3(2a-5) = 12-7a
6a-15=12-7a
6a+7a=12+15
13a= 27
27=3c-3(6-2c)
27=3c-18+6c
27+18=3c+6c
45=9c
C=45/9
C=5
6c-8-2c=-16
6c-2c=-16+8
4c=-8
C=-8/4
C=-2
I hope this helps you
(-2x^2). (-5x^2)+(-2x^2)(4x^3)
10x^4-8x^5
We are given the points
<span>(0,-4),(1,0),(2,2)
The standard form a quadratic function (in terms of x) is
y = Ax2 + Bx + C
Subsitute the points
-4 = 0 + 0 + C
C = -4
0 = A + B - 4
2 = 4A + 2B - 4
Solve for A and B
A = -1
B = 5
The function is
y = -x2 + 5x - 4</span>
Please clarify, is it f(x)=2 to solve it?