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2 years ago

Which of these descriptions matches the graph?

Mathematics

1 answer:

o-na [289]2 years ago

3 0

Answer:

If you had included the graph, maybe someone could answer.

Step-by-step explanation:

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Please help, any help would be appreciated.

k0ka [10]

Answer:

160,170 and 240,250

Step-by-step explanation:

sorry if wrong

3 0

2 years ago

The length of the rectangular tennis court at wimbledon is 6 feet longer than twice the width. if the court's perimeter is 240 f

KATRIN_1 [288]

Length (L): 2w + 6

width (w): w

Perimeter (P) = 2L + 2w

240 = 2(2w + 6) + 2(w)

240 = 4w + 12 + 2w

240 = 6w + 12

228 = 6w

38 = w

Length (L): 2w + 6 = 2(38) + 6 = 76 + 6 = 82

Answer: width = 38 ft, length = 82 ft

8 0

2 years ago

Factor and add steps please 81u^6+192

attashe74 [19]

ANSWER : (9u - 8)2

STEPS:

Step-1 : Multiply the coefficient of the first term by the constant 81 • 64 = 5184

Step-2 : Find two factors of 5184 whose sum equals the coefficient of the middle term, which is -144 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -72 and -72
81u2 - 72u - 72u - 64

Step-4 : Add up the first 2 terms, pulling out like factors :
9u • (9u-8)

Step-5 : Add up the four terms of step 4 :
(9u-8) • (9u-8)
Which is the desired factorization

5 0

1 year ago

The train ride at the zoo covers a distance of 2 1/2 miles in 1/3 of an hour. How many miles per hour does the train go? How far

Gala2k [10]

7.5 mph in 3 hours the train can travel 22.5 miles. Take 2.5 and times it by 3 to get the mph, then multiply that answer by 3 again to get the total miles in 3 hours

6 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:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$

∴ we have evaluated the given limit.

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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