algorithms
- Thread starter mcheytan
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mcheytan said:
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This is not actually an equation (since there is no "equals" sign), and, by nature of the real numbers, there is no one number "closest to" this value, other than the value itself. So the exercise statement makes very little sense.mcheytan said:
Your subject line refers to "algorithms". I would guess that this means that you were given some algorithm by which to evaluate or approximate the expression. If this is correct, kindly please reply with this information, along with a clear listing of what you have tried so far. If, on the other hand, my guess is incorrect: my apologies.
Actually, one closer value would be 10.001, so your book must have something specific in mind. For us to know what that is, you will need to provide us with the rules, assumptions, and/or other information under which you are operating.mcheytan said:
Have you not studied exponent rules yet...?mcheytan said:
When you reply, please be complete. Thank you!
Eliz.
(10^8 - 10^2)/(10^7 - 10^3)
\(\displaystyle \frac{10^8 - 10^2}{10^7 - 10^3} = \frac{10^2(10^6 - 1)}{10^3(10^4 - 1)}\) and of course 10^6 - 1 is real close to 10^6 and 10^4 - 1 is real close to 10^4, so you might just disregard the "-1's" and simplify from there.
\(\displaystyle \frac{10^8 - 10^2}{10^7 - 10^3} = \frac{10^2(10^6 - 1)}{10^3(10^4 - 1)}\) and of course 10^6 - 1 is real close to 10^6 and 10^4 - 1 is real close to 10^4, so you might just disregard the "-1's" and simplify from there.
Hello, mcheytan!
I found a devious approach . . .
. . Maybe you can scare your teacher . . . LOL!
\(\displaystyle \text{Factor: }\;\frac{10^2\overbrace{(10^6 - 1)}^{\text{diff. of cubes}}}{10^3\underbrace{(10^4-1)}_{\text{diff. of squares}}}\)
\(\displaystyle \text{Factor: }\;\frac{10^2(10^2-1)(10^4+10^2+1)}{10^3(10^2-1)(10^2+1)}\)
\(\displaystyle \text{Reduce: }\;\frac{10^4+10^2+1}{10(10^2+1)}\)
Here's where it gets "cute" . . .
\(\displaystyle \text{We have: }\:\frac{10^2(10^2 + 1) + 1}{10(10^2+1)}\)
\(\displaystyle \text{Make two fractions: }\:\frac{10^2(10^2+1)}{10(10^2+1)} + \frac{1}{10(10^2+1)}\)
\(\displaystyle \text{Reduce: }\;10 + \frac{1}{10(10^2+1)}\)
\(\displaystyle \text{The fraction is slightly less than }\frac{1}{1000} \:=\:0.001\)
\(\displaystyle \text{Therefore, the value of the expression is close to }10.\)
I found a devious approach . . .
. . Maybe you can scare your teacher . . . LOL!
\(\displaystyle \text{Factor: }\;\frac{10^2\overbrace{(10^6 - 1)}^{\text{diff. of cubes}}}{10^3\underbrace{(10^4-1)}_{\text{diff. of squares}}}\)
\(\displaystyle \text{Factor: }\;\frac{10^2(10^2-1)(10^4+10^2+1)}{10^3(10^2-1)(10^2+1)}\)
\(\displaystyle \text{Reduce: }\;\frac{10^4+10^2+1}{10(10^2+1)}\)
Here's where it gets "cute" . . .
\(\displaystyle \text{We have: }\:\frac{10^2(10^2 + 1) + 1}{10(10^2+1)}\)
\(\displaystyle \text{Make two fractions: }\:\frac{10^2(10^2+1)}{10(10^2+1)} + \frac{1}{10(10^2+1)}\)
\(\displaystyle \text{Reduce: }\;10 + \frac{1}{10(10^2+1)}\)
\(\displaystyle \text{The fraction is slightly less than }\frac{1}{1000} \:=\:0.001\)
\(\displaystyle \text{Therefore, the value of the expression is close to }10.\)
Thank you guys!
Firstly, I did not get first reply to my post, because it just showed my problem and an answer without explanation. I have studied exponent rules 3 years ago and basically i am reviewing now.
Second, the first time I posted the problem this is all complete information I am given, and have to solve it for 2 minutes, without calculator. This is part of GMAT test and I was trying to get the idea/logic to it.
Thanks for the ideas!
Firstly, I did not get first reply to my post, because it just showed my problem and an answer without explanation. I have studied exponent rules 3 years ago and basically i am reviewing now.
Second, the first time I posted the problem this is all complete information I am given, and have to solve it for 2 minutes, without calculator. This is part of GMAT test and I was trying to get the idea/logic to it.
Thanks for the ideas!