Answer:
Explanation:
We can factor the numerator and denominator as;
(
x
−
2
)
(
x
−
1
)
2
x
(
x
−
1
)
We can now cancel common term in the numerator and denominator:
(
x
−
2
)
(
x
−
1
)
2
x
(
x
−
1
)
⇒
x
−
2
2
x
However, we cannot divide by
0
so we must exclude:
2
x
=
0
⇒
x
=
0
and
x
−
1
=
0
⇒
x
1
x
2
−
3
x
+
2
2
x
2
−
2
x
=
x
−
2
2
x
Where:
x
≠
0
and
x
≠
1
Or
x
2
−
3
x
+
2
2
x
2
−
2
x
=
x
2
x
−
2
2
x
=
1
2
−
1
x
Where:
x
≠
0
and
x
≠
1
Step-by-step explanation:
Hey what number do you need help with
Answer:
a. 6
b. 9
Step-by-step explanation:
a. The product modulo 7 can be found from the product of the individual numbers modulo 7:
(88·95·36·702) mod 7 = (88 mod 7)·(95 mod 7)·(36 mod 7)·(703 mod 7) mod 7
= (4·4·1·3) mod 7 = 48 mod 7 = 6
__
b. Powers of 4 mod 11 repeat with period 5:
4 mod 11 = 4
4^2 mod 11 = 5
4^3 mod 11 = 9
4^4 mod 11 = 3
4^5 mod 11 = 1
So, 4^83 mod 11 = 4^3 mod 11 = 9
The first one simplified is 2j-2k
30% of 520 is 156
I’m sorry that’s not my level but i would love to help
