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Helga [31]
3 years ago
11

An artist follows this formula for making a special tricolored bracelet: 8 orange beads + 3 red beads + 4 green beads + 1 string

â 1 bracelet how many special tricolored bracelets could be formed from 15 orange beads, 86 red beads, 92 green beads, and 17 pieces of string?
Mathematics
1 answer:
natali 33 [55]3 years ago
3 0
Given that a<span>n artist uses 8 orange beads, 3 red beads, 4 green beads and 1 string for making a special tricolored bracelet.

Given that there are </span><span>15 orange beads, 86 red beads, 92 green beads, and 17 pieces of string</span>, the number of special tricolored bracelets that can be made is 1.

This is because, since 8 orange beads is needed for the special tricolored bracelets and there are only 15 orange beads the available, there will only be 7 orange beads left for the second tricolored bracelet.
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What is the solution of the linear-quadratic system of equations?<br> y=x^2+5x−3<br> y−x=2
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Answer:

x =  1, y = 3

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Step-by-step explanation:

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(1,3) , (-5, -3)

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2 years ago
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Help pls. I have been stuck on this for 15 minutes now.​
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Please help very urgent
Alik [6]

Answer:

32, <u>16</u>, <u>8</u>, <u>4</u>, <u>2</u>, 1

Explanation:

The geometric mean can be represented by \sqrt[n]{x_{1} • x_{2} • x_{3} • .. x_{n}}.

Which is the mean of the product of n numbers, used to find the average of a geometric progression.

Don't get confused by geometric mean, it is only asking you about the next numbers in the geometric sequence given the first and sixth term.

The explicit rule for a geometric sequence can be modeled by:

a_{n} = a_{1} • r^{n-1}

Where a_{n} is the nth term, a_{1} is the first term in the sequence, n is the term number, and r is the common ratio.

Since we already know the first term, a_{1} will simply be 32.

Since we know it's geometric, there will be an exponential relationship, which means that we will use the geometric mean to find the common ratio.

There are 6 total terms, r is raised to the n – 1 so 6 – 1 = 5, and that will be the degree of this root.

\sqrt[5]{\frac{a_{6}}{a_{1}}} =

\sqrt[5]{\frac{1}{32}} =

\frac{1}{2}.

Therefore: r = \frac{1}{2}.

Using all the information we have, we can find the explicit rule:

a_{n} = a_{1} • r^{n-1}

  • a_{1} = 32
  • r = \frac{1}{2}

a_{n} = a_{1} • r^{n-1} →

\boxed{a_{n} = 32 • (\frac{1}{2})^{n-1}}

________________________________

We can test that this works by substituting the number location of the term you want to find.

For instance:

a_{1} = 32 • (\frac{1}{2})^{1-1}

a_{1} = 32 • (\frac{1}{2})^{0}

a_{1} = 32 • 1

a_{1} = 32

a_{6} = 32 • (\frac{1}{2})^{6-1}

a_{6} = 32 • (\frac{1}{2})^{5}

a_{6} = 32 • \frac{1}{32}

a_{6} = 1

3 0
2 years ago
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