Homework help?

Question

There is a 50-inch line. On this line, there are circles 2 inches in diameter, with 1 inch between them. The center of each circle is on the segment.

How many circles could fit on the segment? I've estimated 20 but I don't know how to do this problem. I missed math class today.

This isn't my actual homework question, but it's something like it.

PLEASE KEEP IN MIND THAT I MISSED MATH CLASS TODAY. With a good reason.

I had to leave school because I think I got food poisoning. Whatever it is, it means that I spent half the lesson -which was mostly review- trying not to puke. Then I couldn't hold it in any longer, blew chunks over everyone and got sent home.

So I have no idea how to solve this. If I did, I'd do it myself.

Answer 1

do it yourself its the only way to learn!!!!!!

Answer 2

Oddly enough, nobody has answered the original QUESTION. You are inquiring, not the solution itself, but the method of solving problems such as these.

Here's your answer:

What you need to identify, in problems like these, is a pattern. In this case, you know that the system alternates the Circle and then the Line. In a different representation:

A = the circle
b = the line

AbAbAbAbAbAb...

Now, the smallest unique pattern that repeats, is Ab:

Ab Ab Ab Ab Ab...

You where given the limiting characteristics of your system:

50 Inches (A Length)

THEN, you need to determine how your smallest unique pattern applies to that limitation. What's the LENGTH of Ab? 3 Inches (2 inch diameter + 1 inch line)

Divide the limiting characteristic by the representative value of your unique pattern:

50 / 3 = 16 2/3 basic patterns (Ab)

Now you know that 16 of your basic pattern Ab will fit on the line. What do you do with the remainder of 2/3? Convert that back into your basic characteristic (length):

3 Inches (from Ab) * 2/3 = 2 Inches

The next object in your pattern is a circle (2 Inches), so it fits. 16 basic patterns (Ab) each with one circle, plus the extra circle on the end, makes 17.

This type of logic can be followed for any limiting dimension.

For example, the same problem, expressed as money.

Two people, Sam and John, are placing money on a table. Sam has nothing but Two Dollar Bills, and John has Singles. They alternate placing money on the table, one, then the other. How many Two Dollar Bills can they place on the table, without exceeding 50 Dollars total?

Same logic, identify your limiting characteristic (Dollars), identify your pattern ($2, then 1$), divide your limiting characteristic by your pattern ($50 / $3), and you have an extra 2/3...

The math is the same, but the Characteristic you are talking about changes.

There's the THEORY behind that process. I know my answer was long, but I hope the few minutes reading helped to replace the hour of class that you missed.

And, I hope you feel better!

Answer 3

17 circles. for every 2in circle there has to be a one inch space,
so 3 in. total for every circle. 50/3= 16 2/3, the extra 2/3 is enough for another circle so 16+1=17

Answer 4

Okay. I am going to think out loud. if you allow 3 inches then every 10 inches you will have 3 circles and then and extra inch. Then by 50 inches you would have 15 circles and then 5 extra inches. So yeah-- 20 should be the right answer.

Answer 5

Well take the 50 inches and divide it by 3 (2 for the circle diameter and 1 for the inch between) Hint: Your computer's calculator should be able to help find the answer ;)

Answer 6

distance between the centre of one circle and another=3"
so no of circles 2 in every 5" so 50 circles
another approach is 50/(2 1/2)=20

Answer 7

17

Answer 8

Circles diameters are 2in + 1in separation = 3in per circle w/ spacing.
50/3 is 16 remainder 2in.
16 circles + 1 (because there is 2in leftover, which could fit another circle) = 17 circles

Answer 9

16½ if the 50" line starts w/a space; 17 if it starts with a circle.

Answer 10

For two circles distance=2+1+2 =5
For three circles distance=2+1+2+1+2=8
For Four circles distance=2+1+2+1+2+1+2=11
For five circles distance=2+1+2+1+2+1+2+1+2=14
For six circles distance=2+1+2+1+2+1+2+1+2+1+2=17
For seven circles distance=2+1+2+1+2+1+2+1+2+1+2+1+2=20
Therefore seven circles are drawn.
Please note here are n circle and distance is one less than number of circles.
2xn +1x(n-1)=20
2n +n-1=20
3n=21
n=7

Answer 11

let me expalin it my way..for evry 20 inches u will get 7 circles...so in 40 inches it will be 14 and in rest 10 inch..u can make 3 more....so 14+3=17

ya seventeen circles