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Alexeev081 [22]
3 years ago
15

You randomly select a marble from a bag.

Mathematics
1 answer:
AysviL [449]3 years ago
3 0

Answer:

b is the answer

Step-by-step explanation:

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12.00$ change

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7/8, 3/5, 0.75, 0.65 ordering from least to greatest
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Least 3/5, 0.65, 0.75, 7/8 greatest
7 0
2 years ago
PLEASE HELP! I’m so confused. Explain your answer, thanks!
kakasveta [241]

Answer:

y+2 = 3/4 (x-1)

Step-by-step explanation:

We have a slope and a point, we can use the point slope form of the equation for a line

y-y1 = m(x-x1)

where m is the slope and (x1,y1) is the point

y--2 = 3/4(x-1)

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So the equation 12 y equals 132 the equation 12 y equals 132
OlgaM077 [116]
12y = 132

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4 0
2 years ago
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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