In rhombus ABCD, DB=16 and AC=12. What is the perimeter of rhombus ABCD?
1 answer:
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The bases are both 2, so we would subtract the exponents. This is because the rule is
(a^b)/(a^c) = a^(b-c)
In this case,
a = 2
b = 3/4
c = 1/2
So this means
b - c = (3/4) - (1/2) = (3/4) - (2/4) = 1/4
After subtracting the exponents, the final exponent is 1/4
So the expression simplifies to 2^(1/4) which is the same as![\sqrt[4]{2}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B2%7D)
(fourth root of 2)
(a^b)/(a^c) = a^(b-c)
In this case,
a = 2
b = 3/4
c = 1/2
So this means
b - c = (3/4) - (1/2) = (3/4) - (2/4) = 1/4
After subtracting the exponents, the final exponent is 1/4
So the expression simplifies to 2^(1/4) which is the same as