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Fittoniya [83]
2 years ago
15

Nhan and Phuong both left town A for town B at 5 am . After 5 hours , Nhan and Phuong were 102 km and 52 km away from town B res

pectively . The ratio of Nhan's speed to Phuong's speed was 3 : 4 and their speeds remained the same the whole journey . At what time did Nhan reach town B ? Show your full work.​
Mathematics
1 answer:
Pepsi [2]2 years ago
3 0

Answer:

Nhan reach town B at 13:24 pm

Step-by-step explanation:

Let

Speed of Nhan=3x km /hour

Speed of Phoung=4x km/hour

After 5 hours,

Distance covered by Nhan=3x(5)=15 xkm

Distance covered by Phuong=4x(5)=20x km

Total distance covered by Nhan to reach town B=15x+102 km

Total distance covered by Phuong to reach town B=20x+52 km

Distance between town A and town B for both person is same therefore,

15x+102=20x+52

102-52=20x-15x

50=5x

x=10

Speed of Nhan=3(10)=30 km/h

Speed of Phuong=4(10)=40 km/h

Total distance between town A and town B=102+15(10)=252 km

Time taken by Nhan to reach town B from town A=\frac{distance}{speed}

Time taken by Nhan to reach town B from town A=\frac{252}{30}=8.4 hours

8.4 hours=8+0.4\times 60

=8 hours 24 minutes

1 hour=60 minutes

Nhan reach town B at 13:24 pm

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Answer:

a) P (x) = (x + 3) (x-1) (x-4)

b) P (x) = (2x + 5) (5x - 4) (x-6)

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The factors are

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---------------------------------------

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--------------------------------------

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--------------------------------------

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Multiplicity 2 in -1

x = -3\\\\(x-3) = 0

--------------------------------------

x = 1\\\\(x-1) = 0

--------------------------------------

x = 4\\\\(x-4) = 0

----------------------------------------

x = -1\\\\(x + 1) = 0

-----------------------------------------

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