![\frac{3}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B9%7D%20)
+
![\frac{3}{27}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B27%7D%20)
equals
![\frac{4}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B9%7D%20)
.
First, simplify
![\frac{3}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B9%7D%20)
to
![\frac{1}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B3%7D%20)
and also
![\frac{3}{27}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B27%7D%20)
to
![\frac{1}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B9%7D%20)
. Your problem should look like:
![\frac{1}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B3%7D%20)
+
![\frac{1}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B9%7D%20)
.
Second, find the least common denominator of
![\frac{1}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B3%7D%20)
and
![\frac{1}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B9%7D%20)
to get 9.
Third, make the denominators the same as the least common denominator (LCD). Your problem should look like:
![\frac{1x3}{3x3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1x3%7D%7B3x3%7D%20)
+
![\frac{1}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B9%7D%20)
.
Fourth, simplify to get the denominators the same. Your problem should look like:
![\frac{3}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B9%7D%20)
+
![\frac{1}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B9%7D%20)
.
Fifth, join the denominators. Your problem should look like:
![\frac{3+1}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%2B1%7D%7B9%7D%20)
.
Sixth, simplify. Your problem should look like:
![\frac{4}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B9%7D%20)
, which is the answer.
(3a-7)^2
1. 9a^2-21a-21a+49
2. 3a(3a-7)-7(3a-7)
3. (3a-7)(3a-7)
4. Combine like terms
6,050 before 6010 then 5080 after 5060 or basically any numbers that are greater than the one to the left of it :))
4. The Jar Has More Coins. This is because to go from 10% to 8% the number of coins besides penny's has to increase
Please Make Me The Brainlyist!
Answer: A compound event is the combination of two or more simple events (with two or more outcomes).
Step-by-step explanation: