-9a to the second power+12a-4
Case a)
f(x)=[x-1]/[x+5]
step 1
f(x)=y
y=[x-1]/[x+5]
step 2
exchange x for y and y for x
y=[x-1]/[x+5]------> x=[y-1]/[y+5]----> x*[y+5]=[y-1]----> xy+5x=y-1
step 3
clear the variable y
xy+5x=y-1-----> y-xy=5x+1----> y*[1-x]=[5x+1]----> y=[5x+1]/[1-x]
step 4
f(x)-1= [5x+1]/[1-x]
the function and the inverse function are not the same
case b)
g(x)=[x-2]/[x-1]
step 1
g(x)=y
y=[x-2]/[x-1]
step 2
exchange x for y and y for x
y=[x-2]/[x-1]------> x=[y-2]/[y-1]----> x*[y-1]=[y-2]----> xy-x=y-2
step 3
clear the variable y
xy-x=y-2-----> xy-y=-2+x----> y*[x-1]=[x-2]----> y=[x-2]/[x-1]
step 4
g(x)-1= [x-2]/[x-1]
the function and the inverse function are the same
case c)
h(x)=[x+3]/[x-2]
step 1
h(x)=y
y=[x+3]/[x-2]
step 2
exchange x for y and y for x
y=[x+3]/[x-2]------> x=[y+3]/[y-2]----> x*[y-2]=[y+3]----> xy-2x=y+3
step 3
clear the variable y
xy-2x=y+3-----> xy-y=3+2x----> y*[x-1]=[2x+3]----> y=[2x+3]/[x-1]
step 4
h(x)-1= [2x+3]/[x-1]
the function and the inverse function are not the same
case d)
k(x)=[x+1]/[x-1]
step 1
k(x)=y
y=[x+1]/[x-1]
step 2
exchange x for y and y for x
y=[x+1]/[x-1]------> x=[y+1]/[y-1]---> x*[y-1]=[y+1]----> xy-x=y+1
step 3
clear the variable y
xy-x=y+1-----> xy-y=x+1----> y*[x-1]=[x+1]----> y=[x+1]/[x-1]
step 4
k(x)-1= [x+1]/[x-1]
the function and the inverse function are the same
For this question, all you have to do is subtract 6 from 10 to get 4. 4 + 6 is 10. z = 4.
(3 + (2/3))(2+(1/2) = 3*2+2*(2/3)+3*(1/2)+((2/3)(1/2)) = 6+(4/3) +(3/2) + (2/6) = 6 + ((8+9+2)/6) = 6 + (19/6) = 6 + 3 + 1/6 = 9 + 1/6 miles = D
D and E. If I remember right that should be the answer.
