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almond37 [142]
3 years ago
6

A sporting goods store manager was selling a ski set for a certain price. The manager offered the markdowns shown below, making

the one-day sale price of the ski set. The ski set was 10% off and with an extra 30% off.
Mathematics
1 answer:
Akimi4 [234]3 years ago
8 0

The complete question is:

A sporting goods store manager was selling a ski set for a certain price. The manager offered the markdowns​ shown, making the​ one-day sale price of the ski set ​$325. Find the original selling price of the ski set. It was marked down 10% and 30%

Answer:

  • <u>original price = $ 515.87</u>

Explanation:

<u>1. Data:</u>

a) Final price of the ski set: $ 325

b) First mark down: 10%

c) Second mark down: 30%

d) Original selling price: ?

<u>2. Procedure:</u>

<u />

<u>First mark down</u>:

  • Final price = original price - first mark down - second mark down
  • First mark down = 10% × original price = 0.10 × original price
  • Price after first mark down = original price - 0.10 × original price = 0.90 × original price

<u>Second mark down</u>:

  • Second mark down = 30% × price after first mark down = 0.30 × price after first mark down = 0.30 × 0.90 × original price
  • Price after second mark down = price after first mark down - second mark down
  • Price after second mark down = 0.90 × original price - 0.30 × 0.90 × original price = (1 - 0.30) × 0.90 × original price = 0.70 × 0.90 × original price

<u>Final price</u>:

  • Final price = price after second mark down
  • $ 325 = 0.70 × 0.90 × original price

<u>Simplify and solve for the original price</u>:

  • $325 = 0.63 × original price
  • original price = $325/0.63 = $515.87

<u></u>

<u>Answer: original price = $ 515.87</u>

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\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}

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\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle  \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}

Plugging in <em>x</em> = 0 again, we would get:

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