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Yes , it is possible because you can put the tips of each pencil together almost creating a star but they all touch within the center.

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Q: How do you arrange 6 pencils so that each one touches the other five?
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How do you answer Ray bought 15 pencils for 7.75 Some cost 45 each the rest 65 each How many of each did Ray buy. Answer step by step please?

x= # of pencils that cost .45 y= # of pencils that cost .65 So now you need 2 equations for 2 variables: 15= x+y This equation is saying: 15 total pencils bought= pencils that cost .45 + pencils that cost .65 The second equation is: 7.75= .45x + .65y This equation is saying: total money spent (7.75)= price of pencils (.45) times # of pencils + price of pencils (.65) times # of pencils Then you combine these two equations, but first switch around the first equation to look like: y=15-x Then you replace the "y" in the second equation by putting in what y equals the first equation: 7.75= .45x + .65(15-x) Then distribute and solve for x: x = 10 Then enter 10 into the first equation for x to figure out y: y= 15- 10 y= 5 So your answer is-- Ray bought 10 pencils that cost .45 and 5 pencils that cost .65


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i think it is

Related questions

Is it possible to arrange seven pencils in a manner such that each pencil touches every other pencil?

Yes it is. First, they must be sharpened to sharp points, then lay them in a circle with all of the points all joined in the center. Use a magnifying glass if you must ensure that each is touching all of the other six.


If pencils cost 0.26 each could you buy four pencils with 1.00?

If pencils cost 0.26 each could you buy four pencils with 1.00?


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288 pencils in 36 boxes, would mean 8 pencils in each box. If you had 288 pencils in each of 36 boxes, you would a total of 10,368 pencils.


What is the cost of p pencils if pencils cost 20 each and pens cost 30 each?

The cost of p pencils is 20p.


Are two pencils actually pulling on each other?

Yes, two pencils will pull on each other. But the force between them doing the pulling, gravity, is so small (because the mass of the pencils is small) that it would be somewhere between difficult and impossible to measure it. I would be something that an investigator would calculate (based on the mass of the pencils) rather than measure.


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