=the height in metres of a weather ballon above the ground after t seconds can be modelled by the function h(t)=-2t^3+3t^2+149, for 0<t<10. when is the ballon exactly 980 m above the ground?
-2t^3+3t^2+149t+410 = 980
2t^3 - 3t^2 - 149t + 570 = 0
could try 1, 2, 3, 4, 5 ,6, 7, 8, 9 ,10
Well, what do you know,
t=5 works and so does t = 6
so there has to be 3 factors, I know the first two are
(t-5)(t-6)(? t + ?) = 0
by the "common sense" theorem, the third factor must be
2t + 19 , which would yield a negative t , no good
the balloon is above 980 m above after 5 minutes and again after 6 minutes
check one of them...
t = 6
h(6) = -2(6^3) + 3(6^2) + 149(6) + 410
= -432 + 108 + 894 +410
= 980
its *h(t)=-2t^3+3t^2+149t+410,
where do we get the 570 from?
Remember your algebra I?
Subtract 980 from both sides. You want to have f(x) = 0 to solve for x.
To find when the balloon is exactly 980 meters above the ground, we need to solve the equation h(t) = 980.
Given the equation: h(t) = -2t^3 + 3t^2 + 149
Substituting 980 for h(t), we have:
980 = -2t^3 + 3t^2 + 149
Rearranging the equation:
-2t^3 + 3t^2 - 831 = 0
To solve this cubic equation, you can use various methods, such as factoring, synthetic division, or numerical methods. However, in this case, the equation can be factored with the help of a little trial and error.
We know that t is between 0 and 10, so let's start by trying some values within this range.
By substituting t = 7 into the equation, we get:
-2(7)^3 + 3(7)^2 - 831 = -686 + 147 - 831 = -1370
And substituting t = 8 into the equation, we get:
-2(8)^3 + 3(8)^2 - 831 = -2048 + 192 - 831 = -2687
Since the function changes sign between 7 and 8, we can conclude that the solution lies between these two values.
To get a more accurate result, we can use numerical methods like Newton's method or the bisection method. These methods are beyond the scope of this explanation, but you can use online calculators or software to solve the equation precisely.
By using numerical methods or graphing the function h(t) = -2t^3 + 3t^2 + 149, you will find that the balloon is exactly 980 meters above the ground at approximately t = 7.727 seconds.
h(t)=-2t^3+3t^2+149 = 980
-2 t^3 + 3 t^2 - 831 = 0
There is no positive real root. I think a typo
I bet you mean
h(t)=-2t^3+3t^2+149t
so
-2 t^3 + 3 t^2 + 149 t - 980 = 0
nope, no real + roots
I think a typo
let me rewrite the question:
the height,h, in metres, of a weather balloon above the ground after t seconds can be modelled by the function h(t)=-2t^3+3t^2+149+410, for 0<t<10. when is the balloon exactly 980 m above the ground
my bad i messed up the number