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Ann [662]
3 years ago
7

...what are consecutive multiples ​

Mathematics
1 answer:
Korolek [52]3 years ago
5 0

Answer:

H

Step-by-step explanation:

Let represent A and B

A and B is the sum of the first 50 consecutive multiples of 3 and 6, this is the same as a arithmetic sequence because arithmetic sequences have a common sum.

We can represent this as

a _{n} = a {}^{1}  + d(n - 1)

where a^1 is the first term, d is the common sum, n is the number of multiples we adding.

The first term for both A and B respectively is 3 and 6.

Multiples implies that the Difference between A and B are 3 and 6 respectively. There are 50 consecutively. multiples.

Plug in the known parts for both.

3 + 3(49) = 150

6 + 6(49) = 300

We need to what percent of find 150% of 300.

B is twice as A, so this means that we need to multiply 2 by it original or 100% of it self.

2 \times 100 = 200

So 200% is the answer.

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Answer:

A is the answer.

Step-by-step explanation:

Transferring them into 2 decimals behind the point, we have 4.06, 4.59, 4.60, and 4.72. THE ONE THAT MATCHES THIS IS CHOICE A

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Find the volume of a right circular cone that has a height of 5.7 ft and a base with a
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56652.3 ft^3

Step-by-step explanation:

The formula for the volume of a right circular cone is V=13pi*r^2*h where r is the radius of the base.

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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:

  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

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