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blsea [12.9K]
3 years ago
8

Alaina is selling wreaths and poinsettias for her chorus fundraiser. Wreaths cost $27 each poinsettia cost $20 each. If she sold

15 poinsettias and made $543, how many wreaths did she sell?
Mathematics
1 answer:
marishachu [46]3 years ago
4 0

Answer:

The number of wreaths Alaina sells is 9 .

Step-by-step explanation:

As given

Alaina is selling wreaths and poinsettias for her chorus fundraiser.

Wreaths cost $27 each poinsettia cost $20 each.

If she sold 15 poinsettias and made $543.

Let us assume that the number of wreaths Alaina sells = x

As the number of poinsettias  Alaina sells = 15

Than the equation becomes

27x + 15 × 20 = 543

27x + 300 = 543

27x  = 543 - 300

27x  = 243

x = \frac{243}{27}

x = 9

Therefore the number of wreaths Alaina sells is 9 .

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Answer:

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C. PEMDAS - parentheses, exponents, multiplication, division, addition, subtraction

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4 0
3 years ago
Find the length of Arc YWV, if line XZ=12 meters. (Round to the nearest hundredth)
Oksana_A [137]

Answer:

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Step-by-step explanation:

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3 years ago
Someeee one?????????????????
Ad libitum [116K]

Answer:

Option B) a_{n} = 2\cdot 4^{n-1}

Step-by-step explanation:

The given geometric sequence is

2, 8, 32, 128,....

The general form of a geometric sequence is given by

a_{n} = a_{1}\cdot r^{n-1}

Where n is the nth term that we want to find out.

a₁ is the first term in the geometric sequence that is 2

r is the common ratio and can found by simply dividing any two consecutive numbers in the sequence,

r=\frac{8}{2} = 4

You can try other consecutive numbers too, you will get the same common ratio

r=\frac{32}{8} = 4

r=\frac{128}{32} = 4

So the common ratio is 4 in this case.

Substitute the value of a₁ and r into the above general equation

a_{n} = 2\cdot 4^{n-1}

This is the general form of the given geometric sequence.

Therefore, the correct option is B

Note: Don't multiply the first term and common ratio otherwise you wont get correct results.

Verification:

a_{n} = 2\cdot 4^{n-1}

Lets find out the 2nd term

Substitute n = 2

a_{2} = 2\cdot 4^{2-1} = 2\cdot 4^{1} = 2\cdot 4 = 8

Lets find out the 3rd term

Substitute n = 3

a_{3} = 2\cdot 4^{3-1} = 2\cdot 4^{2} = 2\cdot 16 = 32

Lets find out the 4th term

Substitute n = 4

a_{4} = 2\cdot 4^{4-1} = 2\cdot 4^{3} = 2\cdot 64 = 128

Lets find out the 5th term

Substitute n = 5

a_{5} = 2\cdot 4^{5-1} = 2\cdot 4^{4} = 2\cdot 256 = 512

Hence, we are getting correct results!

6 0
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How do work this out
BabaBlast [244]
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(5r^2 - 5r) + (7r - 7)

Factor both groups.

5r(r-1) + 7(r-1)

The factors of (r-1) can be added together to get the answer.

(r-1)(5r+7)
6 0
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