Question

applications of quadratics Exploring Non-Linear Systems and Growth question

Answer 1

@thomaster @hartnn do either of you know how to do this?

Answer 2

replace \(s\) by \(100\) and \(v\) by \(60\)

Answer 3

for part a?

Answer 4

yes

Answer 5

so i it -16t^2+60t+100

Answer 6

*is

Answer 7

yes

Answer 8

Ok what about part B

Answer 9

you know how to find the maximum value of a quadratic?

Answer 10

in this case it has a max because the leading coefficient is negative
(it is \(-16\))
the max is the second coordinate of the vertex

Answer 11

the first coordinate of the vertex of \(y=ax^2+bc+c\) is always \(-\frac{b}{2a}\)
in your case this is \(-\frac{60}{2\times (-2)}=\frac{15}{8}\)

Answer 12

the second coordinate of the vertex is what you get when you replace \(t\) by \(\frac{15}{8}\)

Answer 13

would it go on and be \[100-\frac{ 15 }{ 8 }\] @satellite73

Answer 14

@radar @robtobey could any of you help me finish this problem