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ivann1987 [24]
3 years ago
14

K/6.6 + 44.1 = 21.4 The solution is k =

Mathematics
1 answer:
kakasveta [241]3 years ago
6 0

the interior angles of a triangle when added together will equal 180 degrees

which means 44.1 +47.6 +x = 180

can you find x?

Answer:  m∡K = 88.3° .

__________________________________________

m∡K = 180 − (44.1 + 47.6)

         = 180 − 91.7 = 88.3 .

___________________________________

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Plzz help me im going to most more like this because im not understanding it
lisov135 [29]
Hi there!

In order to solve, you can use substitution. This means that you use one equation and solve for one variable, then use that one equation and plug it into the other equation. Here's how we'd do it:

WORK:
x = 12 - y (since x is already solved for, we'll use that to plug into the other given equation.
2(12 - y) + 3y = 29 (using substitution)
24 - 2y + 3y = 29
24 + y = 29
y = 5
Plug the value of y back into the first equation
x = 12 - 5
x = 7

ANSWER:
A - x = 7, y = 5

It seems as though you're having some trouble with these questions, PLEASE MESSAGE ME!! I'd be more than happy to help you work through as many problems as you need until you get the hang of it (plus, it doesn't cost you any points!)

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!
5 0
3 years ago
Slope: -5, y-intercept:3
Sati [7]
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What is the period of the parent cosine function, y = cos(x)? degrees What is the period of the cosine function shown in the gra
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2 years ago
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katrin2010 [14]

Answer:

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Step-by-step explanation:

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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
2 years ago
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