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Masteriza [31]
2 years ago
13

A baseball measures approximately 74 mm in diameter. What is the volume of a baseball? (Round your answer to the nearest tenth i

f needed)
Mathematics
2 answers:
Inga [223]2 years ago
8 0

Solution:

<u>We know that:</u>

V_{Baseball}  = \frac{4}{3} \pi r^{3} \\ \\ Diameter = 74 \space\ mm\\\\Radius = \frac{Diameter}{2}

<u>Finding the area of the baseball:</u>

V_{Baseball}  = (\frac{4}{3})( \pi )(r^{3})

V_{Baseball}  = [\frac{4}{3}][ 3.14 ][(\frac{74}{2}) ^{3}]

V_{Baseball}  = [\frac{4}{3}][ 3.14 ][(37) ^{3}]

V_{Baseball}  = [\frac{4}{3}][ 3.14 ][50653}]

V_{Baseball}  = 212067.227 \space\ mm^{3}   \space\ (Using\ calculator)

<u>Rounding the volume to the nearest tenth:</u>

V_{Baseball}  = 212067.227 \space\ mm^{3}  = 212067.2 \space\ mm^{3}

Thus, 212067.2 mm³ is the volume of the baseball.

lutik1710 [3]2 years ago
5 0

Answer:

As Per Given Information

Diameter of of baseball = 74 mm

We've been asked to find the volume of baseball .

As we know

Radius = Diameter/2

Radius = 74/2

Radius = 37 mm ( 1 mm = 0.1 cm)

Radius = 37/10 cm

Radius = 3.7 cm

Now let's calculate the volume of baseball

volume of baseball = 4/3 πr³

Put the given value we obtain

→ volume of baseball = 4/3 × 3.14 × (3.7)³

→ volume of baseball = 4/3 × 3.14 × 50.653

→ volume of baseball = 4/3 × 159.05042

→ volume of baseball = 636.20168/3

→ volume of baseball = 212.06

→ volume of baseball = 212 cm³ ( approx)

So, the volume of baseball is 212 cm³.

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What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?
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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)}

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\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:
    \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 <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

3 0
2 years ago
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