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Fudgin [204]
3 years ago
5

A man walking on a railroad bridge is 2/5 of the way along the bridge when he notices a train at a distance approaching at the c

onstant rate of 45 mph . The man can run at a constant rate in either direction to get off the bridge just in time before the train hits him. How fast can the man run?
Mathematics
2 answers:
shusha [124]3 years ago
5 0

Answer: 9 mph

Step-by-step explanation:

Given that a man walking on a railroad bridge is 2/5 of the way along the bridge when he notices a train at a distance approaching at the constant rate of 45 mph .

If the man tend to run in the forward direction, he will cover another 2/5 before the train reaches his initial position. The distance covered by the man will be 2/5 + 2/5 = 4/5

The remaining distance = 1 - 4/5 = 1/5

If the man can run at a constant rate in either direction to get off the bridge just in time before the train hits him, the time it will take the man will be

Speed = distance/time

Time = 1/5d ÷ speed

The time it will take the train to cover the entire distance d will be

Time = d ÷ 45

Equate the two time

1/5d ÷ speed = d ÷ 45

Speed = d/5 × 45/d

Speed = 9 mph

makkiz [27]3 years ago
3 0

Answer:

The Man needs to run at 9 mph

Step-by-step explanation:

Let M stand for the man's speed in mph.  When the man  

runs toward point A, the relative speed of the train with respect  

to the man is the train's speed plus the man's speed (45 + M).  

When he runs toward point B, the relative speed of the train is the  

train's speed minus the man's speed (45 - M).

When he runs toward the train the distance he covers is 2 units.  

When he runs in the direction of the train the distance he covers  

is 3 units. We can now write that the ratio of the relative speed  

of the train when he is running toward point A to the relative speed  

of the train when he is running toward point B, is equal to the  

inverse ratio of the two distance units or

              (45 + M)          3

              -----------  =      ---

              (45 - M)          2

          90+2 M=135-3 M

⇒5 M = 45

⇒ M = 9 mph

The Man needs to run at 9 mph

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Kelli swam upstream for some distance in one hour. She then swam downstream the same river for the same distance in only 6 minut
solmaris [256]

Answer:

6.11km/hr

Step-by-step explanation:

Let the speed that Kelli swims be represented by Y

Speed of the river = 5km/hr

Distance = Speed × Time

Kelli swam upstream for some distance in one hour

Swimming upstream takes a negative sign, hence:

1 hour ×( Y - 5) = Distance

Distance = Y - 5

She then swam downstream the same river for the same distance in only 6 minutes

Downstream takes a positive sign

Converting 6 minutes to hour =

60 minutes = 1 hour

6 minutes =

Cross Multiply

6/60 = 1/10 hour

Hence, Distance =

1/10 × (Y + 5)

= Y/10 + 1/2

Equating both equations we have:

Y - 5 = Y/10 + 1/2

Collect like terms

Y - Y/10 = 5 + 1/2

9Y/10 = 5 1/2

9Y/ 10 = 11/2

Cross Multiply

9Y × 2 = 10 × 11

18Y = 110

Y = 110/18

Y = 6.1111111111 km/hr

Therefore, Kelli's can swim as fast as 6.11km/hr still in the water.

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2 years ago
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It's b. %99.7

Step-by-step explanation:

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Pleeeeese help me as fast as you can!!!
Aleonysh [2.5K]

Answer:

A. y=3x-10

C. y+6=3(x-15)

Step-by-step explanation:

Given:

The given line is 6x+18y=5

Express this in slope-intercept form y=mx+b, where m is the slope and b is the y-intercept.

6x+18y=5\\18y=-6x+5\\y=-\frac{6}{18}x+\frac{5}{18}\\y=-\frac{1}{3}x+\frac{5}{18}

Therefore, the slope of the line is m=-\frac{1}{3}.

Now, for perpendicular lines, the product of their slopes is equal to -1.

Let us find the slopes of each lines.

Option A:

y=3x-10

On comparing with the slope-intercept form, we get slope as  m_{A}=3.

Now, m\times m_{A}=-\frac{1}{3}\times 3=-1. So, option A is perpendicular to the given line.

Option B:

For lines of the form x=a, where, a is a constant, the slope is undefined. So, option B is incorrect.

Option C:

On comparing with the slope-point form, we get slope as  m_{C}=3.

Now, m\times m_{C}=-\frac{1}{3}\times 3=-1. So, option C is perpendicular to the given line.

Option D:

3x+9y=8\\9y=-3x+8\\y=-\frac{3}{9}x+\frac{8}{9}\\y=-\frac{1}{3}x+\frac{8}{9}

On comparing with the slope-intercept form, we get slope as  m_{D}=-\frac{1}{3}.

Now, m\times m_{D}=-\frac{1}{3}\times -\frac{1}{3}=\frac{1}{9}. So, option D is not perpendicular to the given line.

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3 years ago
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