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Lady bird [3.3K]
3 years ago
10

A ship traveled for 4 hours heading east and for 3 hours heading north. If the total distance traveled was 149 miles, and the sh

ip traveled 3 miles per hour faster heading north, at what speed was the ship traveling east?
Mathematics
1 answer:
Elina [12.6K]3 years ago
5 0
Just like the previous one

if the ship travelled a total of 149 miles, then let's say it travelled East "d" miles and thus it travelled North the slack of 149 and "d", thus " 149 - d".

now, if the rate travelling East is say "r", then the faster rate going North is " r + 3".

\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
East&d&r&4\\
North&149-d&r+3&3
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=4r\\
149-d=(r+3)3\\
----------\\
149-\boxed{4r}=3(r+3)
\end{cases}

solve for "r".
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Answer:

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Step-by-step explanation:

<em><u>The question is</u></em>

What is the volume of the open top box as a function of the side length x in cm of the square cutouts?

see the attached figure to better understand the problem

Let

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we know that

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L=(10-2x)\ cm

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substitute

V(x)=(10-2x)(10-2x)x\\\\V(x)=(100-40x+4x^{2})x\\\\V(x)=(4x^{3}-40x^{2}+100x)\ cm^3

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we know that

(10-2x) > 0\\10> 2x\\ 5 > x\\x < 5\ cm

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The domain is the interval (0,5)

The domain is all real numbers greater than zero and less than 5 cm

therefore

The volume of the open top box as a function of the side length x in cm of the square cutouts is

V(x)=(4x^{3}-40x^{2}+100x)\ cm^3

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Answer:

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Step-by-step explanation:

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a^2 + b^2 = c^2

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Answer:

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