I would start off by taking away 1a. That would make the problem be 56ab3-35b.I only took away 1 because each have at least 1a and is okay to do.
Next I would deal with the variable b. I would cross of 1 b. That's because both sides have at least 1b. Now, it's shortened to be 56ab2-35.
Since you cannot take away anymore variables, you have to deal with 56 and 35. I start small with dividing each by 2. I am trying to see what the greatest number could be while making the numbers still be whole. That turns 56 into 28 when it's cut in half. The 35 now turns into 17.5.
I would assume your teacher would want the numbers to be whole. seeing as though when 35 is cut in half and makes a decimal number, I would leave them. What I mean by that is to leave the numbers as 56 and 35.
So, that means the answer is 56ab2-35.
I hope this helps!! (And makes sense)
<h3><u>S</u><u>olution</u></h3>
1). 1/3 + 1/5 = (5 + 3)/15
= 8/15. Ans.
2).
==> 3/5 - 1/6
= (3×5 - 5)/30
= (15 - 5)/30
= 10/30
= 1/3 Ans.
3).
==> 2/3 ÷ 3/4
= 2/3 × 4/3
= 8/9 Ans.
4).
==> 3/8 + 5/12
= (3×3 + 10)/24
= (9+10)/24
= 19/24 Ans.
5).
==> 5/16 - 3/32
= (5×2 - 3)/32
= (10-3)/32
= 7/32 Ans.
Answer:
217 is the perfect swuare i think
C because the square root of 8 is 64
