac{6}{3} ) ^{3}  =  \\ \frac{216}{27}  = 8
The exponents are subtracted one from another when divided.
\frac{ a ^{b} }{ {a}^{c} } =  {a}^{b - c}
We can look at the problem this way:
( \frac{6^{7} }{6 ^{6} }  \times  \frac{3^{3} }{ {3}^{4} } ) = (6^{7 - 6}  \times  {3}^{3 - 4} ) =  \\ ({6}^{1}  \times  {3}^{ - 1} )
Since we have the power of -1 on the 3, we apply this rule:
{a}^{ - b}  =  \frac{1}{ {a}^{b} }
Also this rule because we have the power of 1 on the 6:
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We apply the rule:
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\frac{{6}^{3} }{ {3}^{3} } =  \frac{216}{27}  = 8
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