chasenkcastro chasenkcastro
  • 01-12-2020
  • Mathematics
contestada

This assignment is hard as heck lol!

This assignment is hard as heck lol class=

Respuesta :

jimrgrant1 jimrgrant1
  • 01-12-2020

Answer:

z = 3[tex]\sqrt{13}[/tex]

Step-by-step explanation:

Using the altitude- on- hypotenuse theorem

( leg of large Δ )² = ( part of hypotenuse below it ) × ( whole hypotenuse )

z² = 9 × (9 + 4) = 9 × 13 = 117 ( take square root of both sides )

z = [tex]\sqrt{117}[/tex]

  = [tex]\sqrt{9(13)}[/tex]

  = [tex]\sqrt{9}[/tex] × [tex]\sqrt{13}[/tex]

  = 3[tex]\sqrt{13}[/tex]

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