Answer:
• point C
Step-by-step explanation:
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Answer:
Step-by-step explanation:
Answer:
The answer is A.
Step-by-step explanation:
Lets call f(x)=y, so y= 4*(3*x-5), we want to find 'x', using 'y' as a the variable.

Now lets change the name of 'y' to 'x', and 'x' to f^-1(x).
f-1(x) = (x+20)/12
4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.
We construct a number
such that taking it mod 4, 5, and 7 leaves the desired remainders:

- Taken mod 4, the last two terms vanish and we have

so we multiply the first term by 3.
- Taken mod 5, the first and last terms vanish and we have

so we multiply the second term by 2.
- Taken mod 7, the first two terms vanish and we have

so we multiply the last term by 7.
Now,

By the CRT, the system of congruences has a general solution

or all integers
,
, the least (and positive) of which is 27.
24 pie and now they want me to explain but I don’t want to