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IgorLugansk [536]

3 years ago

13

50 POINTS PLEASE HELP FAST DUE TODAY: Explain why any of the four operations (+ - × ÷) placed between the two terms 5 and -3√8 w

ill result in an irrational number.

1 answer:

marin [14]3 years ago

7 0

Answer:

Because -3-/8 itself is an irrational number. No matter the operation it wil always be irrational

Step-by-step explanation:

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If an endpoint is included—shown by a closed dot, such as in the graph to the right—use = instead of ≤ or ≥ to indicate the numb

kodGreya [7K]

Answer:

<4

Step-by-step explanation:

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8 0

2 years ago

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A girl can run 6 2/3 metres in one second. How long will it take her to run 100 metres?

Crank

Answer:

15 seconds

Step-by-step explanation:

How did i get this answer? It is apparent that the girl is capable of running 6 2/3 metres per one second and she must run 100 metres.

First, let's convert 2/3 into a decimal so that it is easier to calculate later. 2/3=0.667 (you can also just do 2 divide by 3 and will end up with the same number)

Now our numbers are 6.667 and 100. Let's divide 100 by 6.667 which estimates to 15

7 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

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:

f(x) = cxⁿ
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

2 years ago

Khan academyyyy ⠀⠀⠀⠀⠀⠀⠀⠀

matrenka [14]

Answer:

option A, C , E

Step-by-step explanation:

- 4 + ( 4 + 5 ) = ( - 4 + 4 ) + 5 [ associative property ]

5 0

3 years ago

Read 2 more answers

How much in cash is 41/50% of 28800.

guajiro [1.7K]

41/50%=82
28800-82= 28718.
That may be the answer.

3 0

2 years ago