No, they forgot to switch variable labels after solving for the independent variable...
y=-8x+4
y-4=-8x
(y-4)/-8=x
Now that you have solved for the independent variable x, you switch the variable labels...
y=(x-4)/-8
f^-1(x)=(x-4)/-8 which should really be rewritten as:
f^-1(x)=(4-x)/8 :P
Answer:
C
Step-by-step explanation:
x²+2x-15 = x²-3x+5x-15 = x(x-3)+5(x-3)=(x+5)(x-3)
or delta:
∆=2²-4*1*(-15)=4+60=64
√∆=8
x1 = (-2+8)/2 = 3
x2=(-2-8)/2=-5
and 4x+20 = 4(x+5)
The same is (x+5), so it is GCF. answer C
Answer:
R = 5t
Step-by-step explanation:
given s is three times t then s = 3t and
R = 3t + 2t = 5t
One way is by using the same number
