Original price: $32 Sale Price: $8 So what is the percent if discount?

Freddy’s savings account went from $520 to $321 after he purchased a new bike. What was the percent decrease?

Vinil7 [7]

Step-by-step explanation:

The bike cost $199 to get $199 you do a number line then label 520 then at the end label 321 so then you do -20 then -100 then -50 then -20 then -9. so now you add them up 20+100+50+20+9 which equals =199

so bike cost $199

then 520%×199= 1034.8

or 520%×321=1669.2

so the percentage is 1669.2 or 1034.8 probably 1669.2 but they bot count.

5 0

2 years ago

A caterer prepared a turkey that weighed 22 1/2 pounds.Each serving of the turkey will be 1/3 pound.How many whole servings will

AlladinOne [14]

1/2 1/2 1/2 1/3
$1 \times 2 = 3$

8 0

2 years ago

Read 2 more answers

Ricky is filling an 8 inch by 4 inch by 4 1/2 inch rectangular box with packing peanuts. The peanuts are cubes that come in two

RideAnS [48]

First find the volume of the box: 8•4•4.5=144 cubic inches.
8 of the 1/8 inch=1 of the 1 cubic inch
He could do 144 and 0
143 and 8
142 or 16
Or just keep following the pattern... Hope that helped

8 0

2 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

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

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

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

$=5x^4y$

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

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

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