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Mathematics 7 Online
OpenStudy (kaylak12345):

help

OpenStudy (kaylak12345):

OpenStudy (kaylak12345):

@sooobored

OpenStudy (kaylak12345):

OpenStudy (kaylak12345):

OpenStudy (kaylak12345):

1. is still 5/4 right

OpenStudy (sooobored):

is there any additional information above question 1, that gives the height of the building?

OpenStudy (kaylak12345):

you haveto come up with a height of a building

OpenStudy (sooobored):

oh, then what height did you decide on?

OpenStudy (kaylak12345):

25

OpenStudy (kaylak12345):

ft

OpenStudy (sooobored):

ok, from this point on h0 =25ft for the rest of the problem then 1. yes 5/4 would be correct

OpenStudy (kaylak12345):

yay

OpenStudy (sooobored):

wont help you graph, 2. -16t^2 since its negative it opens downwards 16t^2 would open upwards we already went over this

OpenStudy (sooobored):

you should remember this type of parabola behavior as its very useful in future problems

OpenStudy (sooobored):

ok 3. \[h=-16t^2 +v_0 t + h_0\] vo is given as 150 upwards so vo=150 we defined h0 = 25 from your first problem so plug those in and you're get your equation

OpenStudy (kaylak12345):

-16t^2+150t+25?

OpenStudy (sooobored):

dont forget to put h= infront of it but yes

OpenStudy (sooobored):

3b. you set h=0 since h=0 is the height of the ground then solve for t \[0=-16t^2+150 t +25\]

OpenStudy (kaylak12345):

-16t^2 =-25? 150t=187.5

OpenStudy (sooobored):

uhhhh i have no idea what you just did but it doesnt look right you need to factor the binomial

OpenStudy (sooobored):

or use the quadratic formula to determine t

OpenStudy (kaylak12345):

9.54

OpenStudy (sooobored):

sorry, i honestly dont know since its not something i can do in my head

OpenStudy (kaylak12345):

oh well i did it in a quadratic calcultor thats what it gave me

OpenStudy (sooobored):

should be about 9.5388

OpenStudy (sooobored):

so yes

OpenStudy (kaylak12345):

now i have to make a table and thats what it gave me but we usually cut off after 2 numbers

OpenStudy (sooobored):

you can make the table easily yourself, just pick t values and get the solution by "plugging in" skipping to 3f

OpenStudy (sooobored):

why cant you have negative t and h(t) values?

OpenStudy (kaylak12345):

It wouldnt make sense because the point of fireworks is to go up and negative height wouldnt make any sense at all. Its how tall something is

OpenStudy (sooobored):

on the right track, h=1 means 1 ft off the ground h=-1 ,means 1 ft under the ground doesnt make a lot of sense now what does a negative t mean?

OpenStudy (kaylak12345):

-t means less time you cant subtract time

OpenStudy (sooobored):

sorta of, it means going back in time from an event

OpenStudy (sooobored):

and time travel is not possible yet

OpenStudy (kaylak12345):

lol

OpenStudy (sooobored):

ok quick overview of 4. its asking you to change vo so instead of using 150 like in problem 3, you can choose something like 80

OpenStudy (kaylak12345):

so any number

OpenStudy (kaylak12345):

ok 81 it is

OpenStudy (sooobored):

sorry, trying to think of a nice number that would work

OpenStudy (sooobored):

399 is a very nice number

OpenStudy (kaylak12345):

so i have to pick a firework?

OpenStudy (sooobored):

4. is choosing another initial velocity, which was vo=150 ft/s in problem 3

OpenStudy (kaylak12345):

ok so 79 then so how would i set this up the same way in question 3

OpenStudy (sooobored):

no no use 399, its easier

OpenStudy (kaylak12345):

ok how would i set this up to solve?

OpenStudy (sooobored):

h=-16t^2 +399 t +25 look familiar?

OpenStudy (kaylak12345):

ok and yes

OpenStudy (sooobored):

oh, actually 399 doesnt help as much since you dont need to calculate when it lands on the ground

OpenStudy (sooobored):

anyways, 4b is asking when t=3, what is the height? so plug in 3

OpenStudy (sooobored):

are you familiar with how to find the symmetry point?

OpenStudy (kaylak12345):

kind of

OpenStudy (sooobored):

ok quick run through then

OpenStudy (sooobored):

you're familiar with the quadratic formula \[\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

OpenStudy (kaylak12345):

yes

OpenStudy (sooobored):

we can divide that into 2 parts \[\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}\]

OpenStudy (kaylak12345):

144+1197+25 i think im off on the height for 4b

OpenStudy (sooobored):

now parabola are always symmetrical so the center line "x" is always \[\frac{-b}{2a}\]

OpenStudy (sooobored):

and the other component of the quadratic equation, is the distance the roots are away from the symmetry point

OpenStudy (sooobored):

-144

OpenStudy (sooobored):

all other numbers are correct, so just add them all together

OpenStudy (kaylak12345):

1222 for b

OpenStudy (sooobored):

no, 1197+25-144

OpenStudy (kaylak12345):

oh crap

OpenStudy (kaylak12345):

1078

OpenStudy (sooobored):

right well hopefully you can do the rest on your own since i gotta run

OpenStudy (kaylak12345):

ok thank you for the help

OpenStudy (sooobored):

though ive covered everything you need to finish the rest

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