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Sophie [7]
3 years ago
8

Steven used the composition f(f^-1(x)) to show that the functions f(x)=(x+2)^3 and f^-1(x)=3 sqrt x-2 are inverses.

Mathematics
1 answer:
patriot [66]3 years ago
6 0
Your smart please answer mine
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The equation a=1/2(b^1+b^2)h can be determined the area, a, of a trapezoid with height, h, and base lengths, b^1 and b^2 Which a
Evgesh-ka [11]

The complete question is as follows.

The equation a = \frac{1}{2}(b_1 + b_2 )h can be used to determine the area , <em>a</em>, of a trapezoid with height , h, and base lengths, b_1 and b_2. Which are equivalent equations?

(a) \frac{2a}{h} - b_2 = b_1

(b) \frac{a}{2h} - b_2 = b_1

(c) \frac{2a - b_2}{h} = b_1

(d) \frac{2a}{b_1 + b_2} = h

(e) \frac{a}{2(b_1 + b_2)} = h

Answer: (a) \frac{2a}{h} - b_2 = b_1; (d) \frac{2a}{b_1 + b_2} = h;

Step-by-step explanation: To determine b_1:

a = \frac{1}{2}(b_1 + b_2 )h

2a = (b_1 + b_2)h

\frac{2a}{h} = b_1 + b_2

\frac{2a}{h} - b_2 = b_1

To determine h:

a = \frac{1}{2}(b_1 + b_2 )h

2a = (b_1 + b_2)h

\frac{2a}{(b_1 + b_2)} = h

To determine b_2

a = \frac{1}{2}(b_1 + b_2 )h

2a = (b_1 + b_2)h

\frac{2a}{h} = (b_1 + b_2)

\frac{2a}{h} - b_1 = b_2

Checking the alternatives, you have that \frac{2a}{h} - b_2 = b_1 and \frac{2a}{(b_1 + b_2)} = h, so alternatives <u>A</u> and <u>D</u> are correct.

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How to tell if vector is unit vector?
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Triangle X Y Z is shown. Angle X Y Z is 75 degrees and angle Y Z X is 50 degrees. The length of X Y is 2 and the length of X Z i
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Answer:

2.5

Step-by-step explanation:

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