To find the roots of the function, set y = 0. The equation is 0 = 4x2 + 2x – 30.

Factor out the GCF of .

x = - 3, x =

Step-by-step explanation:

Given

4x² + 2x - 30 = 0 ( divide through by 2 )

2x² + x - 15 = 0

Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 2 × - 15 = - 30 and sum = + 1

The factors are + 6 and - 5

Use these factors to split the x- term

2x² + 6x - 5x - 15 = 0 ( factor the first/second and third/fourth terms )

2x(x + 3) - 5(x + 3) = 0 ← factor out (x + 3) from each term

(x + 3)(2x - 5) = 0

Equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

2x - 5 = 0 ⇒ 2x = 5 ⇒ x =

when rotating a figure 180 degrees, imagine that you are able to take the figure and turn it clockwise or counterclockwise around a center point. to rotate a figure 180 degrees, you will need to apply the rule (x, y) → (-x, -y). start by using a coordinate grid with coordinates for each vertex of the figure. the center point of the coordinate grid is located at (0, 0), which is what you will rotate the figure around. write down the original coordinates of the shape you are going to rotate. then, apply the rule. for example, coordinate (1, 2) becomes (-1, -2). another example would be, coordinate (-4, -2) becoming (4, 2). after you change each original coordinate to the rotated coordinate, you will draw your new figure.

step-by-step explanation:

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