A student estimated based on the video that the ball left my hand 1.65 meters off the ground, and after 0.58 seconds the ball reached the maximum height of 3.26 meters. Use this information to find an equation of the form h = a ( t − t 1 ) 2 + h 1 for the height of the ball, in meters, after t seconds. h =

9514 1404 393

  h = -4.79(t -0.58)^2 +3.26

Step-by-step explanation:

The coordinates (t1, h1) are the time and height at the maximum. Then 'a' can be found from ...

  h = a(t -t1)^2 +h1

  1.65 = a(0 -0.58)^2 +3.26

  -1.61 = 0.3364a . . . . . subtract 3.26

  -4.786 = a . . . . . . . divide by the coefficient of a

The equation is ...

  h = -4.79(t -0.58)^2 +3.26

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.

65

step-by-step explanation: