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vlada-n [284]
3 years ago
12

Hey hun can any of y'all help me?

Mathematics
1 answer:
11111nata11111 [884]3 years ago
6 0
\bf \cfrac{5}{\frac{1}{6}+\frac{1}{x+1}}\qquad \cfrac{}{\impliedby LCD\textit{ will be 6(x+1)}}\implies \cfrac{5}{\frac{1(x+1)+1(6)}{6(x+1)}}
\\\\\\
\cfrac{5}{\frac{x+1+6}{6(x+1)}}\implies \cfrac{\frac{5}{1}}{\frac{x+7}{6(x+1)}}\implies \cfrac{5}{1}\cdot \cfrac{6(x+1)}{x+7}\implies \cfrac{30(x+1)}{x+7}

and you can expand the numerator if you wish, it won't be simplified further though.
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In triangle △ABC, ∠ABC=90°, BH is an altitude. Find the missing lengths. AC=5 and BH=2, find AH and CH.
Yakvenalex [24]

Answer:

  AH = 1 or 4

  CH = 4 or 1

Step-by-step explanation:

An altitude divides a right triangle into similar triangles. That means the sides are in proportion, so ...

  AH/BH = BH/CH

  AH·CH = BH²

The problem statement tells us AH + CH = AC = 5, so we can write

  AH·(5 -AH) = BH²

  AH·(5 -AH) = 2² = 4

This gives us the quadratic ...

  AH² -5AH +4 = 0 . . . . in standard form

  (AH -4)(AH -1) = 0 . . . . factored

This equation has solutions AH = 1 or 4, the values of AH that make the factors be zero. Then CH = 5-AH = 4 or 1.

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