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3 years ago

Write the equations for the lines parallel and perpendicular to the given line j that passes through Q.

Mathematics

1 answer:

Anna007 [38]3 years ago

3 0

Answer:

the third option

i think this is right but I could be wrong, sorry.

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Which represents the polynomial written in standard form? 4m – 2m4 – 6m2 + 9 9 + 4m + 2m4 – 6m2 2m4 – 6m2 – 4m + 9 9 – 6m2 + 4m

Alenkasestr [34]

Answer:

$- 2m^4 - 6m^2 + 4m + 9$

Step-by-step explanation:

Given

$4m - 2m^4 - 6m^2 + 9$

Required

Express in standard form

To do this, we simply reorder the terms of the polynomial in descending order of power.

So, we have:

$- 2m^4 - 6m^2 + 4m + 9$

4 0

3 years ago

Read 2 more answers

..the cat and the hat to the pat in the rug on the hug

swat32

Answer:

uh get it i guess?

Step-by-step explanation:

yes

8 0

2 years ago

HELPPPPPP MEEEEE y=6x-1 and -2x-3y=-2 USING SUBSTITUTION

bazaltina [42]

Answer:
x = 4
y =23

work:
plug in the value of Y to -2x-3y=-2. -2x - 3(6x-1) = -2
multiply 3 and the numbers in parenthesis. -2x -18x - 3 =-2
combine like terms (-2x -18x). -20x - 3 = -2
subtract 3 from both sides to isolate the variable. -20x = -5
divide to isolate the variable (-20/-5) x = 4
-
now that we know the value of x, it'll be easy to find Y.
Plug it in to the first equation.
y=6(4)-1
y = 24 - 1
y = 23

so x = 4
and y = 23

hope this helps! :D

7 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:

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

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

∴ we have evaluated 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

3 0

1 year ago

There are three separate, equal-size boxes, and inside each box there are four separate small boxes, and inside each of the smal

katen-ka-za [31]

Answer:

There are 51 boxes all together.

Step-by-step explanation:

Since there are three separate, equal-size boxes, and inside each box there are four separate small boxes, and inside each of the small boxes there are three even smaller boxes, to determine how many boxes are there all together, the following calculation must be done:

3 + 3 x 4 + 3 x 4 x 3 = X

3 + 12 + 36 = X

51 = X

Therefore, there are 51 boxes all together.

6 0

2 years ago