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.)
The first bullet is correct.
No. because
__________________________
are there two middles to that line segment?? no.
what if i divided it equally like so...:
______|______|_____
no. you've just divided it into 3rds.
so the answer is no. you can only have one midpoint, one middle, in a single line segment.
Answer:
5/6 (i think im not good at math soz)
-Mina
Step-by-step explanation:
Branily stop deleting my awnsers
Answer:
3
5
÷
j
=
3
8
Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.
(
3
5
)
⋅
8
=
j
⋅
3
Step-by-step explanation:Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.
(
3
5
)
⋅
8
=
j
⋅
3