An object is thrown upward from a height of 6 feet at a velocity of 63 feet per second. t1 = 0, t2 = 3 (a) Use the position equation s = −16t2 + v0t + s0 to write a function that represents the situation. s = (c) Find the average rate of change of the function from t1 to t2.

(d) Describe the slope of the secant line through t1 and t2.
The slope of the secant line is positive.
The slope of the secant line is negative.

(e) Find the equation of the secant line through t1 and t2.
s =

(f) Graph the secant line in the same viewing window as your position function.

someone please help me i cant figure this out and its due soon

225x²+4=0

225x²=0-4

225x²=-4

x²=-4/225

x=±√-4/225

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