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3 years ago

Simplify. -9c-(5-3c)

Mathematics

1 answer:

Sidana [21]3 years ago

7 0

Answer:

-12c-5

Step-by-step explanation:

Step 1- Distribute into the parenthesis(Since there's nothing next to the parenthesis to multiply, there is a 1).

-9c-1(5-3c)

-9c-5(1)-3(1)c

Step 2- Multiply,

-9c-5-3c

Step 3- Add common variables.

(-9c-3c)-5

-12c-5

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Hi i need help with this

Illusion [34]

Answer:

2 (n + 2) = 2n + 4

n + n + n + n = 4n

2 (3n) = 6n

4n + n = 5n

Step-by-step explanation:

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3 years ago

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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:

Calculus

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 x = 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 x = 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 x = 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

Find the range of the function for the fiven domain. f(x)= 3x-8; {-2, -1, 0, 1, 2}

ryzh [129]

f(x)= 3x-8

x = -2, f(x)= 3(-2) -8 = -6 - 8 = -14
x = -1, f(x)= 3(-1) -8 = -3 - 8 = -11
x = 0, f(x)= 3(0) -8 = 0 - 8 = -8
x = 1, f(x)= 3(1) -8 = 3 - 8 = -5
x = 2, f(x)= 3(2) -8 = 6 - 8 = -2
Domain {-2, -1, 0, 1, 2}
Range: {-14, -11, -8, -5, -2}

answer
Range: {-14, -11, -8, -5, -2}

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4 years ago

Double checking my answer for this equation

TEA [102]

4a^2-2ab-5b^2 :) that is simplified

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den301095 [7]

Answer:

1/4