What is the average of 3.28+6.90+5.30+12.00+4.77

A store can buy 8 cartons of milk for $24 or 7 cartons of milk for $28.

USPshnik [31]

Answer:

A. $1 less

Step-by-step explanation:

24 divided by 8 (unit rate) = $3

28 divided by 7 = $4

4-3=1

3 0

2 years ago

6 is added to the product of 7 and a number

notka56 [123]

The value of the expression will be equal to 55.

<h3>What is an expression?</h3>

The mathematical expression combines numerical variables and operations denoted by addition, subtraction, multiplication, and division signs.

Mathematical symbols can be used to represent numbers (constants), variables, operations, functions, brackets, punctuation, and grouping. They can also denote the logical syntax's operation order and other properties.

Given that 6 is added to the product of 7 and a number 7. The expression will be written as,

E = 7 x 7 + 6

E = 49 + 6

E = 55

Therefore, the value of the expression will be equal to 55.

To know more about an expression follow

brainly.com/question/11847301

#SPJ2

5 0

1 year ago

Read 2 more answers

H+3h+4h=100 how do i get h?

Dominik [7]

Answer:

h = 12.5

Step-by-step explanation:

Simplifying

h + 3h + 4h = 100

Combine like terms: h + 3h = 4h

4h + 4h = 100

Combine like terms: 4h + 4h = 8h

8h = 100

Solving

8h = 100

Solving for variable 'h'.

Move all terms containing h to the left, all other terms to the right.

Divide each side by '8'.

h = 12.5

Simplifying

h = 12.5

7 0

3 years ago

Read 2 more answers

When combining like terms are x and 3x the same

Maurinko [17]

x+3x= 4x

Because there is a invisable 1 in front of the X. Hope this helps!

8 0

3 years ago

Using the pattern, give the coefficients of (x + y)^5 and (x + y)^6

musickatia [10]

Answer:

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

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

Step-by-step explanation:

In order to find the values of $(x+y)^5$ and $(x+y)^6$ , you need to apply the binomial theorem (high-level math you most likely don't need to worry about, it's easier than multiplying all the binomials together).

$(x+y)^5$ = $\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$ = $\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$ = $x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$ .

$(x+y)^6$ = $\sum _{i=0}^6\binom{6}{i}x^{\left(6-i\right)}y^i$ = $\frac{6!}{0!\left(6-0\right)!}x^6y^0+\frac{6!}{1!\left(6-1\right)!}x^5y^1+\frac{6!}{2!\left(6-2\right)!}x^4y^2+\frac{6!}{3!\left(6-3\right)!}x^3y^3+\frac{6!}{4!\left(6-4\right)!}x^2y^4+\frac{6!}{5!\left(6-5\right)!}x^1y^5+\frac{6!}{6!\left(6-6\right)!}x^0y^6$ = $x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$ .

5 0

3 years ago