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avanturin [10]
3 years ago
8

Peter travels to his uncle's home 30km away from his place.He cycles for two thirds of the journey before the bicycle develops a

mechanical problem and he has to push it for the rest of the journey. If his cycling speed is 10km/hr faster than his walking speed and he completes the journey in 3hr 20minutes, determine his cycling speed?
Mathematics
1 answer:
emmasim [6.3K]3 years ago
8 0

Answer:

Cycling Speed is 15km/hr

Step-by-step explanation:

Let the cycling speed of Peter be xkm/hr

His walking speed is (x-10km/hr)

Distance cycling is 2/3 x 30 = 20km

Distance walking 1/3 x 30 = 10km

Time taken in cycling = distance/ speed

Time = 20/x

This gives 20/x km/hr

Time walking is 10 / (x-10)

= 10/x-10) hrs

Total time = (20/x + 10/x+10)

Therefore 10/x+10/x-10 = 10/3

60(x-10)+30(x)=10x(x-10)

10x² - 190x+600=0

x²-19+60=0

x=<u>19±√361-240</u>

        2

x=15 0r x =4

His cycling speed is 4

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There are 3 islands A,B,C. Island B is east of island A, 8 miles away. Island C is northeast of A, 5 miles away and northwest of
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The bearing needed to navigate from island B to island C is approximately 38.213º.

Step-by-step explanation:

The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:

AC^{2} = AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \theta (1)

Where:

AC - The distance from A to C, measured in miles.

AB - The distance from A to B, measured in miles.

BC - The distance from B to C, measured in miles.

\theta - Bearing from island B to island C, measured in sexagesimal degrees.

Then, we clear the bearing angle within the equation:

AC^{2}-AB^{2}-BC^{2}=-2\cdot AB\cdot BC\cdot \cos \theta

\cos \theta = \frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC}

\theta = \cos^{-1}\left(\frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC} \right) (2)

If we know that BC = 7\,mi, AB = 8\,mi, AC = 5\,mi, then the bearing from island B to island C:

\theta = \cos^{-1}\left[\frac{(7\mi)^{2}+(8\,mi)^{2}-(5\,mi)^{2}}{2\cdot (8\,mi)\cdot (7\,mi)} \right]

\theta \approx 38.213^{\circ}

The bearing needed to navigate from island B to island C is approximately 38.213º.

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kenny6666 [7]
The answer is letter b

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