Answer:
See the attached figure which represents the problem.
As shown, AA₁ and BB₁ are the altitudes in acute △ABC.
△AA₁C is a right triangle at A₁
So, Cos x = adjacent/hypotenuse = A₁C/AC ⇒(1)
△BB₁C is a right triangle at B₁
So, Cos x = adjacent/hypotenuse = B₁C/BC ⇒(2)
From (1) and (2)
∴ A₁C/AC = B₁C/BC
using scissors method
∴ A₁C · BC = B₁C · AC
Answer: $23..50 see attachment for the explanation.
I don’t know how to do these
Answer:
D
Step-by-step explanation:
(1/7)^(x)=A^(-9x)
(1/7)^(x)=(A^(-9))^x
A^(-9)=1/7
A=(1/7)^9