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3 years ago

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A triangle has vertices at r(1,1), s(-2,-4) and t(-3,-3). The triangle is transformed according to the rule r0,270.what are the

coordinates of s

1 answer:

Blababa [14]3 years ago

3 0

It is on he other side of the #

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Complete each statement. An event with a probability of 0 is An event with a probability of 1 is

BaLLatris [955]

Answer:

An event with a probability of 0 is impossible.

An event with a probability of 1 is certain.

Step-by-step explanation:

Probability is typically expressed in terms of a fraction between 0 and 1 where the denominator is the total number of outcomes and the numerator is the number of desired outcomes. Since probability is expressed as a fraction, if the probability is 0, that means it is impossible, or there is no chance that the event can happen. However, if the probability is 1, that means that the event is certain to happen and the odds are completely in your favor that the event will happen.

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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]:

f(x) = cxⁿ
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:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
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:

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

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

Unit: Limits

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2 years ago

Find the equation of a line through B(5,2) which is perpendicular to the line 3x+5=0. Hence find the coordinate of the foot of t

Oduvanchick [21]

Answer:

y=2, the equation of a line which is perpendicular to the line 3x+5=0

A(-5/3,2) the foot of the perpendicular from B to the line

Step-by-step explanation:

d1 : 3x+5=0, so 3x=-5, x=-5/3

y=2, the equation of a line which is perpendicular to the line 3x+5=0

A(-5/3,2) the foot of the perpendicular from B to the line

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3 years ago

A pizza parlor sold 38 pizzas during a dinner hour. If each pizza contained 8 slices, how many slices of pizza were sold?

AlladinOne [14]

38 x 8 = 304
So the answer is b

4 0

3 years ago

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For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity

NARA [144]

Answer: (a). at x = 0, its a removable discontinuity

and at x = 1, it is a jump discontinuity

(b). at x = -3, it is removable discontinuity

also at x = -2, it is an infinite discontinuity

(c). at x = 2, it is a jump discontinuity

Step-by-step explanation:

in this question, we would analyze the 3 options to determine which points gave us discontinuous in the category of discontinuity as jump, removable, infinite, etc.

(a). given that f(x) = x/x² -x

this shows a discontinuous function, because we can see that the denominator equals zero i.e.

x² - x = 0

x(x-1) = 0

where x = 0 or x = 1.

since x = 0 and x = 1, f(x) is a discontinuous function.

let us analyze the function once more we have that

f(x) = x/x²-x = x/x(x-1) = 1/x-1

from 1/x-1 we have that x = 1 which shows a Jump discontinuity

also x = 0, this also shows a removable discontinuity.

(b). we have that f(x) = x+3 / x² +5x + 6

we simplify as

f(x) = x + 3 / (x + 3)(x + 2)

where x = -3, and x = -2 shows it is discontinuous.

from f(x) = x + 3 / (x + 3)(x + 2) = 1/x+2

x = -3 is a removable discontinuity

also x = -2 is an infinite discontinuity

(c). given that f(x) = │x -2│/ x - 2

from basic knowledge in modulus of a function,

│x│= │x x ˃ 0 and at │-x x ∠ 0

therefore, │x - 2│= at │x - 2, x ˃ 0 and at │-(x - 2) x ∠ 2

so the function f(x) = at│ 1, x ˃ 2 and at │-1, x ∠ 2

∴ at x = 2 , the we have a Jump discontinuity.

NB. the figure uploaded below is a diagrammatic sketch of each of the function in the question.

cheers i hope this helps.

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3 years ago

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