Since O is inscribed in triangle ABC, you know that AB, AC, and BC are all tangent to the circle. The tangent segments theorem asserts that AD is congruent to A F; BD is congruent to BE; and CF is congruent to CE. (NOTE: E is the point where the circle touches BC - I've taken the liberty of labeling it as such.)
Then the perimeter 
of triangle ABC will be twice the sum of the labeled edges:
Answer: x = -0.377
Step-by-step explanation:
We have the equation:
4^(5*x) = 3^(x - 2)
Now we can use the fact that:
Ln(A^x) = x*Ln(A)
Then we can apply Ln(.) to both sides of the equation to get:
Ln(4^(5*x)) = Ln(3^(x - 2))
(5*x)*Ln(4) = (x - 2)*Ln(3)
(5*x)*Ln(4) - x*Ln(3) = -2*Ln(3)
x*(5*Ln(4) - Ln(3)) = -2*Ln(3)
x = -2*Ln(3)/(5*Ln(4) - Ln(3)) = -0.377
Step-by-step explanation:
3 - 2(b - 2) = 2 - 7b
To solve this first distribute -2 to (b - 2)
3 + (-2b + (-2) x -2) = 2 - 7b
when we simplify this it becomes:
3 + (-2b + 4) = 2 - 7b
We take (-2b + 4) out of the parenthesis:
3 - 2b + 4 = 2 - 7b
Now simplify again.
7 - 2b = 2 - 7b
Now send all the b's to one side and constants on the other.
7 - 2 = -7b +2b
5 = 5b
b = 1
These two quantities obviously refer to a picture or drawing that shows how they're related to each other. Without the picture, they're not related, and there's no answer.
