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lisabon 2012 [21]
3 years ago
13

3x + 4 213 or 2x -15 -11

Mathematics
1 answer:
ehidna [41]3 years ago
6 0

Answer:

743979734

Step-by-step explanation:

qdwwsffsfsc

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salantis [7]

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:

  1. f(x) = cxⁿ
  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)

Unit: Limits

6 0
3 years ago
I have no clue what the answer is please help me?
dedylja [7]

Answer:

C

Step-by-step explanation:

in the quadratic equation: ax^2+bx+c,

in order to factor when a=1, then we need to find values that add to b and multiply to c.

8 0
3 years ago
Read 2 more answers
PLZ HELP NO WRONG ANSWERS
Ray Of Light [21]

Answer :

to find area of the square = 6 to the power 2

to find the area of the rectangle = (5x + 12) * 5x

therefore,

square's area = 36

rectangle's area = 5x*5x + 12 * 5x

= 25x * 60x

Now we have to find the sum of both the areas,

so, 36 + 1500 = 1536

Step-by-step explanation:

look I know my answer is wrong but I just wanted points! thnx for the free points! :)))))))))

NANRI! VANAKKAM!

3 0
3 years ago
What is the area of the following shape​
Ad libitum [116K]

Answer:

30 ft^2

Step-by-step explanation:

Break the figure into smaller pieces:

5*2=10

10*2=20

Add the areas

20+10=30

The are of the figure is 30 ft^2

8 0
3 years ago
6 lesson 10 the value of x unit
denis-greek [22]

Answer:

what do you want me to help you

5 0
2 years ago
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