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ipn [44]
2 years ago
11

Write the sentence as an equation. 62 more than the product of 150 and r is 40

Mathematics
1 answer:
Elenna [48]2 years ago
6 0

Answer:

150 • r + 62 = 40

Step-by-step explanation:

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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
Simplify the expression<br> 26 · 268
Gre4nikov [31]

Answer:

26 * 26^8 = 5.4295037e+12.

Step-by-step explanation:

26^8 = 208,827,064,576.\\26 * 208,827,064,576 = 5.42950373e+12.

hope this helps.

6 0
3 years ago
suppose you have 8 pairs of pants, 8 shirts and 7 pairs of shoes how many different outfits can you wear assuming you take one o
Tom [10]

Answer:

Step-by-step explanation:

8(8)7=448 different outfit combinations

4 0
3 years ago
Is the following a proportion? Yes or no? 6/9 = 14/21
tresset_1 [31]
I think it's not, because for proportion, you use the given clues to solve for a value or an unknown number.
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4 0
3 years ago
What is 5/12 Plus 5/12?
mestny [16]
\sf\dfrac{5}{12}+\dfrac{5}{12}

Since we already have like denominators, just add the numerators and keep the denominator.

\sf\dfrac{5+5}{12}\rightarrow\dfrac{10}{12}

Simplify:

\boxed{\sf\dfrac{5}{6}}
3 0
3 years ago
Read 2 more answers
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