An old-fashioned chinese restaurant offers a family dinner where you get to choose one dish from “column a” (which has 8 dishes), one dish from “column b” (which has 10 dishes) and one dish from “column c” (which has 5 dishes). how many different family dinners can be chosen?

 400

Step-by-step explanation:

Given : The number of dishes in column A = 8

The number of dishes in column B = 10

The number of dishes in column C = 5

Since for dinner we need to choose one dish from each column, then

The total number of combinations of dishes for dinner will be :-

Hence, the number of family dinners =400

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.

the square root of what number? if you add that, then someone will probably answer the question.