(4n-3n^3)-(3n^3+4n) answer

Nicole sold 1 cherry pie and 9 pumpkin pies for $60. Lisa sold 11 cheery pies and 4 pumpkin pies for $90. what's the cost each o

zubka84 [21]

X= cost per cherry pie
y= cost per pumpkin pie

NICOLE
1x + 9y= $60

LISA
11x + 4y= $90

STEP 1
multiply Nicole's equation by -11

-11(1x + 9y)= -11($60)
multiply -11 by all terms

(-11 * x) + (-11 * 9y)= (-11 * 60)

-11x - 99y= -660

STEP 2
add Nicole's new equation from step 1 to Lisa's equation to solve for y (using the elimination method)

-11x - 99y= -660
11x + 4y= 90
the x terms "cancel out"

-95y= -570
divide both sides by -95

y= $6 per pumpkin pie

STEP 3
substitute y value into either original equation to solve for x

x + 9y= $60

x + 9(6)= 60

x + 54= 60
subtract 54 from both sides

x= $6 per cherry pie

CHECK
11x + 4y= $90
11(6) + 4(6)= 90
66 + 24= 90
90= 90

ANSWER: Each cherry pie costs $6 and each pumpkin pie costs $6.

Hope this helps! :)

3 0

3 years ago

12 times the number of feet

weqwewe [10]

Answer:

12 feet

Step-by-step explanation:

7 0

2 years ago

Jerry’s loan had a principal of $22,000. He made quarterly payments of $640 for nine years until the loan was paid in full. How

Leto [7]

Answer: $1040

A = P + I

A = Total

P = Principal

I = Interest

First find the total amount

A = 640(4)(9)

A = 23040

Plug In the Numbers

23040 = 22000 + I

Substract 22000 on both sides

I = $1040

8 0

2 years ago

Sam bought 2gallons of milk at the supermarket how many quarts of milk did he buy

Scilla [17]

2 gallons is equal to 8 Us quarts (i need 20 words to answer, that is what this is)

5 0

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

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

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

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

$=5x^4y$

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

2 years ago