6. A 70 kg pole-vaulter converts the kinetic energy, of running at ground level, into
potential energy to clear the crossbar at a height of 4 m above the gound. What is the
minimum velocity that he must have had at takeoff (while on the ground) to clear the bar
(minimum would be when velocity at the top was zero - you could still be moving
forward at the top, but this would require more velocity at the beginning)?

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.

the correct answers i think are b. and d.

give me brainliest if it is correct!

really really sorry if i'm

-12 -5(x + 4) = 3x

-12 - 5x - 20 = 3x

-12 - 20 - 5x = 3x

-32 - 5x = 3x

    + 5x   + 5x

-32 = 8x

8         8

- 4= x