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?

Point used by Harold was:

(7, 0)

Step-by-step explanation:

Given that

Equation of linear function used by Harold:

We know that linear equation in point slope form can be represented as:

Where are the coordinates of a given point.

is the slope of line.

Formula for Slope, m is given as:

Where and are the two points on the line.

If slope and a point with coordinates is know, the equation of a line can be represented in linear form as:

....... (1)

Now, the given equation is:

Re-writing the equation with a slight modification:

Now, comparing the above equation with equation (1):

We get that:

So, the point used by Harold is (7, 0).

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse

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