Question

Поезд был задержан с отправлением на 16 минут,Чтобы прибыть в пункт назначения вовремя он увеличил скорость на 10км/ч.С какой скоростью двигался поезд , если пройденное им расстояние равно 80 км?????

Answer 1

решим систему: 5х+4у= -6      х+3у=1

из второго ур-я: х=-6-3у

5*(-6-3у)+4у=-6

-30-15у+4у=-6

-11у=24

у=-24/11<0

х=1-3*(-24/11)=1+72/11>0

Когда x>0 y<0 это IV четверть

Answer 2

Calculation of Train Speed

To calculate the speed at which the train was moving, we can use the formula:

Speed = Distance / Time

Given that the train was delayed by 16 minutes and the distance traveled is 80 km, we need to find the time it took for the train to cover this distance.

To do this, we need to consider that the train increased its speed by 10 km/h to arrive at its destination on time. Let's assume the original speed of the train is x km/h.

The time it took for the train to travel 80 km at the original speed can be calculated as:

Time = Distance / Speed

The time it took for the train to travel 80 km at the increased speed can be calculated as:

Time + 16 minutes = Distance / (Speed + 10)

Now we can solve these equations to find the original speed of the train.

Solution

Let's solve the equations:

1. Time = Distance / Speed - Time = 80 km / x km/h

2. Time + 16 minutes = Distance / (Speed + 10) - Time + 16/60 hours = 80 km / (x + 10) km/h

To solve these equations, we can set them equal to each other:

80 km / x km/h = 80 km / (x + 10) km/h + 16/60 hours

Now we can solve for x:

80 / x = 80 / (x + 10) + 16/60

To simplify the equation, we can multiply both sides by x(x + 10) to eliminate the denominators:

80(x + 10) = 80x + 16(x + 10)/60

Simplifying further:

80x + 800 = 80x + 16(x + 10)/60

80x - 80x = 16(x + 10)/60 - 800

0 = 16(x + 10)/60 - 800

16(x + 10)/60 = 800

16(x + 10) = 800 * 60

16x + 160 = 48000

16x = 48000 - 160

16x = 47840

x = 47840 / 16

x = 2990

Therefore, the original speed of the train was 2990 km/h.

Please note that this calculation assumes a constant speed throughout the journey and does not take into account factors such as acceleration or deceleration.