Answer:
a cyclist covers a distance of 15km in 2hours calculate his speeda cyclist covers a distance of 15km in 2hours calculate his speeda cyclist covers a distance of 15km in 2hours calculate his speed
Answer:
Distance between A and B is 5400 meters
Step-by-step explanation:
Consider "D" the letter to identify distance between A and B
Let's use "t" to identify the time of the first encounter (Devi and Kumar), and create an equation that states that the distance covered by Devi (at 100 m/min) in time "t", is equal to the total distance D minus what Kumar has covered at his speed (80 m/min) in that same time:
Recall that distance equals the speed times the time:
distance= speed * time
First encounter:
100 * t = D - 80 * t
180 * t = D Equation (1)
Not, 6 minutes later (at time t+6) , Devi and Li Ting meet .
Then for this encounter the distance covered by Devi equals total distance d minus the distance covered by Li Ting:
100 *(t+6) = D - 75 * (t+6)
100 t + 600 = D - 75 t - 450
175 T + 150 = D Equation (2)
Now, let's equal equation (1) to equation (2), since D should be the same:
180 t = 175 t + 150
5 t = 150
t = 30
Then the time t (first encounter) is 30 minutes. Knowing this, we can use either equation to find D:
From Equation (1) for example: D = 180 * t = 180 * 30 = 5400 meters
Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.
First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.
Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:
1, 3, 5, 7, 9, 11, 13, 15 .
Number of items in the set . . . 8
Sum of all the items in the set . . . 64
Hmmm. That's interesting. 64 happens to be the square of 8 .
Do you think that might be all there is to it ?
Let's check it out:
Even-numbered lists of odd numbers:
1, 3 Items = 2, Sum = 4
1, 3, 5, 7 Items = 4, Sum = 16
1, 3, 5, 7, 9, 11 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .
Amazing ! The sum is always the square of the number of items in the set !
For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.
I slogged through it on my calculator, and it's true.
I never knew this before. It seems to be something valuable
to keep in my tool-box (and cherish always).
Answer:
Hi! I hope this is correct please inform me if this is incorrect :)
a = 1 will have.
2a + 3 < 9 - 3a.
2a + 3 > 10 -3a.
3a - 2a + 1>2.
a = 2 will have
ANY others I did not include in a = 1 :)
Please once again tell me if this is incorrect.
Step-by-step explanation:
