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3 years ago

The perimiter of a rectangle is 90 the length is 27 what is the width

Mathematics

2 answers:

nikdorinn [45]3 years ago

8 0

Answer:

Step-by-step explanation:

Aleks04 [339]3 years ago

5 0

Answer: Your answer is 18 have a good day

Step-by-step explanation: Can I get brainliest

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Mrs.Fielder decides to build a small snow shelter for her children to wait in before the school bus arrives in the morning she h

sergejj [24]

Answer:

The answer is below

Step-by-step explanation:

Mrs. Fielder decides to build a small snow shelter for her children to wait in before the school bus arrives in the morning. She has only enough wood for a total perimeter of 20 feet.

a. Make a table of all the whole number possibilities for the length and width of the shelter. Find the area of each shelter.

b. What dimensions should Mrs. Fielder choose to have the greatest area in her shelter?

c. What dimensions should Mrs. Fielder choose to have the least area in her shelter?

d. Township building codes require 3 square feet for each child in a snow shelter. Which shelter from part (a) will fit the most children? How many children is this? Explain your reasoning.

Solution:

a) Let W represent the width of the school shelter and let L represent the length of the school shelter. Therefore:

Perimeter of the school shelter = 2(length + breadth)

20 = 2(L + W)

L + W = 10

Also, the area of the school shelter = L * W

Length (ft) Width (ft) Area(ft²) = length * width

1 9 9

2 8 16

3 7 21

4 6 24

5 5 25

b) The shelter with a length of 5 ft and width of 5 ft has the largest area.

c) The shelter with a length of 1 ft and width of 9 ft has the least area.

d) The 4 by 6 ft shelter can hold 8 children (24 ft² / 3 ft² = 8) and the 5 by 5 ft shelter can hold 8 children with an extra space (25 ft² / 3 ft² = 8.33).

8 0

2 years ago

Sahra is three times as old as her daughter. If the sum of their age is 32. what is the age of the daughter?

Andrews [41]

So Sahra is 3 times as old as her daugter. That means that the sum of their age is the daughters age times 4.
We can prove that by saying the daughters age is x. Sahra's age must be 3x. Their sum must be x+3x = 4x.
So to get the daughters age we divide the sum by 4, which is 8.

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3 years ago

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Daniel wants to order shirts on the internet, shirts cost $20.00 and shipping on all oders is $10.99. Daniel has $50 to spend

MatroZZZ [7]

He spent $30.99 and has $10.01 left

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2 years ago

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The dimensions in inches of a shipping box at we ship 4 you can be expressed as width x, length x+5, and height 3x-1. The volume

MakcuM [25]

Answer:

Connexus: C,D,C,A,A.

Step-by-step explanation:

6 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.

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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