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3 years ago

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A bag contains white marbles and red marbles, 22 in total. The number of white marbles is 2 more than 3 times the number of red

marbles. How many white marbles are there?

1 answer:

koban [17]3 years ago

8 0

the answer is 47 marblez

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Orly uses 2 cups of raisins for every 8 cups of trail mix she makes. How many cups of trail mix will she make is she uses 12 cup

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You have a 2:8 ratio, or 1:4. Since you have 12 cups of raisins, you're going to get the answer 48.

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3 years ago

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Compute 41, 42, 43, 44, 45, 46, 47, and 48. Make a conjecture about the units digit of 4n where n is a positive integer. Use str

padilas [110]

Answer:

The conjecture is that the unit digit of 4^n = 4 when n = odd also 4^n = 6 when n = even

Step-by-step explanation:

$4^1 = 4\\4^2 = 16\\4^3 = 64\\4^4 = 256 \\4^5 = 1024\\4^6 = 4096\\4^7 = 16384\\4^8 = 65536$

The conjecture is that the unit digit of 4^n = 4 when n = odd also 4^n = 6 when n = even

To prove this conjecture

$4^1 = 4\\4^2 = 16$ unit digit = 6

hence the property is true for ; n = 1 and n = 2 and also for every odd and even number ( i.e. from 1 to 8 )

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3 years ago

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The answer is false

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3 years ago

Find the missing side.<br> 17 m<br> 15 m

Sonja [21]

Is there a image supposed to be attached?

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<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

3 years ago