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3 years ago

Write the equation of a line that is PERPENDICULAR to the line y=3x-4 and passes through the point (-6,1)

Mathematics

1 answer:

Verdich [7]3 years ago

4 0

Answer: y = -1/3x - 1

Step-by-step explanation:

Slope of perpendicular line is opposite reciprocal --> -1/3
Using point given (-6, 1), find b.
1 = -6(-1/3) + b
1 = 2 + b
-1 = b

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IF ANSWERED CORRECTLY WILL GIVE BRAINLIEST BUT IF YOU GIVE ME PDFS OR LINKS I WILL REPORT YOU

Illusion [34]

Answer:

a is the answer

Step-by-step explanation:

i got it right

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3 years ago

Read 2 more answers

Do the following side lengths form a triangle? 6, 18, 14

iren2701 [21]

Answer:

Yes

Step-by-step explanation:

Conditions for the sides to form a triangle is that sum of any two sides should be greater than the third side.

6 + 18 = 24

24 > 14

18 + 14 = 32

32 > 6

6 + 14 = 20

20 > 18

Therefore,

sides 6, 18 and 14 forms a triangle.

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3 years ago

Two of your friends, Matt and Karen, both run to you to settle a dispute. They were working on a math problem, and got different

Anna11 [10]

So in matt's equation, he made a mistake in the a transision from line 2 to line 3
in line 2: -4(-2)2
in line 3: -4(4)
the mistake is that -2 times 2 is not equal +4 it is equal to -4
also from lines 5 to 6 he made a mistake in order of opperations (mulit division then addition and subtract)
line 5: -10+30/5
line 6: 20/5

so he first subtracted 10 then divided, he should have divided then subtracted
so the equation should have equaled

Karen used the correct (-) times (+) property and the order of operations
so Karen is correct and Matt is wrong.

5 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:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

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:

Step 1: Define

Identify given limit.

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

Step 2: Find Limit

Let's start out by directly 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:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$