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gogolik [260]
3 years ago
10

In Example 3, the track has 6 lanes that are each 1 meter in width. a. What is the outer perimeter of the track? Round your answ

er to the nearest meter. The outer perimeter is about meters.
Mathematics
1 answer:
son4ous [18]3 years ago
5 0

Answer:

Incomplete question, check attachment for the necessary diagram

Step-by-step explanation:

Note in the attachment,

We have two identical straight line of lenght

L1 = L2 = 84.39m

We also have two identical semicircle or radius 36.5m to the first track lane

But this is not the radius of the circle, the radius of the circle will now be 36.5 plus the 6 track lane and we are told that one track lane is 1m, then, the track lane is 6m

So, radius = 36.5+6

r = 42.5m

Then, we need to calculate the perimeter of the semicircle using the formula of perimeter of a circle and dividing by2

P = 2πr/2

P =πr

P = 22/7 × 42.5

P = 133.57 m

Then, the arc 1 is equal to arc 2 which is equal to 133.57 m

A1 = A2 = 133.57 m

Now we have all the dimensions,

Then, the perimeter can be calculated by adding the length of the sides

The perimeter of the field = Lenght of the two straight lines plus the length of the two semicircle arc

P = L1 + L2 + A1 + A2

P = 84.39 + 84.39 + 133.57 + 133.57

P =435.923 m

So, to the nearest meter

P ≈ 436m

The perimeter of the track is 436m

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3 years ago
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Marina CMI [18]

Answer:

\frac{-13}{15} x-52

Step-by-step explanation:

\frac{-5}{6}x-\frac{7}{30}x+\frac{1}{5}x-52

=\frac{-5}{6}x+\frac{-7}{30}x+\frac{1}{5}x+-52

Combine Like Terms:

=\frac{-5}{6}x+\frac{-7}{30}x+\frac{1}{5}x+-52

=(\frac{-5}{6}x+\frac{-7}{30}x+\frac{1}{5}x)+(-52)

=\frac{-13}{15}x+-52

Therefore, \frac{-13}{15}x+-52 will be your final answer.

7 0
2 years ago
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12345 [234]
7x-2(x+3)
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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:

<u />\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }

General Formulas and Concepts:

<u>Calculus</u>

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]:

  1. f(x) = cxⁿ
  2. 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:

<u>Step 1: Define</u>

<em>Identify given limit</em>.

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

<u>Step 2: Find Limit</u>

Let's start out by <em>directly</em> evaluating the limit:

  1. [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}
  2. 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:

  1. [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}
  2. [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}
  3. [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}
  4. Evaluate:
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∴ we have <em>evaluated</em> the given limit.

___

Learn more about limits: brainly.com/question/27807253

Learn more about Calculus: brainly.com/question/27805589

___

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

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Answer:

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Step-by-step explanation:

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