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nikdorinn [45]
3 years ago
10

There are 16 entrees available at a restaurant. From these, Archie is to choose 6 for his party. How many groups of 6 entrees ca

n he choose, assuming that the order of the entrees chosen does not matter? )If necessary, consult a list of formulas.)
Mathematics
1 answer:
Vikki [24]3 years ago
3 0

Answer:

8008

Step-by-step explanation:

This is the selection of 6 elements out of 16 elements. Applying the combination formula, the total number of selections is nCr with n =16 and r =6.

16C6 = 8008

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Consider the set {4, 5, 7, 7, 8, 8, 12, 13, 13, 15, 18, 20, 22, 24, 26, 27, 27, 37, 43}. What would be an appropriate interval t
adelina 88 [10]
Proper interval would be 5
8 0
3 years ago
It takes a hose 2 minutes to fill a rectangular aquarium 8 inches long, 9 inches wide, and 13 inches tall. How long will it take
nirvana33 [79]
To make it easier, you calculate the volume of the first aquarium.

1st aquarium:
V = L x W x H
V = 8 x 9 x 13
V = 72 x 13
V = 936 in.
Rate: 936 in./2 min.

Now that you've got the volume and rate of the first aquarium, you can find how many inches of the aquarium is filled within a minute, which is also known as the unit rate. To do that, you have to divide both the numerator and denominator by their least common multiple, which is 2. 936 divided by 2 is 468 and 2 divided by 2 is 1.
So the unit rate is 468 in./1 min. Now that you've got the unit rate, you can find out how long it'll take to fill the second aquarium up by finding its volume first.

2nd aquarium:
V = L x W x H
V = 21 x 29 x 30
V = 609 x 30
V 18,270 inches

Calculations:
Now, you divide 18,270 by 468 to find how many minutes it will take to fill up the second aquarium. 18,270 divided by 468 is about 39 (the answer wasn't exact, so I said "about").
 
2nd aquarium's rate:
18,270 in./39 min.

As a result, it'll take about 39 minutes to fill up an aquarium measuring 21 inches by 29 inches by 30 inches using the same hose. I really hope I helped and that you understood my explanation! :) If I didn't, I'm sorry. I tried. :(
7 0
3 years ago
Jean has $280 in her savings account. Starting next week, she will deposit $30 in her account every week.
Sophie [7]
It’s not proportional
8 0
2 years ago
<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7
salantis [7]

Answer:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]:                                                                     \displaystyle \lim_{x \to c} x = c

L'Hopital's Rule

Differentiation

  • Derivatives
  • Derivative Notation

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:

We are given the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}

When we directly plug in <em>x</em> = 0, we see that we would have an indeterminate form:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle  \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}

Plugging in <em>x</em> = 0 again, we would get:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}

Substitute in <em>x</em> = 0 once more:

\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}

And we have our final answer.

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

Unit: Limits

6 0
3 years ago
Which expressions are equivalent to 647 x 39
stellarik [79]
25233 is your answer for this problem
8 0
3 years ago
Read 2 more answers
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