In triangle △abc, ∠abc=90°, bh is an altitude. find the missing lengths. ac=5 and bh=2, find ah and ch.

  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.

what level of math? also i don't need to be paid if i can you.

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol