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anonymous
 one year ago
A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = 16t^2 + 32t + 6. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?
A) 1 s; 22 ft
B) 2 s; 22 ft
C) 2 s; 6 ft
D) b1 s; 54 ft
anonymous
 one year ago
A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = 16t^2 + 32t + 6. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height? A) 1 s; 22 ft B) 2 s; 22 ft C) 2 s; 6 ft D) b1 s; 54 ft

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1find the vertex of the parabola

freckles
 one year ago
Best ResponseYou've already chosen the best response.1oh wait your function isn't a parabola

freckles
 one year ago
Best ResponseYou've already chosen the best response.1and it should be interesting

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\text{ maybe you meant } h=16t^2+32t+6 ?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah sorry forgot the ^2

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[h=at^2+bt+c \\ h=a(t^2+\frac{b}{a}t)+c \\ h=a(t^2+\frac{b}{a}t+(\frac{b}{2a})^2)+ca(\frac{b}{2a})^2 \\ h=a(t+\frac{b}{2a})^2+ca (\frac{b}{2a})^2 \text{ is \in vertex form } \\ \text{ so the vertex is } (\frac{b}{2a},ca (\frac{b}{2a})^2)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so if a<0 then you have a max and if a>0 then you have a min we have the first case so \[ca(\frac{b}{2a})^2 \text{ the max } \\ \text{ and } \\ \frac{b}{2a} \text{ will be where the max occurs }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm kinda confused.. I don't really get it

freckles
 one year ago
Best ResponseYou've already chosen the best response.1do you need an example of writing in vertex form?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there's too much letters, I really got confused

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[h=2t^2+5t+5\] we are going to follow the same steps above: \[h=(2t^2+5t)+5 \\ h=2(t^2+\frac{5}{2}t)+5 \\ h=2(t^2+\frac{5}{2}t+(\frac{5}{2 \cdot 2})^2)+5 2 \cdot (\frac{5}{2 \cdot 2})^2 \\ h=2(t+\frac{5}{2 \cdot 2 })^2+52 (\frac{5 }{ 2 \cdot 2})^2 \\ \text{ so the vertex here is } (\frac{5}{2 \cdot 2 },52 (\frac{5}{2 \cdot 2})^2) \\ \text{ you can simplify } (\frac{5}{4},52 (\frac{25}{16})) \\ \text{ still simplifying } (\frac{5}{4},5\frac{25}{8}) \\ \text{ sill simplifying } (\frac{5}{4}, \frac{4025}{8}) \\ \text{ and finally } (\frac{5}{4},\frac{15}{8})\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1i used completing the square to do this if it wasn't obvious \[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[h=16t^2+32t+6 \\ h=(16t^2+32t)+6\] first step look at the terms in ( ) and factor out what is in front of your first term inside

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[h=16(t^22t)+6 \\ h=16(t^22t+?)+6+16(?) \text{ this step notice I added in a zero } \\ \text{ this zero it to keep the equation equivalent and to complete the square } \\ \text{ the zero I added in was of the form } 16(?)+16(?) \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1compare \[t^22t+? \text{ to the left hand side of } \\ t^2+kt+(\frac{k}{2})^2=(t+\frac{k}{2})^2\] what do we need to make k and therefore the ? mark is ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1oops I should have ended that question with punctuation because that might be confusing :p

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you know since I'm already using a ? for a constant
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