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cricket20 [7]
3 years ago
6

In Jana's closet she has 5 pairs of black shoes and 11 pairs of leather shoes. Two of her pairs of shoes are both black and leat

her. Determine the total number of shoes that are either black, leather, or both.
A) 5 pairs
B) 11 pairs
C) 12 pairs
D) 14 pairs
Mathematics
2 answers:
Mila [183]3 years ago
8 0
I just did the test 14 is the answer

iren2701 [21]3 years ago
8 0

Answer: D) 14 pairs

Step-by-step explanation:

Let B represents black shoes and L represents leather shoes.

Then, n(B)=5

n(L)=11

n(B\cap L)=2

Then the total number of shoes that are either black, leather, or both. is given by :-

n(B\cup L)=n(B)+n(L)-n(B\cap L)=5+11-2=14

Hence, the total number of shoes that are either black, leather, or both =14

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Answer:

14 weeks. the equation would be a.

Step-by-step explanation:

120-50=70   70/5=14 it will take him 14 weeks to get 120 if he saves 5 a week on top of the additional 50 he already has.

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(x +y)^5<br> Complete the polynomial operation
Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

\left(x+y\right)^5

\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i

a=x,\:\:b=y

=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i

so expanding summation

=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5

solving

\frac{5!}{0!\left(5-0\right)!}x^5y^0

=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5

=1\cdot \:1\cdot \:x^5

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also solving

=\frac{5!}{1!\left(5-1\right)!}x^4y

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similarly, the result of the remaining terms can be solved such as

\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2

\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3

\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4

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so substituting all the solved results in the expression

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\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

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