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vitfil [10]
3 years ago
7

Of the 50 U.S. states, 4have names that start with the letter W . What percentage of U.S. states have names that start with the

letter W?
Mathematics
1 answer:
Verizon [17]3 years ago
4 0
8% of US states start with W
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Zen deposited some money in a savings account. The graph below shows the value of Zen's investment y, in dollars, after x months
kifflom [539]

The y-intercept of 1000 in the graph is at the point where x is equal to 0.

At x=0, it means no month have passed. It is the initial point where Zen deposited the money.

What is the amount (y axis)? 1000!

So y represents the initial deposit of Zen, which is 1000.


ANSWER: Answer choice D (Amount of money Zen deposited in the savings account)

6 0
3 years ago
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A book store offers a sale on a new book. The sale is "buy two of the books and get the third book free." Tommy pays $26 for thr
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Salvo en cada library 4.33

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3 years ago
Melanie's food bill at a restaurant is $28.45. The sales tax rate is 6%. If she added 20% of the amount for a tip, what is the t
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The total cost of the meal would be $36.19
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3 years ago
(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

=x^5

also solving

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

=\frac{5}{1!}x^4y

=\frac{5}{1!}x^4y

=\frac{5x^4y}{1}

=\frac{5x^4y}{1}

=5x^4y

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

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

so substituting all the solved results in the expression

=\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

=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

Therefore,

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

6 0
3 years ago
Simplify<br> 6^7 /6^5<br> leaving your answer in index form.
Agata [3.3K]

Answer:

36

Step-by-step explanation:

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