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Rus_ich [418]
3 years ago
11

I need help asap!! please!!

Mathematics
2 answers:
Triss [41]3 years ago
6 0
H> 75.5

use the third symbol ⇒ (open) start at 75.5 (right in the middle of 75 and 76) and draw it to the right.

andre [41]3 years ago
4 0
Wouldnt it be a dot on the line right in between 75 and 76 pointing towards the left? im not sure 
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vitfil [10]

Answer:

repost it, I cant see all of the question

3 0
3 years ago
If a university wants to maintain a 14:1 ratio between students and teachers, how many teachers would be needed to accommodate 8
kolezko [41]
The answer is C. 64 teachers. 

14 x 64= 896 
so 1 x 64= 64
4 0
3 years ago
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What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?
scoray [572]

Answer:

<u />\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }

General Formulas and Concepts:

<u>Calculus</u>

Limits

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

Special Limit Rule [L’Hopital’s Rule]:
\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}

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)]
Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. 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:

<u>Step 1: Define</u>

<em>Identify given limit</em>.

\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}

<u>Step 2: Find Limit</u>

Let's start out by <em>directly</em> evaluating the limit:

  1. [Limit] Apply Limit Rule [Variable Direct Substitution]:
    \displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}
  2. Evaluate:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

  1. [Limit] Apply Limit Rule [L' Hopital's Rule]:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}
  2. [Limit] Differentiate [Derivative Rules and Properties]:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}
  3. [Limit] Apply Limit Rule [Variable Direct Substitution]:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}
  4. Evaluate:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}

∴ we have <em>evaluated</em> the given limit.

___

Learn more about limits: brainly.com/question/27807253

Learn more about Calculus: brainly.com/question/27805589

___

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

Unit: Limits

3 0
2 years ago
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ZanzabumX [31]
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If you were to add them together(.77+.25) it would be $1.02
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3 years ago
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Lera25 [3.4K]
x^{2} is equal to 2x, since it is the derivative of x^{2}.
3 0
3 years ago
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