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Charra [1.4K]
2 years ago
8

Jack and jill exercise in a 25.0-m-long swimming pool. jack swims nine lengths of the pool in 2 minutes and 34.5 seconds, while

jill, the faster swimmer, covers ten lengths in the same time interval. find the average velocity and average speed of each swimmer.
Mathematics
1 answer:
grin007 [14]2 years ago
5 0
Jack swims a distance of 9*25.0m=225.0 meters is 2 min and 34.5 seconds.

2 min and 34.5 seconds are 2*60sec +34.5 sec= 154.5 s


Jill covers 10*25.0m = 250 m in 154.5 s.


\displaystyle{ Average\ Speed= \frac{Distance\ traveled}{Time \ of \ travel}\\\\



\displaystyle{ Average \ Velocity =   \frac{Displacement}{time}



thus,

the speed of Jack is : 225.0 meters /154.5 s =(225/154.5) m/s = 1.46 m/s

the speed of Jill is : 250.0 meters /154.5 s =(250/154.5) m/s = 1.62 m/s


Assuming Jack and Jill depart from point 0 towards the positive direction, 

each even number of lengths, means Displacement = 0, and each odd number of lengths means Displacement = 25 m 


So, average Velocity of Jill is 0, 

average velocity of Jack is 25/ 154.5 = 0.16 m/s

Answer: 

Average Speed of Jack : 1.46 m/s

Average Speed of Jill : 1.62 m/s

Average Velocity of Jack : 0

Average Velocity of Jill : 0.16 m/s

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Answer:

30 2/3

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Then multiply 16 times 3

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I hope this helps :)

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6x2y – 2x2y – 10x2y + 8x2y =
svp [43]

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6 0
2 years ago
Which equation is the inverse of y = 2x2 – 8?​
natita [175]

The inverse of the function is y=\pm \sqrt{\frac{x+8}{2}}

Explanation:

To find the inverse of the equation y=2x^{2} -8, we need to interchange the variables x and y for the variables y and x.

Thus, the equation becomes

x=2y^{2} -8

Now, we shall find the value of y.

Now, adding 8 to both sides of the equation, we have,

x+8=2y^{2}

Interchanging the sides,

2y^{2} =x+8

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y^{2} =\frac{x+8}{2}

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Thus, the inverse of the function is y=\pm \sqrt{\frac{x+8}{2}}

5 0
2 years ago
I've been stuck on this for so long and I have an exam soon, anybody who can help me :'( ?
san4es73 [151]
(i)  speed = distance / time
so time =  distance / speed
here we have

time t = 1080/x  hours

(ii) return flight  time  = 1080 / (x + 30)  hours

(a)  1080/x - 1080/(x + 30) = 1/2

Multiplying  through by the LCD 2x(x + 30) we get:-

1080*2(x + 30) - 2x*1080 = x(x+30)
2160x + 64800 - 2160x = x^2 + 30x
x^2 + 30x - 64800  = 0

(b)  factoring;  -64800 = 270 * -240  ans 270-240 = 30 so we have

(x + 270)(x - 240) = 0   so x = 240  ( we ignore the negative -270)

So the speed for outward journey is 240 km/hr

(c) time ffor outward flight = 1080 / 240 =  4 1/2  hours

(d) average speed for whole flight = distance / time
   Time for outward journey = 4.5 hours and time for  return journey = d / v
= 1080 / (240+30) =  4 hours
 Therefore the average speed for whole journey =  2160 / 8.5 = 254.1 km/hr
8 0
2 years ago
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