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ziro4ka [17]
3 years ago
14

What is the length of YZ?​

Mathematics
1 answer:
Marianna [84]3 years ago
3 0
40
Subtract 48 from 88
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Rina8888 [55]
To solve this we are going to use the formula for speed: S= \frac{d}{t}
where
S is the speed
d is the distance 
t is the time 

Let S_{l} be the speed of the boat in the lake, S_{a} the speed of the boat in the river, t_{l} the time of the boat in the lake, and t_{a} the time of the boat in the river. 

We know for our problem that <span>the current of the river is 2 km/hour, so the speed of the boat in the river will be the speed of the boat in the lake minus 2km/hour:
</span>S_{a}=S_{l}-2
We also know that in the lake the boat<span> sailed for 1 hour longer than it sailed in the river, so:
</span>t_{l}=t_{a}+1
<span>
Now, we can set up our equations.
Speed of the boat traveling in the river:
</span>S_{a}= \frac{6}{t_{a} }
But we know that S_{a}=S_{l}-2, so:
S_{l}-2= \frac{6}{t_{a} } equation (1)

Speed of the boat traveling in the lake:
S_{l}= \frac{15}{t_{l} }
But we know that t_{l}=t_{a}+1, so:
S_{l}= \frac{15}{t_{a}+1} equation (2)

Solving for t_{a} in equation (1):
S_{l}-2= \frac{6}{t_{a} }
t_{a}= \frac{6}{S_{l}-2} equation (3)

Solving for t_{a} in equation (2):
S_{l}= \frac{15}{t_{a}+1}
t_{a}+1= \frac{15}{S_{l}}
t_{a}=\frac{15}{S_{l}}-1
t_{a}= \frac{15-S_{l}}{S_{l}} equation (4)

Replacing equation (4) in equation (3):
t_{a}= \frac{6}{S_{l}-2}
\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}

Solving for S_{l}:
\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}
(15-S_{l})(S_{l}-2)=6S_{l}
15S_{l}-30-S_{l}^2+2S_{l}=6S_{l}
S_{l}^2-11S_{l}+30=0
(S_{l}-6)(S_{l}-5)=0
S_{l}=6 or S_{l}=5

We can conclude that the speed of the boat traveling in the lake was either 6 km/hour or 5 km/hour.
3 0
4 years ago
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