Looks to be adjacent and virtually all of them are there
Answer:
In order to solve this problem, you have to find the lowest common denominator. Take multiples of both 6 and 4 and see which is the lowest you have in common.
4: 4, 8, 12, 16, 20
6: 6, 12, 18, 24
So 12 is the lowest common multiple....
5/6 x ?/? = ?/12 (you have to multiply the numerator and the denominator by the same number)
3/4 x ?/? = ?/12 (you have to multiply the numerator and the denominator by the same number)
When the denominators are the same, you can subtract the numerators. The denominator will stay 12!
Step-by-step explanation:
YAY! I want a brain
Answer:
its 2
Step-by-step explanation:
because 2*2*2 is 8
Answer:
In the operation, simplify
9-a² = (3-a)(3+a)
Factorize
4a²-4a-24-------------------divide each term by 4
a²-a-6-------------------------factorize
a²-3a+2a-6
a(a-3)+2(a-3)
(a+2)(a-3)
Factorize
a²-6a+9
a²-3a-3a+9
a(a-3)-3(a-3)
(a-3)(a-3)
Rewrite operation as
![\frac{8}{(3-a)(3+a)} /\frac{(a+2)(a-3)}{(a-3)(a-3)}](https://tex.z-dn.net/?f=%5Cfrac%7B8%7D%7B%283-a%29%283%2Ba%29%7D%20%2F%5Cfrac%7B%28a%2B2%29%28a-3%29%7D%7B%28a-3%29%28a-3%29%7D)
Multiply by the recipricol
![\frac{8}{(3-a)(3+a)} *\frac{(a-3)(a-3)}{(a+2)(a-3)}](https://tex.z-dn.net/?f=%5Cfrac%7B8%7D%7B%283-a%29%283%2Ba%29%7D%20%2A%5Cfrac%7B%28a-3%29%28a-3%29%7D%7B%28a%2B2%29%28a-3%29%7D)
cancel the similar terms
![\frac{8}{(3-a)(3+a)} *\frac{(a-3)}{(a+2)}](https://tex.z-dn.net/?f=%5Cfrac%7B8%7D%7B%283-a%29%283%2Ba%29%7D%20%2A%5Cfrac%7B%28a-3%29%7D%7B%28a%2B2%29%7D)