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Katyanochek1 [597]
1 year ago
6

When you brush your wig will all the hair come off the wig completely or will it still be the same in a year

Mathematics
2 answers:
miv72 [106K]1 year ago
4 0

Answer:

no

Step-by-step explanation:

Tomtit [17]1 year ago
3 0
Yes, brush it… not too much though. Try a wide comb/wig brush.
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4 parts of water is to 1 part of glue

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Mary bought a 7lb bag of flour. She used 1/3 of the bag to bake wedding treats.
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she is left with 4.67 LB of flour.

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2 years ago
2.) You are buying gifts for 10 people. You decide to buy each person either a CD or a
Katena32 [7]

Answer:

a. The number of DVDs is 10 - c

b. The cost of the CDs is 12c

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d. The total cost of all gifts is 200 - 8c

Step-by-step explanation:

* <em>Lets explain how to solve the problem</em>

- You are buying gifts for 10 people

# You will buy 10 gifts

- The gifts are CD or DVD

# The total numbers of CDs and DVDs is 10

- The cost of each CD is $12 and the cost of each DVD is $20

a.

# c is the number of CDs

∵ You buy c CDs

∵ The total numbers of CDs and DVDs is 10 as we said up

∴ c + the numbers of DVDs = 10

- Subtract c from both sides

∴ The number of DVDs = 10 - c

* The number of DVDs is 10 - c

b.

∵ The number if CDs is c

∵ The cost of each CD is $12

∴ The cost of the CDs = 12 × c = 12c

* The cost of the CDs is 12c

c.

∵ The number of DVDs is 10 - c

∵ The cost of each DVD is $20

∴ The cost of the DVDs = 20(10 - c)

* The cost of the DVDs is 20(10 - c)

d.

∵ The cost of the CDs is 12c

∵ The cost of the DVDs is 20(10 - c)

- The total cost of all the gifts is the sum of the costs of CDs and DVDs

∴ The total cost of all gifts = 12c + 20(10 - c)

- Multiply the bracket by 20

∵ 20(10 - c) = 100(2) - 20(c) = 200 - 20c

∴ The total cost of all gifts = 12c + 200 - 20c

- Add like terms

∴ The total cost of all gifts = 200 - 8c

* The total cost of all gifts is 200 - 8c

3 0
2 years ago
Our school's girls volleyball team has 14 players, including a set of 3 triplets: Missy, Lauren, and Liz. In how many ways can w
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\displaystyle |\Omega|=\binom{14}{6}=\dfrac{14!}{6!8!}=\dfrac{9\cdot10\cdot11\cdot12\cdot13 \cdot14}{2\cdot3\cdot4\cdot5\cdot6}=3003\\ |A'|=\binom{11}{3}=\dfrac{11!}{3!8!}=\dfrac{9\cdot10\cdot11}{2\cdot3}=165\\ |A|=3003-165=2838\\\\ P(A)=\dfrac{2838}{3003}=\dfrac{86}{91}\approx95\%

3 0
3 years ago
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