Naturally, any integer

larger than 127 will return

, and of course

, so we restrict the possible solutions to

.
Now,

is the same as saying there exists some integer

such that

We have

which means that any

that satisfies the modular equivalence must be a divisor of 120, of which there are 16:

.
In the cases where the modulus is smaller than the remainder 7, we can see that the equivalence still holds. For instance,

(If we're allowing

, then I see no reason we shouldn't also allow 2, 3, 4, 5, 6.)
I'm assuming this question is 3(1-4)(5-7)
3(-3)(-2)
-9(-2)
18
1/2 because 2/4 is just half of 4/5
Answer:
x is all real numbers
Step-by-step explanation:
5(x-7) + 42 = 5x+7
Distribute
5x - 35 +42 = 5x+7
Combine like terms
5x +7 = 5x+7
Subtract 5x from each side
7=7
This is always true, so x can be any number
Rose petal with 5 petals
Domain: all real numbers
Range: [-2,2]
Symmetric about the x-axis
Continuous
Bounded
Maximum r-value: 2
No asymptotes