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? (7, 3) (0, 7) (7, 0) (3, 7)

The answer to your question is (7, 0)

Step-by-step explanation:

Data

A (7, 3)

B (0, 7)

C (7, 0)

D (3, 7)

Process

1.- Substitute each point in the equation to find the answer

Point A           y - 3 = m(x - 7)

Point B           y - 7 = m(x - 0)

Point C           y - 0 = m(x - 7)

Point D           y - 7 = m(x - 3)

2.- Conclusion

The only possibility is that m = 3 and the answer will be the third point (7,0)

The point Harold used is (7, 0) to represent point-slope form equation.

where

m = slope

x₁ = coordinates on the x axis

y₁ = coordinates on the y axis

Harold correctly wrote the equation as follows using a point-slope form;

y = 3(x - 7)

The point Harold used is (7, 0).

y - 0 = m(x - 7)

y = m(x - 7)

m = 3

y = 3(x - 7)

learn more on linear function here: link

maybe showing the graph would

step-by-step explanation:

-387129 iq

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v