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alina1380 [7]
3 years ago
15

Harini has twice as many sweet as Dora. After eating 18 sweets each she has 5 times as much as Dora. How many sweets did Harini

start with?
Mathematics
1 answer:
Nezavi [6.7K]3 years ago
6 0

9514 1404 393

Answer:

  48

Step-by-step explanation:

Let d represent the number of sweets Dora has to start with. Then 2d is the number Harini had. After they each ate 18, the ratio of their numbers is ...

  (2d -18)/(d -18) = 5/1

  2d -18 = 5(d -18) . . . . . multiply by d-18

  72 = 3d . . . . . . . . . . subtract 2d-90

  24 = d . . . . . . . . . divide by 3

  2d = 48

Harini started with 48 sweets.

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Answer:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]:                                                                     \displaystyle \lim_{x \to c} x = c

L'Hopital's Rule

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

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  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                    \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

We are given the limit:

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

When we directly plug in <em>x</em> = 0, we see that we would have an indeterminate form:

\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:

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

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}

Substitute in <em>x</em> = 0 once more:

\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

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