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lookagain
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lookagain
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08-22-2010
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07:06 PM
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06-20-2018,
06:03 PM
lookagain
replied to a thread
How to solve this type of problems in easy way?
in
Arithmetic
Let's see: One one thousand Two one thousand Three one thousand No, after seeing Jomo's suggested solution, which would include the time...
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12 replies | 417 view(s)
06-13-2018,
03:46 PM
lookagain
replied to a thread
Preparing for ASVAB [Help]
in
Arithmetic
If that section of that ASVAB test allows the use of a basic functions calculator, then I am okay with this problem/question. However, if no...
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2 replies | 172 view(s)
06-12-2018,
03:03 PM
lookagain
replied to a thread
Is 4a^3+5a always a multiple of 3 for positive integers?
in
Intermediate/Advanced Algebra
Each of the following are divisible by 3: (a - 1)(a)(a + 1) \ = a(a^2 - 1) \ = a^3 - a
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11 replies | 393 view(s)
06-03-2018,
01:58 PM
lookagain
replied to a thread
Use four 4's to make 5 with limitations
in
Math Odds & Ends
\dfrac{\sqrt{4}*\sqrt{4}}{4} \ + \ 4 \ = \ 5
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6 replies | 559 view(s)
06-03-2018,
08:07 AM
lookagain
replied to a thread
Use four 4's to make 5 with limitations
in
Math Odds & Ends
\dfrac{4\sqrt{4} \ + \ \sqrt{4}}{\sqrt{4}} \ = \ 5
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6 replies | 559 view(s)
06-02-2018,
10:54 PM
lookagain
replied to a thread
Use four 4's to make 5 with limitations
in
Math Odds & Ends
4 + 4/(sqrt(4)*sqrt(4)) = 5 (sqrt(4) + 4)/sqrt(4) + sqrt(4) = 5
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6 replies | 559 view(s)
06-02-2018,
03:05 PM
lookagain
replied to a thread
Use four 4's to make 5 with limitations
in
Math Odds & Ends
\sqrt{4}\bigg( \sqrt{4} \ + \ \dfrac{ \sqrt{4}}{4}\bigg) \ = \ 5
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6 replies | 559 view(s)
05-30-2018,
11:50 PM
lookagain
replied to a thread
logarithmic inequality: log_0.5(2x - 1) - log_2(5 - 3x) <= 0
in
Intermediate/Advanced Algebra
No, those highlighted inequalities of yours in the quote box are not equivalent to the original inequality. Yours have a different solution set. ...
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21 replies | 802 view(s)
05-30-2018,
03:18 PM
lookagain
replied to a thread
logarithmic inequality: log_0.5(2x - 1) - log_2(5 - 3x) <= 0
in
Intermediate/Advanced Algebra
\dfrac{log(2x - 1)}{log(1/2)} \ - \ \dfrac{log(5 - 3x)}{log(2)} \ \le \ 0 \dfrac{log(2x - 1)}{log(2^{-1})} \ - \ \dfrac{log(5 - 3x)}{log(2)} \ \le...
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21 replies | 802 view(s)
05-29-2018,
01:46 PM
lookagain
replied to a thread
integral problem, please respond soon: int 4 dx / x ln(5x)
in
Calculus
\displaystyle 4 \int \dfrac{du}{u} \ = \ 4 ln|u| \ + C
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4 replies | 220 view(s)
05-29-2018,
01:35 PM
lookagain
replied to a thread
Explain Integration by Parts: why you need to find u and dv in order to int. by parts
in
Calculus
The exponents must be in grouping symbols because of the Order of Operations. lookagain edit: You have the integral dx. You can make ANY...
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6 replies | 259 view(s)
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