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Can anyone solve this without picking numbers?
In the formula a=b^2c, if b is multiplied by 2 and c is multiplied by 3, then a must be multiplied by
a. 4
b. 6
c. 9
d. 12
e. 18
In the formula a=b^2c, if b is multiplied by 2 and c is multiplied by 3, then a must be multiplied by
a. 4
b. 6
c. 9
d. 12
e. 18
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iruns0metimes
Joined: 09 Apr 2008
Last visit: 12 Feb 2013
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Concentration: Strategy, Operations
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20 Apr 2008, 12:07
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From working through this, that must be a = b^2 * c rather than a = b^(2*c)...
the way to do this without picking numbers is to take the b-multiplier (2) and square it (since b is being squared) and multiply this new number by 3 (the c-multipier) since it is being multiplied by c in the equation.
a = (b)^2 * (c)
becomes
a = (2*b)^2 * (3*c)
-OR-
a = 2^2 * b^2 * 3 * c
a = 4 * b^2 * 3 * c
a = 4*3 * b^2 * c
a = 12 * b^2 * c
[D]
the way to do this without picking numbers is to take the b-multiplier (2) and square it (since b is being squared) and multiply this new number by 3 (the c-multipier) since it is being multiplied by c in the equation.
a = (b)^2 * (c)
becomes
a = (2*b)^2 * (3*c)
-OR-
a = 2^2 * b^2 * 3 * c
a = 4 * b^2 * 3 * c
a = 4*3 * b^2 * c
a = 12 * b^2 * c
[D]
