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How many total beads are there. Are the beads only red or blue?

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15y ago

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Related Questions

How many blue beads would there be in a pattern of 15 red beads?

None, because the beads in the pattern are RED


How many blue beads would there be in a pattern of 13 red beads?

36


How many blue beads would there be in a pattern with ten red beads?

None, because there are only ten red beads!


Look at the pattern. How many blue beads would there be in a pattern with 13 red beads?

There would be 36 beads.


How many red beads would there be in a pattern with 90 blue beads?

32 I think


A bottle contains blue beads and red beadsthe number of red beads is 4 times the number of blue beads if there are 3568 red beads how many more red beads than blue beads are there?

2676


How many blue beads would there be in a pattern with red beads?

It really depends on how many beads you are using if u was using 12 beads there will be 6 beads.


If you alternate red beads and blue beads how many red and blue beads can you fit on 4 in of string?

8


How many blue beads would there be in a pattern with 10 black beads?

There is not enough information in order to answer this question.The amount of blue beads would depend on the size of the blue beads.The amount of blue beads would depend on the size of the red beads, too.It would also depend on the size of the bracelet.It would also depend on how complex or simple the bracelet design is.


How many blue beads did Stacy use if she made a necklace using four times as many blue beads as red beads and she used a total of forty beads?

32 blue, 8 red


If Fran has 250 red beads she has twice as many blue beads as red beads Fran has half as many yellow beads as red beads How many beads does Fran have?

250 red beads 2x250 = 500 blue beads (1/2)X250 = 125 yellow beads 250+500+125=875 Fran has 875 beads.


What is the nth term of 4 red beads and 3 blue beads in a pattern increasing by two red beads and 3 blue beads each time?

The pattern starts with 4 red beads and 3 blue beads, and increases by 2 red beads and 3 blue beads for each subsequent term. Therefore, the nth term can be expressed as: ( \text{Red beads} = 4 + 2(n-1) ) and ( \text{Blue beads} = 3 + 3(n-1) ). Simplifying these gives: ( \text{Red beads} = 2n + 2 ) and ( \text{Blue beads} = 3n ). Thus, the nth term consists of ( (2n + 2) ) red beads and ( 3n ) blue beads.