Answer: 1045 cm of red ribbon and 1045 cm of yellow ribbon
<u>Step-by-step explanation:</u>
Let R represent the amount red ribbon and Y represent the amouont of yellow ribbon.
- bought the same lengths of red and yellow: R = Y
- after used 286 of red, yellow is 5 times the remaining red; 5(R - 836) = Y
Use substitution method to solve:
5(R - 836) = R
5R - 4180 = R <em>distributed 5 on the left side</em>
4R - 4180 = 0 <em>subtracted R from both sides</em>
4R = 4180 <em>added 4180 to both sides</em>
R = 1045 <em>divided both sides by 4</em>
Answer:
Step-by-step explanation:
Answer: 
<u>Step-by-step explanation:</u>
Term-to-term rule is also called the recursive rule.
The recursive rule for a geometric sequence is:
where

- r is the common ratio
The common ratio (r) is the second term divided by the first term <em>(which is the same ratio for each term divided by its previous term) </em>
In the given sequence {64, 32, 16, 8, 4}, 
So, the recursive rule is: 
Answer:
Step-by-step explanation:

Answer:
2(x + y)² - 9( x + y ) -5 = 0
⇒2(x + y)² - 10 (x+y) +1(x+y) -5 = 0
⇒2(x+y)(x + y - 5 ) + 1(x + y -5 ) = 0
taking (x + y -5 ) common ,
⇒(x + y -5 )[2(x + y) + 1] =0
⇒(x + y -5)(2x + 2y +1) =0
hope , you got this