The equation of a linear function in point-slope form is y – y1 = m(x – x1). harold correctly wrote the equation y = 3(x – 7) using a point and the slope. which point did harold use? when harold wrote his equation, the point he used was (7, 3). when harold wrote his equation, the point he used was (0, 7). when harold wrote his equation, the point he used was (7, 0). when harold wrote his equation, the point he used was (3, 7).

For this case we must find the point that Harold used to arrive at the following equation:

Starting from the fact that the equation of the point-slope form of a line is given by:

If we compare the standard equation with Harold's, we see that the slope of the line is

In addition, it is observed that and

Then, the correct option is: Harold used the point (7,0)

When Harold wrote his equation, the point was used (7,0).

step-by-step explanation:

from the graph attached below:

domain of graph is the value of x for which the given function profit function p(x) is defined.

(a) the domain represents the number of years for which the outlet will earn a profit.

range of profit function defines the lowest to highest value of profit earned.

(b) the range of this function is (0 , 160 ). for this function, the range represents the outlet's profit earned.

the answer is b.

step-by-step explanation:

identity property is where 0+whatever number=whatever number. basically 0+ a number.