Y=8
your teacher is introducing algebra
Are your numbers right? Diana worked on her project for 613 hours, Gabe worked on it for 134 times as long as Diana, and Paula worked on her science project 34 times as long as Diana? Assuming they are, It is false that Diana worked longer on her science project than Gabe worked on his. Tell me if I read the question wrong, I am happy to help.
Answer:
0.55
Step-by-step explanation:
diference= 25/100*2.20=0.55
Your question seems a bit incomplete, but for starters you can write

Expanding where necessary, recalling that

, you have

and you can stop there, or continue to rewrite in terms of the reciprocal functions,

Now, since

, the final form could also take

or
Answer:
The answer can be calculated by doing the following steps;
Step-by-step explanation: