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Lera25 [3.4K]
2 years ago
4

Two points move along a circle of length 120m with a constant speed. If they move in different directions, then they meet every

15 seconds. When going in the same direction, one point catches up to the second every 60 seconds. Find the speeds of the points.
Mathematics
1 answer:
AVprozaik [17]2 years ago
6 0
Total Distance covered in one round of circle = 120 m

Let x and y be the speeds of the two points.

When two points are moving in opposite directions, they meet in every 15 seconds. This means they cover 120 m in 15 seconds. They are moving in opposite directions, so the relative speed will be equal to the sum of speeds of both points. So we can set up the equation as:

x + y = 120/15
x + y = 8 

This means, sum of their speeds is 8 m/s

When moving in the same direction, one point catches the other point every 60 seconds. We can set up the equation for this case as:

x - y = 120/60
x - y = 2

This means, difference of their speeds is 2m/s

Adding both equations we get:

x+ y +x - y = 8 + 2

2x = 10

x = 5 m/s

x + y =8
This means, y = 3 m/s

Thus, the speed of the two points is 5 m/s and 3 m/s
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ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
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Therefore, for the two expressions to be conjugates, we must satisfy the two conditions. 

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the 

   x²y = -4 ... (I)

Condition 2: Real parts are the same

   x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

   x²y = -4 ... (I)
   x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

   y = -3 - x² ... (II)

Substituting into equation (I)

   x²y = -4 ... (I)
   x²(-3 - x²) = -4
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   (x² + 4)(x² - 1) = 0
   (x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

   y = -3 - x² ... (II)
   y = -3 - (±1)²
   y = -3 - 1
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So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

   -3 + ix²y 
   = -3 + i(±1)²(-4)
   = -3 - 4i

   x² + y + 4i
   = (±1)² - 4 + 4i
   = 1 - 4 + 4i
   = -3 + 4i

They result in conjugates
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y=-2000\,\frac{ft}{min} \,x\,+\,33000\,ft

Part 4)

The information that was not used to write the descent equation was the initial details about how long the plane traveled (4 hours) and for 1800 miles.

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