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Ganezh [65]
3 years ago
13

Allison is twice as old a bob. The sum of their ages is 54. How old is Bob?

Mathematics
1 answer:
frez [133]3 years ago
7 0

Answer:

27

Step-by-step explanation:

There ages when multiplied by 2, have to equal 54. Simply put, just divide 54 by 2 to get your answer. Thus, the answer is 27

Hope this helps.

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Step-by-step explanation:

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3 years ago
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Y equals the quotient of the quantity x squared plus 4 times x and the quantity x cubed minus 5.
koban [17]

Given:

Consider the completer question is "Find the derivative \dfrac{dy}{dx} for y=\dfrac{x^2-4x}{x^3-5}."

To find:

The derivative \dfrac{dy}{dx}.

Solution:

Chain rule: \dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)

Quotient rule: \dfrac{d}{dx}\dfrac{f(x)}{g(x)}=\dfrac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}

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\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{x^2-4x}{x^3-5}\right)

Using chain rule and quotient rule, we get

\dfrac{dy}{dx}=\dfrac{(x^3-5)\dfrac{d}{dx}(x^2-4x)-(x^2-4x)\dfrac{d}{dx}(x^3-5)}{(x^3-5)^2}

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\dfrac{dy}{dx}=\dfrac{2x^4-4x^3-10x+20-3x^4+12x^3}{(x^3-5)^2}

\dfrac{dy}{dx}=\dfrac{-x^4+8x^3-10x+20}{(x^3-5)^2}

Therefore, the required answer is \dfrac{dy}{dx}=\dfrac{-x^4+8x^3-10x+20}{(x^3-5)^2}.

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