A ball is thrown into the air from height 1.6 m above the ground. It initially gains height at 16.4 ms-1. After 2 seconds, it is falling at 3.2 ms-1. The height of the ball over time can be
modelled by the quadratic function h(t) = at+ bt+c metres, where t is the time in seconds.
a State h(0) and hence find c.
b State h'(0) and hence find b.
c Use the remaining information to find a.
d What was the maximum height reached by the ball, and when did it reach this height?

answer: the 4 kilogram bag

step-by-step explanation: this is because if you divide the number of kilograms into the total it will tell you how much each individual kilogram would be worth. 2 kilos is 7 and change 3 would be around 6 ish and the 4 kilo would be 4.6 and change and the 5 kilos would be like 4.88

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

t= 1/3.

hope this .

step-by-step explanation: