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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?

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)

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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

mars1129 [50]

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,

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

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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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3 years ago