Solve: x+2=15
x=13
Simplify: 2y2 * 3y9
6y11
Is this relation a function?
{(1,2), (4,7), (2,7), (-1,2), (7,4)}
Yes - the x-values have exactly one y-value (they do not repeat!)
Factor: 2x+6
2(x+3)
Factor: x2+4x+3
(x+3)(x+1)
Solve: 2x-13=15
x=9
Simplify: 15y3x6/5yz2
3y2z4
Given f(x)=x2-9x+10, find f(5).
f(5)=-10
Factor: 15x+25y
5(3x+5y)
Factor: x2+9x+8
(x+8)(x+1)
Solve: 4x=3x-10
x=-10
Add: (3x+4) + (8x-10)
11x-6
Identify the type of function: y=3x+1
Linear
Factor: 9x2-3x
3x(3x-1)
Factor: x2-10x+25
(x-5)(x-5)
Solve: 3(2x+5)=3
x=-2
Subtract: (2x2-4x+10) - (x2+8x+10)
x2-12x
Identify the type of function: (1,1), (2,4), (3,5), (4,4), (5,1)
Quadratic
Factor: 36xy2-48x2y
12xy(3y-4x)
Factor: x2+11x-60
(x-15)(x+4)
Solve: 5(3x+1)=10x-25
x=-6
Multiply: (3x-5)(2x+7)
6x2+11x-35
Identify the type of function: y=2(0.5)x+7
Factor: 63a3bc2-72a2bc
9a2bc(7ac-8)
Factor Completely: 2x3+10x2+8x
2x(x+4)(x+1)