If (ax^2 + 3x + 2b) - (5x^2+bx-3c)=3x^2-9, what is the value of A + B + C?

We have the equation:

(ax^2 + 3x + 2b) - (5x^2+bx-3c)=  3x^2 - 9

First, move all to the left side.

(ax^2 + 3x + 2b) - (5x^2+bx-3c) - 3x^2 + 9 = 0

Now let's group togheter terms with the same power of x.

(a - 5 - 3)*x^2 + (3 - b)*x + (2b + 3c + 9) = 0.

This must be zero for all the values of x, then the things inside each parenthesis must be zero.

1)

a - 5 - 3 = 0

a = 3 + 5 = 8.

2)

3 - b = 0

b = 3.

3)

2b + 3c + 9 = 0

2*3 + 3c + 9 = 0

3c = -6 - 9 = -15

c = -15/3 = -5

Then we have:

a = 8, b = 3, c = -5

a + b + c = 8 + 3 - 5 = 6

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

we now create space between the x and y terms with parenthesis.

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

we add the squares to both sides.

simplify.

and write the quadratics in factored form.

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.

there would only be 50 odd numbers between 1 and 100 so therefor there would be 50 even so that would only be 50 odd numbers.