0=-4t^2+20t+1 i need to figure out wat t is

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0=-4t^2+20t+1 i need to figure out wat t is

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Have you learned the quadratic formula?
yea but dat aint working it would be a=4 b=-20 c=1 can u try
a = -4 b = 20 c = 1

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Other answers:

yea
im trying 2 figure dis out
(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???
im stuck on the step with the quadratic formula
So what is the initial velocity?
Plug that in for v. What is the height above the ground? Plug that in for c.
v is 80
And c?
by-bye polpok
c is 4
0=-16^2+80t+4 simplified its 0=-4t^2+20t+1
Yep.
but i need t
Ok so using the formula for the quadratic, what is a, b, and c.
\[0 =at^2 + bt + c \implies t = \frac{-(b) \pm \sqrt{(b)^2 - 4(a)(c)}}{2(a)}\]
a=-4 b=20 c=1
yea but idk y i keep getting really big numbers wat do u get
t has 2 be 5.1s or 1s or 2s,3s
I got -0.049509756796392246, and 5.049509756796392
Are you forgetting to divide by -8?
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
Start with the part under the square root
so dont put -20 first
20*20 - 4*(-4)*(1)
What do you get for that part
416
Now take the square root of that
20.3
Ok, now you have 2 solutions: \[x = \frac{-20 \pm 20.3}{2(-4)}\]
Oh, and you didn't round up.. I think it's 20.396 or 20.4
im still getting 50.75
-20 + 20.4 = ? -20 - 20.4 = ?
2(-4) = ?
.4
-8
ok so u dont multiply the -20 to the 20.4/-8
No, you add and subtract them
(-20 plus or minus 20.4) / -8
i got -0.5
There's one more possibility.
nope i got -17.45
and -22.55
(-20 - 20.4)/-8
-0.05 is one option.
But the other one is positive
ok i got u now wow thanks but its weird how i cant put the whole thing in the calculator
Because the order matters
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.
ok can u help me wit another one
probably
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
So how would you set this one up?
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
i got 100t=-16t^2+80t+4
it's just 100, h is the height.
so \[100 = -16t^2 + 80t + 4\]
ok wat next
move the 100 to the other side so it equals 0 and then use the quadratic formula again.
by subtracting 100 from both sides that is
0=-4t^2+20+24
4-100 is -196, not positive
i got 3
err -96 rather
well den dis aint working now idk u do it
You did it right, I don't know how, but 3 is one answer. There is another answer
If you used -4t^2+20+24 You should have gotten (-1 and 6) though
But the correct formula -4t^2+20t-24 = 0 gives one of the solutions being 3 and the other being 2
the answer can only be 1s or 2s,3s
show me ur formula
\[0=-4t^2 + 20t - 24 \implies t = \frac{-20 \pm \sqrt{20^2 - 4(-4)(-24)}}{2(-4)}\]
ok i got dat but now getting big numbers again
Do the part under the square root first
\(20^2 - 4(-4)(-24)\)
16
and the square root of 16 is?
4
Ok, and what's the denominator? 2(-4)
-8
polpak, i have some confusion. check your equation once again
So you have two solutions: (-20 + 4)/-8 and (-20 - 4)/-8
for the first one you have -16/-8 and the second is -24/-8
omg thanks
Hrm India?
oh whoopse
No.. wait, that's right
ok linda dis is my question create ur own i need help 2 get off mines
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 \[-4t ^{2}-20t-24 \]
ah. ok
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
can u check the answer i get
i cant do \[\sqrt{-21}\]
so how do i do dis problem
it is a imaginary number!!!! it will come \[i \sqrt{21}\]
does dat equal -1
or 1
\[-1 = i ^{2}\]
i=1 right
no, it ain't right. i is nothing but an imaginary number. it can not be solved. it has to be written i only.
well it has 2 be possible cuz there is no i answer on dis paper
one thing, time can't be negative.
your solution seems wrong!!!
i have 0=-4t^+5-1
\[-4t ^{5}-1\]
do you want to write this equation???
my bad 0=-4^2+5-1
i know the answer has 2 be 1 or 1.65 but idk y it aint working
still something wrong in the equation. there is no "t"
0=-4t^2+5t-1
yes, the equation is giving t = 1 and o.5
how uget dat did u use the quadratic formula
hmm, yes
how cuz i got -41/-8 + or - 5
you must be wrong cuz i used calculator to solve.
i did 2 show me wat u put in
\[\frac{-b \pm\sqrt{b^2-4ac}}{2a}\]
i know dat wat numbers did u end up wit
1 and 0.5
will be the answer
no i mean like wat was b^2-4ac
0=-4t^2+5t-1 put a = -4 b = 5 and c = -1
i did i keep getting -41 on top
\[b ^{2}-4ac = 5^{2}-4(-4)(-1)\]
you will get 9
i was putting -5^2
-5^2 will also be 25. the sign doesn't matter when there is a square
thanks
my pleasure!!!
0=-16t^2+27+1 need some help

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