when looking back the whole question is so easy. However, its all about the approach to the question. The solution explains that answer needs be taken with a common denomitor so that the answers will fit. How should one know how to approach these questions that stagger you at first but are actually easy questions?
(1/2)^-3 * (1/4)^-2 * (1/16)^-1 =
a. (1/2)^-48
b. (1/2)^-11
c. (1/2)^-6
d. (1/8)^-11
e.(1/8)^-6
qa is d... the math isnt that hard but the approach is staggering. what to do?
strategy..hard problem
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- Neo2000
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Look at the answer options. The numbers are either 2 or 8. So you know straight away that you have to convert to either 2 or 8. Start with 2 because you already have a 2, 4 is the same as 2Square and 16 is the same as 2^4Enginpasa1 wrote:when looking back the whole question is so easy. However, its all about the approach to the question. The solution explains that answer needs be taken with a common denomitor so that the answers will fit. How should one know how to approach these questions that stagger you at first but are actually easy questions?
(1/2)^-3 * (1/4)^-2 * (1/16)^-1 =
a. (1/2)^-48
b. (1/2)^-11
c. (1/2)^-6
d. (1/8)^-11
e.(1/8)^-6
qa is d... the math isnt that hard but the approach is staggering. what to do?
-
- Master | Next Rank: 500 Posts
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- Joined: Fri Jan 04, 2008 8:10 pm
- Thanked: 15 times
Given Question...
(1/2)^-3 * (1/4)^-2 * (1/16)^-1
just remember the following formuale for this kind of questions.
1. a^m*a^n=a^m+n
2. (a^m)^n=a^mn
3. a^m/a^n=a^m-n where a!=0
4. (a*b)=a^m*b^m
5. 1/a^-m=a^m
so the question can be written as
2^3*4^2*16^1
2^3*2^4*2^4
2^11
=1/2^-11
(1/2)^-3 * (1/4)^-2 * (1/16)^-1
just remember the following formuale for this kind of questions.
1. a^m*a^n=a^m+n
2. (a^m)^n=a^mn
3. a^m/a^n=a^m-n where a!=0
4. (a*b)=a^m*b^m
5. 1/a^-m=a^m
so the question can be written as
2^3*4^2*16^1
2^3*2^4*2^4
2^11
=1/2^-11