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notsponge [240]
3 years ago
13

The vagabonds were 200 miles apart at 2 p.M and were headed toward each other if they met at 6 p.M and one was traveling 10 mile

s per hour faster than the other what was the speed of each vagabond
Mathematics
2 answers:
muminat3 years ago
8 0

Answer:

<em>The answer is 200</em>

Step-by-step explanation:

<em>To solve the question given, let us recall the following,</em>

<em>Let r4=200  miles apart</em>

<em>r=50 mph combined speed </em>

<em>Where </em>

<em>2f+10=50 </em>

<em>2f=40 </em>

<em>f=20 </em>

<em>f+10=30 </em>

<em>or we can express it in another way,</em>

<em>which is</em>

<em>f x4+(f+10)x 4=200 </em>

<em>f x4+f x4+40=200 </em>

<em>8f+40=200</em>

<em>Then</em>

<em>8f is =160 </em>

<em>f=20 </em>

<em>f+10=30 </em>

<em>Therefore,</em>

<em>4x 20+4 x 30=200 </em>

<em>80+120=200 </em>

<em>200=200</em>

<em />

Softa [21]3 years ago
8 0

Answer:

First vagabond: 30 miles per hour

Second vagabond: 20 miles per hour

Step-by-step explanation:

Let's call the speed of the first vagabond A, and the speed of the second vagabond B.

They were heading toward each other, so the relative speed is the sum of both speeds: A+B

The total number of hours spent until they meet is 6 - 2 = 4 hours, so we have the first equation:

(A+B) * 4 = 200

A + B = 50

One vagabond was traveling 10 miles faster than the other, so we have the second equation:

A = B + 10

Using the value of A from the second equation in the first one, we have:

B + 10 + B = 50

2B = 40

B = 20 miles per hour

Now, from the second equation, we find the value of A:

A = 20 + 10 = 30 miles per hour

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