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Setler [38]
3 years ago
13

Last one you guys!!!

Mathematics
2 answers:
Delicious77 [7]3 years ago
7 0

Answer:

53 degree

Step-by-step explanation:

180-127=53.......

liberstina [14]3 years ago
4 0

Answer:

53

Step-by-step explanation:

Hi,

When two lines add up to 180 degrees, they are said to be supplementary angles. Since we have one angle measurement, all we have to do is subtract it from 180.

180 - 127 = 53

That means the other angle measures 53 degrees.

Hope this helps :)

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Can you find the limits of this ​
Pavel [41]

Answer:

\displaystyle  \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{-3}{8}

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Constant]:                                                                                             \displaystyle \lim_{x \to c} b = b

Limit Rule [Variable Direct Substitution]:                                                             \displaystyle \lim_{x \to c} x = c

Limit Property [Addition/Subtraction]:                                                                   \displaystyle \lim_{x \to c} [f(x) \pm g(x)] =  \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)

L'Hopital's Rule

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Addition/Subtraction]:                                                         \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

We are given the following limit:

\displaystyle  \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16}

Let's substitute in <em>x</em> = -2 using the limit rule:

\displaystyle  \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{(-2)^3 + 8}{(-2)^4 - 16}

Evaluating this, we arrive at an indeterminate form:

\displaystyle  \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{0}{0}

Since we have an indeterminate form, let's use L'Hopital's Rule. Differentiate both the numerator and denominator respectively:

\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \lim_{x \to -2} \frac{3x^2}{4x^3}

Substitute in <em>x</em> = -2 using the limit rule:

\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{3(-2)^2}{4(-2)^3}

Evaluating this, we get:

\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{-3}{8}

And we have our answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit:  Limits

6 0
2 years ago
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