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sattari [20]
3 years ago
6

Sam used 666 loaves of elf bread on an 888 day hiking trip. He wants to know how many loaves of elf bread (b)(b)left parenthesis

, b, right parenthesis he should pack for a 121212 day hiking trip if he eats the same amount of bread each day.
How many loaves of elf bread should Sam pack for a 121212 day trip?
Mathematics
2 answers:
ipn [44]3 years ago
7 0

Answer:

90909 pieces of elf bread

Step-by-step explanation:

666 divided by 888 is .75

.75 times 121212 is 90909

rusak2 [61]3 years ago
3 0

Answer:

90909 loaves

Step-by-step explanation:

If he used 666 loaves of elf bread in 888 days

That’s 666 loaves = 888 day

How many loaves shouldn’t he pack for 121212 day trip

That’s x loaves = 121212 day

Combine both equations togather

That’s

666 loaves = 888 day

X loaves = 121212 day

Cross multiply

X x 888 = 121212 x 666

X x 888 = 80727192

Divide both sides by 888

X = 80727192/888

X = 90909

Sam will need 90909 Loaves of elf bread for a 121212 day trip

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3. A dish contains 100 candies. Juan removes candies from the dish each day and no
iVinArrow [24]

Answer:

11

Step-by-step explanation:

If i am correct n= the days so with the days adding up day 10 juan would have 55 of the 100 candies, day 11 he would have 66. The statement said at least 64, so it would have to be 64 or the closet amount to 64 but it cannot be lower than 64. So there for 66 is the closest to the given amount, and in order for juan to get 64 he would need to take candies out for 11 days.

7 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:
    \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
Which equation represents a circle that contains the point (-5, -3) and has a center at (-2, 1)?
Evgen [1.6K]

The equation (x + 2)² + (y - 1)² = 25 represents a circle that contain the point (-5 , -3) and has a center at (-2 , 1)

Step-by-step explanation:

The equation of a circle of center (h , k) and radius r is:

(x - h)² + (y - k)² = r²

The given is:

  • The center of the circle is (-2 , 1)
  • The circle passes through point (-5 , -3)

The length of the radius is the distance from the center of the circle

to a point on the circle

∵ The formula of the distance is d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}

∵ The center of the circle is (-2 , 1)

∵ The circle passes through point (-5 , -3)

∴ r=\sqrt{[(-5)-(-2)]^{2}+[(-3)-(1)]^{2}}

∴ r=\sqrt{[-3]^{2}+[-4]^{2}}

∴ r=\sqrt{9+16}

∴ r=\sqrt{25}

∴ r = 5

∵ The equation of the circle is (x - h)² + (y - k)² = r²

∵ The center of the circle is (-2 , 1)

∴ h = -2 and k = 1

∵ r = 5

∴ r² = (5)² = 25

- Substitute the values of h , k , r² in the equation of the circle

∴ (x - -2)² + (y - 1)² = 25

∴ (x + 2)² + (y - 1)² = 25

The equation (x + 2)² + (y - 1)² = 25 represents a circle that contains

the point (-5 , -3) and has a center at (-2 , 1)

Learn more:

You can learn more about the equation of the circle in brainly.com/question/9510228

LearnwithBrainly

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