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OpenStudy (anonymous):
Have you learned the quadratic formula?
OpenStudy (anonymous):
yea but dat aint working it would be a=4 b=-20 c=1 can u try
OpenStudy (anonymous):
a = -4
b = 20
c = 1
OpenStudy (anonymous):
yea
OpenStudy (anonymous):
im trying 2 figure dis out
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OpenStudy (anonymous):
(h(t)=-16t^2+vt+c where h is the approximate height in ft t is the time in motion in seconds v is the initial upward velocity in ft per second and c is the initial height in ft)
A baseball player hits a pitched ball when it is 4ft above the ground the initial upward velocity is 80ft/s how long will it take for the ball to hit the ground???
OpenStudy (anonymous):
im stuck on the step with the quadratic formula
OpenStudy (anonymous):
So what is the initial velocity?
OpenStudy (anonymous):
Plug that in for v. What is the height above the ground? Plug that in for c.
OpenStudy (anonymous):
v is 80
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OpenStudy (anonymous):
And c?
OpenStudy (anonymous):
by-bye polpok
OpenStudy (anonymous):
c is 4
OpenStudy (anonymous):
0=-16^2+80t+4 simplified its 0=-4t^2+20t+1
OpenStudy (anonymous):
Yep.
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OpenStudy (anonymous):
but i need t
OpenStudy (anonymous):
Ok so using the formula for the quadratic, what is a, b, and c.
OpenStudy (anonymous):
\[0 =at^2 + bt + c \implies t = \frac{-(b) \pm \sqrt{(b)^2 - 4(a)(c)}}{2(a)}\]
OpenStudy (anonymous):
a=-4
b=20
c=1
OpenStudy (anonymous):
yea but idk y i keep getting really big numbers wat do u get
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OpenStudy (anonymous):
t has 2 be 5.1s or 1s or 2s,3s
OpenStudy (anonymous):
I got -0.049509756796392246, and 5.049509756796392
OpenStudy (anonymous):
Are you forgetting to divide by -8?
OpenStudy (anonymous):
tell me exactly how 2 put it in my calculator cuz everytime i do i get large #'s and yes i put divide y -8
OpenStudy (anonymous):
Start with the part under the square root
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OpenStudy (anonymous):
so dont put -20 first
OpenStudy (anonymous):
20*20 - 4*(-4)*(1)
OpenStudy (anonymous):
What do you get for that part
OpenStudy (anonymous):
416
OpenStudy (anonymous):
Now take the square root of that
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OpenStudy (anonymous):
20.3
OpenStudy (anonymous):
Ok, now you have 2 solutions:
\[x = \frac{-20 \pm 20.3}{2(-4)}\]
OpenStudy (anonymous):
Oh, and you didn't round up.. I think it's 20.396 or 20.4
OpenStudy (anonymous):
im still getting 50.75
OpenStudy (anonymous):
-20 + 20.4 = ?
-20 - 20.4 = ?
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OpenStudy (anonymous):
2(-4) = ?
OpenStudy (anonymous):
.4
OpenStudy (anonymous):
-8
OpenStudy (anonymous):
ok so u dont multiply the -20 to the 20.4/-8
OpenStudy (anonymous):
No, you add and subtract them
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OpenStudy (anonymous):
(-20 plus or minus 20.4) / -8
OpenStudy (anonymous):
i got -0.5
OpenStudy (anonymous):
There's one more possibility.
OpenStudy (anonymous):
nope i got -17.45
OpenStudy (anonymous):
and -22.55
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OpenStudy (anonymous):
(-20 - 20.4)/-8
OpenStudy (anonymous):
-0.05 is one option.
OpenStudy (anonymous):
But the other one is positive
OpenStudy (anonymous):
ok i got u now wow thanks but its weird how i cant put the whole thing in the calculator
OpenStudy (anonymous):
Because the order matters
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OpenStudy (anonymous):
Plus plugging the whole thing in at once makes it harder to find where you made your mistake. Just put in each piece at a time.
OpenStudy (anonymous):
ok can u help me wit another one
OpenStudy (anonymous):
probably
OpenStudy (anonymous):
using the same formula
a flare is launched from the deck of a life boat 4ft above the water surface the initial upward velocity is 80ft/s after how many seconds will the flare be 100 ft above the water surface
OpenStudy (anonymous):
So how would you set this one up?
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OpenStudy (anonymous):
h(t)=-16t^2+vt+c where h is the approximate height in ft t is the time in motion in seconds v is the initial upward velocity in ft per second and c is the initial height in ft
OpenStudy (anonymous):
i got 100t=-16t^2+80t+4
OpenStudy (anonymous):
it's just 100, h is the height.
OpenStudy (anonymous):
so \[100 = -16t^2 + 80t + 4\]
OpenStudy (anonymous):
ok wat next
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OpenStudy (anonymous):
move the 100 to the other side so it equals 0 and then use the quadratic formula again.
OpenStudy (anonymous):
by subtracting 100 from both sides that is
OpenStudy (anonymous):
0=-4t^2+20+24
OpenStudy (anonymous):
4-100 is -196, not positive
OpenStudy (anonymous):
i got 3
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OpenStudy (anonymous):
err -96 rather
OpenStudy (anonymous):
well den dis aint working now idk u do it
OpenStudy (anonymous):
You did it right, I don't know how, but 3 is one answer. There is another answer
OpenStudy (anonymous):
If you used -4t^2+20+24
You should have gotten (-1 and 6) though
OpenStudy (anonymous):
But the correct formula
-4t^2+20t-24 = 0 gives one of the solutions being 3 and the other being 2
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OpenStudy (anonymous):
\(20^2 - 4(-4)(-24)\)
OpenStudy (anonymous):
16
OpenStudy (anonymous):
and the square root of 16 is?
OpenStudy (anonymous):
4
OpenStudy (anonymous):
Ok, and what's the denominator? 2(-4)
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OpenStudy (anonymous):
-8
OpenStudy (anonymous):
polpak, i have some confusion. check your equation once again
OpenStudy (anonymous):
So you have two solutions:
(-20 + 4)/-8 and (-20 - 4)/-8
OpenStudy (anonymous):
for the first one you have -16/-8 and the second is -24/-8
OpenStudy (anonymous):
omg thanks
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OpenStudy (anonymous):
Hrm India?
OpenStudy (anonymous):
oh whoopse
OpenStudy (anonymous):
No.. wait, that's right
OpenStudy (anonymous):
ok linda dis is my question create ur own i need help 2 get off mines
OpenStudy (anonymous):
no, its alright now. i don't how but the latex code was not properly loaded and the \[-4t ^{2}+20t-24 \] was looking like \[-4t ^{2}-20t-24 \]
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OpenStudy (anonymous):
ah. ok
OpenStudy (anonymous):
a basketball player shoots the ball with an initial velocity of 20ft/s the ball is 6ft above the floor wen it leaves her hands how long will it take for the ball to reach the rim of the basket 10 ft above th floor on its way down
OpenStudy (anonymous):
can u check the answer i get
OpenStudy (anonymous):
i cant do \[\sqrt{-21}\]
OpenStudy (anonymous):
so how do i do dis problem
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OpenStudy (anonymous):
it is a imaginary number!!!! it will come \[i \sqrt{21}\]