I would say about £500
500.000
One. You pay one cent for each bead [unit price is (12 cents) / (12 beads) = 1 cent /bead).Eric
I'm not sure how much it is worth but my guess is not a whole lot. This is because on the Mohs scale of hardness(diamond being 10) it ranks only a 5. This means it can chip. The best quality turquoise is clear blue with no veins or brown spots. Even still u can buy a top quality 16mm bead for $10 each. I usually only make jewelry with gems that have a hardness of 7 or higher. However the colour of turquoise is beautiful and I'm considering using it. Also if u looking to purchase turquoise make sure it is stabilized turquoise.It has been treated to harden it. Also-"African turquoise"is not turquoise- it is jasper.
the first section has a cross then there is one bead Then three beads then one bead then there is a connector, which connects three chains then 10 beads then 1 bead then 10 beads then 1 bead then 10 beads then 1 bead then 10 beads then 1 bead then 10 beads then it reconnects to that three-way connector
1 mass = 1200 beads
7/35 or 1/5
The value of a turquoise stone cannot be measured by its weight. Almost like diamonds there are certain criteria that must be met in order to help determine actual value, such as; Hardness/density , Luster, Color, Matrix which help to determine its rating such as Gem, Very height grade and High grade. However it is good to keep in mind that less than 1 percent of turquoise can legitimately be called "Gem"
250 red beads 2x250 = 500 blue beads (1/2)X250 = 125 yellow beads 250+500+125=875 Fran has 875 beads.
Well, if there are 12 questions on the test and each question is worth the same amount, then each question is worth 1/12 of the total points. So, technically, each question is worth 1/12 of the total score. Math doesn't lie, honey!
Well there are 4 6mm beads in an 1", so there would be 64 6mm beads in 16".
Patan can create rows of equal length using the 84 beads by determining the factors of 84. The possible row lengths are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84. By choosing any of these factors, she can evenly distribute the beads into rows. For example, if she chooses a row length of 12, she would have 7 rows of 12 beads each.
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.