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kondor19780726 [428]
3 years ago
9

Gabriella said a triangle with an angle of 98 degrees was an obtuse angle. Cynthia said a triangle with an angle of 98 degrees w

as an acute triangle. Who is right and why?
Mathematics
1 answer:
antoniya [11.8K]3 years ago
7 0
An obtuse triangle has an angle that is greater than 90 degrees

so if we see gabby's statement, ,98>90 so tecnically, she is correct

cynthya's statement is false because 98>90 so it is an obtuse angle
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4 (x-9) = -48 x= please help me <br>​
valentina_108 [34]

distribute the 4:

4x-36=-48x

subtract 4x

-36=-52x

divide by -52

x=1/2

7 0
3 years ago
Please someone help?
Sati [7]

Answer:

C

Step-by-step explanation:

C is congruent to A

3 0
2 years ago
Given:
matrenka [14]

Answer:

A. If angles XYZ and RST are congruent, then they are vertical angles.

Step-by-step explanation:

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4 0
3 years ago
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
1 year ago
13 + 11s = -15 + 8s − 20
koban [17]

<em>Answer,</em>

<em><u>S = -16</u></em>

<em>Explanation,</em>

<em><u>Step 1: Simplify both sides of the equation.</u></em>

<em>13 + 11s = − 15 + 8s − 20</em>

<em>13 + 11s = − 15 + 8s + − 20</em>

<em>11s + 13 = (8s) + (</em><em><u>− 15</u></em><em> + </em><em><u>− 20</u></em><em>) </em><em>(Combine Like Terms)</em>

<em>11s + 13 = 8s + − 35</em>

<em>11s + 13 = 8s − 35</em>

<em><u>Step 2: Subtract 8s from both sides.</u></em>

<em>11s + 13 − 8s = 8s − 35 − 8s</em>

<em>3s + 13 = − 35</em>

<em>Step 3: Subtract 13 from both sides.</em>

<em>3s + 13 − 13 = − 35 − 13</em>

<em>3s = − 48</em>

<em><u>Step 4: Divide both sides by 3.</u></em>

<em>3s/3 = −48/3</em>

<em>s = -16</em>


<u><em>Hope this helps :-)</em></u>


3 0
3 years ago
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