F(x)=2x+4/3
Replace f(x) by y:
y=2x+4/3
Solving for x:
y-4/3=2x+4/3-4/3
Subtracting the two terms on the left side of the equation:
(3y-4)/3=2x
Dividing both sides of the equation by 2:
[(3y-4)/3]/2=2x/2
[(3y-4)/3]*(1/2)=x
[(3y-4)*1]/[(3)*(2)]=x
(3y-4)/6=x
x=(3y-4)/6
Replace "x" by "f^(-1) (x)" (inverse function) and "y" by "x":
f^(-1) (x) = (3x-4)/6 Inverse function
Then, for x=6
f^(-1) (6) = [ 3(6)-4]/6
f^(-1) (6) = (18-4)/6
f^(-1) (6)= 14/6
f^(-1) (6) = 7/3
Her hair will have grown 6 after 7/3 = 2.33 months
Answer:
I believe the answer is 406 miles.
Let me know of this is the correct answer
If you change both to a fraction you would see that you have 1/6 and 1/4 (.25 as a fraction). So 1/5 would be in between them.
You could also change both as a decimal. 1/6 =.166666 and .25 so any number in between them would be correct such as .2 which is the same as 1/5.
Factor:
3x^2 + 27
= 3(x^2 + 9)
Answer is 3(x^2 + 9), when factored.
A) (3x + 9i)(x + 3i)
= (3x + 9i)(x + 3i)
= (3x)(x) + (3x)(3i) + (9i)(x) + (9i)(3i)
= 3x^2 + 9ix + 9ix + 27i^2
= 27i^2 + 18ix + 3x^2
B) (3x - 9i)(x + 3i)
= (3x + - 9i)(x + 3i)
= (3x)(x) + (3x)(3i) + ( - 9i)(x) + (- 9i)(3i)
= 3x^2 + 9ix - 9ix - 27i^2
= 27i^2 + 3x^2
C) (3x - 6i)(x + 21i)
= (3x + - 6i)(x + 21i)
= (3x)(x) + (3x)(21i) + (- 6i)(x) + ( -6i)(21i)
= 3x^2 + 63ix - 6ix - 126i^2
= - 126i^2 + 57ix + 3x^2
D) (3x - 9i)(x - 3i)
= (3x + - 9)(x + - 3)
= (3x)(x) + (3x)( - 3i) + (- 9)(x) + ( - 9)( - 3i)
= 3x^2 - 9ix - 9x + 27i
= 9ix + 3x^2 + 27i - 9x
Hope that helps!!!
Answer:
1. A 2. C
Step-by-step explanation:
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