Rewrite the expressions in each pair so that they have the same base. c) (1 /2)^2x and (1/ 4)^x - 1

1 answer:

7 0

I'm not completely sure but this is what I would do.
evaluate <span>(1/ 4)^x - 1 </span>as is. But change the (1 /2)^2x to (2/4)^2x. This way both fractions have the same denominator and in this sense, the same base. The 2/4 base still evaluates into 1/2 so nothing, mathematically, is being broken here.

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1-5 I'm not sure what to do on any of these please help!!

IRINA_888 [86]

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7 0

3 years ago

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Points A, B, C, D, and O have coordinates (0, 3), (4, 4), (4, y), (4, 0), and (0, 0), respectively. The area of ACDO is 5 times

belka [17]

Solution:

Area of Quadrilateral ACDO= Area (ΔAOD)+Area(ΔACD)-----(1)

Area of Quadrilateral ABDO= Area (ΔAOD)+Area(ΔABD)------(2)

⇒ (1) = 5 ×(2)→→→→GIven

→Area (ΔAOD)+Area(ΔACD)=5 Area (ΔAOD)+5 Area(ΔABD),

→5 Area (ΔABD)+4 Area (ΔAOD)-Area (ΔACD)=0-----(3)

As, Area of a Triangle $={\text{having vertices}} (x_{1},y_{1}),(x_{2},y_{2}) {\text{and}} (x_{3},y_{3})=\frac{1}{2}\times[x_{1}(y_{2}-y_{3})-x_2(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$