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3 years ago

14

The area of a trapezoid is 864cm². It has a height of 24 cm, and the length of one of its bases is 30 cm. What is the length of

the other base?

1 answer:

Amiraneli [1.4K]3 years ago

5 0

The area <em>A</em> of a trapezoid with height <em>h</em> and bases <em>b</em>₁ and <em>b</em>₂ is equal to the average of the bases times the height:

<em>A</em> = (<em>b</em>₁ + <em>b</em>₂) <em>h</em> / 2

We're given <em>A</em> = 864, <em>h</em> = 24, and one of the bases has length 30, so

864 = (<em>b</em>₁ + 30) 24 / 2

864 = (<em>b</em>₁ + 30) 12

864 = (<em>b</em>₁ + 30) 12

72 = <em>b</em>₁ + 30

<em>b</em>₁ = 42

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Drupady [299]

Answer:

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Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

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Replace y with a'(d)

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Prove that a(d) and a'(d) are inverse functions

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To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

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Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

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$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

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$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

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Answer:

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