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Mandarinka [93]
3 years ago
13

An oarsman can row his boat 3 mph in still water. He sets out on the Illinois River, which flows at 5 mph. We are interested in

what an observer on shore measures. 14) When the man heads the boat directly downstream and rows as fast as he can, how fast does the observer on shore see the boat going
Mathematics
1 answer:
Artyom0805 [142]3 years ago
4 0

Answer:

The speed from the observer is 8 mph.

Step-by-step explanation:

When the boat is going downstream with full force, its speed will be the one the oarsman can row on still water added by the flow of the river. In this case he can row the boat at speed of 3 mph, while the stream is 5 mph, so the speed from the observer standpoint is:

speed = 3 + 5\\speed = 8 \text{ mph}

The speed from the observer is 8 mph.

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value at  

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=

−

1

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2

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x

=

−

2

.

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=

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.

y

=

10

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y

value at  

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2

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y

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20

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(

0

,

5

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,

(

−

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,

2.5

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,

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,

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10

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,

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=

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1.25

−

1

2.5

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5

1

10

2

20

Find the  

y

value at  

x

=

−

1

.

Tap for more steps...

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−

2

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6 0
3 years ago
A carpentry shop currently sells 66000 decoy ducks per year at a cost of 16 dollars each. In the previous year when they raised
Rus_ich [418]

Answer:

19 dollars

Step-by-step explanation:

<h2>Buckle up! This is a good one!</h2>

First we have to find the line that goes trough the two points. And it has the form :

y=mx+b

<em>m is often called the slope.</em>

Given the two points (66000, 16) and (51000, 21) we find m with this formula:

m=\frac{y_{2} -y_{1} }{x_{2} -x_{1} }\\m=\frac{21-16}{51000-66000} \\m=-\frac{5}{15000} \\m=-\frac{1}{3000}

We pick the point (51000,21) and plug it into the equation of the line and we find b:

y= mx + b\\21=(-\frac{1}{3000} )(51000)+b\\21=-17+b\\21+17=b=38

<h3>And we have got our line equation</h3>

y= -\frac{1}{3000}x + 38\\

<u>Where x represents the quantity and y represents the price.</u>

<u />

<h2>The revenue</h2>

The revenue is price times quantity, since price is y and quantity is x, we get

y=-\frac{1}{3000} x+38\\y*x=x*(-\frac{1}{3000} x+38)\\yx=-\frac{1}{3000} x^2+38x\\R(x)=-\frac{1}{3000} x^2+38x

<h3>to maximize the revenue, we get the first derivative:</h3>

R(x)=-\frac{1}{3000} x^2+38x\\R'(x)=-(2)\frac{1}{3000} x+38\\R'(x)=-\frac{2}{3000} x+38\\\\R'(x)=-\frac{1}{1500} x+38\\\\

<h3>and we make R'(x)=0:</h3>

R'(x)=-\frac{1}{1500} x+38=0\\38=\frac{1}{1500} x\\38*1500=x=57000

So the quantity that maximizes revenue is 57000,

<h3>let's find out the price by plugging it into the equation:</h3>

y= -\frac{1}{3000}x + 38\\y= -\frac{1}{3000}(57000) + 38\\y= -19 + 38\\y=19

<h2>And there you go! 19 dollars is the price the carpentry shop should charge per decoy duck.</h2>
7 0
3 years ago
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Lady_Fox [76]

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Step-by-step explanation:

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3 years ago
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