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mihalych1998 [28]
2 years ago
7

Five-sixths of the boats in the marina are white, 2 5 of the remaining boats are blue, and the rest are red. If there are 12 red

boats, how many boats are in the marina?
Mathematics
1 answer:
Inga [223]2 years ago
4 0

Answer: There are 120 boats in the marina.

Step-by-step explanation:

Let's suppose that there are N boats

We know that 5/6 of them are white, then we have: (5/6)*N white boats.

The remaining is: N - (5/6)*N = (1/6)*N

We know that 2/5 of the remaining are blue, then we have:

(2/5)*(1/6)*N blue boats

And the rest are red, the rest is:

(1/6)*N - (2/5)*(1/6)*N = (3/5)*(1/6)*N

Then we have (3/5)*(1/6)*N red boats.

And we know that there are 12 red boats, then:

(3/5)*(1/6)*N = 12

Now we can solve this for N, which is the total number of boats in the marina.

N = 12*(6)*(5/3) = 120

There are 120 boats in the marina.

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10x + 4 = 5x - 4 (Clean up)

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x = -0.8

<CAD = 5x - 4

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3 years ago
A community theater uses the function p (d) = (-4d+40) (d−40) to model the profit (in dollars) expected in a weekend when the ti
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The theater make the maximum profit at d = $25. Then the maximum profit of the theatre is $ 900.

<h3>What is differentiation?</h3>

The rate of change of a function with respect to the variable is called differentiation. It may be increasing or decreasing.

A community theater uses the function P(d) = (− 4d + 40) (d − 40) to model the profit (in dollars) expected on a weekend when the tickets to a comedy show are priced at d dollars each.

Then the maximum profit of the theatre will be

The function is P(d) = (− 4d + 40) (d − 40)

Differentiate the function with respect to d and put it equal to zero for maximum or minimum profit.

\begin{aligned} \dfrac{\mathrm{d} }{\mathrm{d} d}P(d) &= 0\\\\\dfrac{\mathrm{d} }{\mathrm{d} d}(- 4d + 40) (d - 40) &= 0\\\\(-4d+40) -4 (d-40) &= 0\\\\-8d + 200 &= 0\\\\d &= 25 \end{aligned}

Then the checking for maximum or minimum, again differentiate, we have

\begin{aligned} \dfrac{\mathrm{d} }{\mathrm{d} d}P(d) &= \dfrac{\mathrm{d} }{\mathrm{d} d}(- 4d + 40) (d - 40) \\\\\dfrac{\mathrm{d} }{\mathrm{d} d}P(d) &= \dfrac{\mathrm{d} }{\mathrm{d} d}(-8d + 200) \\\\\dfrac{\mathrm{d} }{\mathrm{d} d}P(d) &= -8\\\\ \dfrac{\mathrm{d} }{\mathrm{d} d}P(d) & < 0\end{aligned}

The value is less than zero hence maximum value will occur at d = 25.

Then maximum profit will be

P(d) = (− 4×25 + 40) (25 − 40)

P(d) = (− 100 + 40) (−15)

P(d) = (− 60) (− 15)

P(d) = $ 900

More about the differentiation link is given below.

brainly.com/question/24062595

#SPJ1

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