Two balls A and B rotate along a circular track. Ball A makes 4 full rotations in 120 seconds. Ball B makes 3 full rotation in 6
1 answer:
Find the rate in which both balls make one full rotation around the track.
Ball A:
It takes 30 seconds for Ball A to make a full rotation.
Ball B:
It takes 20 seconds for Ball B to make a full rotation.
Find the least common multiple of 20 and 30 to find the time at which they'll meet.
Multiples of 20: {20,40,60}
Multiples of 30: {30,60}
60 is the least common multiple of both 20 and 30.
The balls will meet again at 60 seconds.
Ball A:
It takes 30 seconds for Ball A to make a full rotation.
Ball B:
It takes 20 seconds for Ball B to make a full rotation.
Find the least common multiple of 20 and 30 to find the time at which they'll meet.
Multiples of 20: {20,40,60}
Multiples of 30: {30,60}
60 is the least common multiple of both 20 and 30.
The balls will meet again at 60 seconds.
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Hello,
Using vectors and scalar product:
![[(a+c)*\vec{i}+(b-0)*\vec{j} ].[(a-c)*\vec{i}+(b-0)*\vec{j}]=0](https://tex.z-dn.net/?f=%5B%28a%2Bc%29%2A%5Cvec%7Bi%7D%2B%28b-0%29%2A%5Cvec%7Bj%7D%20%5D.%5B%28a-c%29%2A%5Cvec%7Bi%7D%2B%28b-0%29%2A%5Cvec%7Bj%7D%5D%3D0%20)
Thus
By the way how can we make text larger in latex \larger{.....} don't work.
Answer A
Using vectors and scalar product:
Thus
By the way how can we make text larger in latex \larger{.....} don't work.
Answer A
Answer:
Step-by-step explanation:
For this case we have the following profit function:
And we are interested on find the profit after selling 700 pretzels, so we just need to replace X=700 into the profit function like this:
The profit function is on the figure attached. We see that for 0<X<217 we have negative profits, and for X>217 we will have positive values for the profit. And also we can see that for X=700 we have a profit of approximately 169.
