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2 years ago

Please help if u get all 3 right ill put u as brainiest!

Mathematics

1 answer:

scoray [572]2 years ago

7 0

Answer: 13

Step-by-step explanation: Alright with the median you take all the numbers "11, 20, 17, 8, 8, 9, 20, 13, 21" and put them in numerical order. "8, 8, 9, 11, 13, 17, 20, 20, 21" Then the middle one is the median, since there are 9 numbers you take the fifth one. In this case it is 13. (Only knew the first one since I have no idea what they mean by quartiles)

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Point B is on the graph of the function f(x) = -3x^2. If the x-coordinate of point B is 2, which ordered pair gives the location

Oksana_A [137]

f(x) = -3x^2
insert x=2
y=-3*2^2
y=-3*4
y=-12
so your pair is (2,-12)

6 0

2 years ago

What -7 x4=36? amogus amogus amogus amogus

77julia77 [94]

Based on the given problem of -7x⁴ = 36, the solution to the equation is x = -1.5.

<h3>What is the solution to this problem?</h3>

The equation to find x is given as:

-7x⁴ = 36

Solving it gives:

-7x⁴ = 36

x⁴ = 36/-7

x⁴ = -36/7

x = ⁴√(-36/7)

x = -1.5

In conclusion, x is -1.5.

Find out more questions on solving for x at brainly.com/question/2910769

#SPJ1

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11 months ago

Montel collected 26Leaves for his science project he wants to put six leaves on a page how it shows how many pages Montello need

andrey2020 [161]

Answer:

4.3 pages

beacause 26 divided by 6 is 4.3

7 0

2 years ago

PLEASE HURRY HELP ME DUE IN 5 MIN

Mars2501 [29]

Answer:

Parallel I checked with a graph

3 0

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