3 years ago

The ______ average ignores compounding.

1 answer:

8 0

The answer is: [B]: "arithmetic" .
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"The arithmetic average ignores compounding."
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Find the square. (1/4A + 1/4B)^2 a) 1/16A^2 + 1/8AB + 1/16B^2 b) 1/4A^2 + 1/8AB + 1/4B^2

alisha [4.7K]

$(a + b)^2 = a^2 + 2ab + b^2$

$(\dfrac{1}{4}A + \dfrac{1}{4}B)^2 =$

$= \dfrac{1}{16}A^2 + \dfrac{1}{8}AB + \dfrac{1}{16}B^2$

Answer: A.

Another way to solve by factoring 1/4.

$(\dfrac{1}{4}A + \dfrac{1}{4}B)^2 =$

$= [\dfrac{1}{4}(A + B)]^2$

$= (\dfrac{1}{4})^2(A + B)^2$

$= \dfrac{1}{16}(A^2 + 2AB + B^2)$

$= \dfrac{1}{16}A^2 + \dfrac{1}{16}2AB + \dfrac{1}{16}B)^2$

$= \dfrac{1}{16}A^2 + \dfrac{1}{8}AB + \dfrac{1}{16}B)^2$

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Using the rule

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