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telo118 [61]
3 years ago
11

PLEASE HELP WILL GIVE GOOD RATINGS AND 15 POINTS!! I will give BRAINLIEST

Mathematics
1 answer:
insens350 [35]3 years ago
3 0

Answer:Really koltan

-Eliezer Ludan

Step-by-step explanation:

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Graph ∆KLM with vertices K(2,2), L(0,- 1), and M(-3,2). Graph ∆K'L'M' with a scale factor of 2. What are the coordinates of the
olasank [31]

Answer:

The coordinates of the new vertices are:

K'(4,4), L'(0,- 2), and M'(-6,4).

Step-by-step explanation:

In order to dilate a shape by 2, you can simply multiply the coordinates of the orignal triangle each by 2 to find the new coordinates of the new vertices.

K(2,2), L(0,- 1), and M(-3,2)

K'(2*2,2*2), L'(0*2,- 1*2), and M'(-3*2,2*2).

Now become...

K'(4,4), L'(0,- 2), and M'(-6,4).

You can now graph ∆K'L'M' using these coordinates.

5 0
2 years ago
50 POINTS!
padilas [110]
To double the principle the formula is
2p=p e^rt
2=e^rt
2=e^0.12t
Solve for t
T=(log(2)÷log(e))÷0.12
T=5.78 years
5 0
3 years ago
Round 2. 873 to the nearest tenth
PtichkaEL [24]
2.873 ≈2.9 (nearest tenth)
6 0
3 years ago
Read 2 more answers
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
WILL MARK BRIANLIEST!
gayaneshka [121]

Answer:

(4, - 2 )

Step-by-step explanation:

Given the 2 equations

x + y = 2 → (1)

x - y = 6 → (2)

Adding the 2 equations term by term will eliminate the term in y, that is

(x + x) + (y - y) = (2 + 6)

2x = 8 ( divide both sides by 2 )

x = 4

Substitute x = 4 in either of the 2 equations and solve for y

Substituting in (1)

4 + y = 2 ( subtract 4 from both sides )

y = - 2

Solution is (4, - 2 )

7 0
3 years ago
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