Which ordered pair is a solution of the equation?

D, neither

Step-by-step explanation:

If you want to solve this through elimination, just substitute the given points in each choice. (a,b) → (x,y)

In a coordinate the first number is the x value, and the second is the y.

All you have to do is substitute the indicated values given in the coordinate into the point slope equation, and determine whether or not the equation is true. If the equation is true (for example 1 = 1) the order pair IS a solution. If the equation is false (for example 1 = 2), than the ordered pair IS NOT a solution.

If you want to look at this linear equation solving for the quantity y, then use slope intercept.

Choice 1 is incorrect because (5,4) →

y - 4 = 7(x - 6) → (4) - 4 = 7((5) - 6) →

0 = 7(-1) → 0 = -7 (0 is never equal to -7) ; contradictory

Choice 2 is incorrect because (6,5) →

y - 4 = 7(x - 6) → (5) - 4 = 7((6) - 6) →

1 = 7(0) → 1 = 0 (1 is never equal to 0) ; contradictory

Therefore Choice 3 must be incorrect as well because neither of these two pairs are solutions, so combining two pairs that do not satisfy the equation does not make it any more true.

Since Choice 4 is the remaining choice, it must be the correct answer.

The points that fit on this line are given by the slope intercept equation: y = 7x - 3.

To convert point slope form to slope intercept form, first distribute everything in the parenthesis. Then remove all the terms besides y from the left side of the equation to the right:

y - 4 = 7(x - 6) → y - 4 = 7(x) + 7(-6) →

y - 4 = 7x - 42 → y - 4 + 4 = 7x - 42 + 4 →

y = 7x - 38.

Just substitute the x or y value into the equation, and find the corresponding y or x variable that fits the line.

yes.. it's proven to be 36.

answer: 5

step-by-step explanation:

since, the range is the difference between the lowest and highest values of the given sample space.

and, here the sample space = { 3, 1, 0, 4, 1, 4, 2, 3, 0, 5, 0, 1, 1, 2, 0, 4, 1,4, 4 ,2 ,3 ,2 ,0 ,1 ,4 }

here, the highest value of the given sample = 5

and, the lowest value = 0

therefore, the range of the sample space = 5 - 0 = 5