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Brut [27]
3 years ago
10

The price of a candy bar is $1.53. If this is three cents more than triple the price ten years ago, how much did the candy bar c

ost ten years ago?
Mathematics
1 answer:
Tamiku [17]3 years ago
6 0

Answer:

$0.51

Step-by-step explanation:

Hello! Using what we know, the price of the candy bar at the moment is $1.53, triple the price from 10 years ago. So, triple cents would technically be $0.03. We divide $1.53 and $0.03 and get 51! But, it would not make sense for the price ten years ago to be 51 then go to 1.53, so we make 51 a decimal, 0.51. We can check to see if this correct, by doing 0.51 times 3, because the price is tripling, and we get $1.53! Have an awesome day! :)

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Find the derivative
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\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dy}\cdot\dfrac{\mathrm dy}{\mathrm dx}

By the power rule,

f(x)=y^2\implies\dfrac{\mathrm df}{\mathrm dy}=2y=\dfrac{2(x+5)}{x^2+3}

By the quotient rule,

y=\dfrac{x+5}{x^2+3}\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{(x^2+3)\frac{\mathrm d(x+5)}{\mathrm dx}-(x+5)\frac{\mathrm d(x^2+3)}{\mathrm dx}}{(x^2+3)^2}

\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{(x^2+3)-(x+5)(2x)}{(x^2+3)^2}

\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3-10x-x^2}{(x^2+3)^2}

So

\dfrac{\mathrm df}{\mathrm dx}=\dfrac{2(x+5)}{x^2+3}\cdot\dfrac{3-10x-x^2}{(x^2+3)^2}

\implies\dfrac{\mathrm df}{\mathrm dx}=\dfrac{2(x+5)(3-10x-x^2)}{(x^2+3)^3}

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ANd you can’t heck from any given point and just try all the positivw possibilties.

Or you can just use rise over run, start form a point, rise which means go up until you reach the second layer of the line and just run (go To The right) until you hit a point.

You’ll get 1/4

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