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3 years ago

Which property would be used to solve the expression x + 12 = 47 ?

Mathematics

2 answers:

KIM [24]3 years ago

3 0

A. Inverse

<h2>Explanation:</h2>

Basically, inverse Property of Addition states that if you add any number added to its opposite the result will be equal zero. So, here we have the following equation:

$x + 12 = 47$

Adding -12 to both sides we get:

$x + 12 - 12= 47-12 \\ \\ \\ But \ 12+(-12)=0 \\ \\ \\ So: \\ \\ x + 0= 35 \\ \\ Finally: \\ \\ \boxed{x=35}$

<h2>Learn more:</h2>

Distributive property: brainly.com/question/12256474

#LearnWithBrainly

Gelneren [198K]3 years ago

3 0

Answer:

A. Inverse

Step-by-step explanation:

The given equation is $x+12=47$

To solve this equation, we use the inverse property of addition.

We add the additive inverse of 12 to both sides to get:

$x+12-12=47-12$

We then simplify to obtain:

$x+0=35$

This gives us x=35.

The correct answer is A

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Answer:

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Step-by-step explanation:

8x+5=35

Subtract 5 from the left side to cancel it out, do the same thing to 35.

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Divide 8x by 8 to get x alone, do the same thing to 30.

x=3.75, or once you simplify 30/8, you get 15/4.

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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:

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

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