Answer:
I'm not sure what the answer is but, I'm thinking the answer w
The inscribed circle has its center at the point of intersection of the angle bisectors, which also happen to be the medians. Hence the altitude of the triangle is 3 times the radius, or 12 inches.
The side length of this triangle is 2/√3 times the altitude, so the area is
... Area = (1/2)·b·h = (1/2)·(24/√3 in)·(12 in)
... Area = 48√3 in² ≈ 83.1384 in²
X^2 + 5x = -2
x^2 + 5x + 2 = 0
x = -b (+-) sqrt (b^2 - 4ac) / 2a
a = 1, b = 5, and c = 2
x = -5 (+-) sqrt (5^2 - 4(1)(2)) / 2(1)
x = -5 (+-) sqrt (25 - 8) / 2
x = -5 (+-) sqrt (17) / 2
answer is : negative 5 plus or minus the square root of 17 divided by 2
here we have to find the quotient of '(16t^2-4)/(8t+4)'
now we can write 16t^2 - 4 as (4t)^2 - (2)^2
the above expression is equal to (4t + 2)(4t - 2)
there is another expression (8t + 4)
the expression can also be written as 2(4t + 2)
now we have to divide both the expressions
by dividing both the expressions we would get (4t + 2)(4t - 2)/2(4t + 2)
therefore the quotient is (4t - 2)/2
the expression comes out to be (2t - 1)