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Well cubic equations are not one-one functions in general. So it seems to me that your function cannot possibly have an inverse function unless you restrict the domain. If the domain IS restricted, there's probably a way but I feel that this requires solving a cubic equation, which I think isn't in our syllabus. Do you have a past paper question like this?
can u tell me the paper year ??
I) s=45, v=0, a= -10
v^2 = u^2 + 2as
0 = u^2 - 900
u = 30
ii) s = ut + 1/2*a*t^2
s = 30t - 5t^2
For ball to be above cliff:
s > 40
30t - 5t^2 > 40
-5t^2 + 30t - 40 > 0
t^2 - 6t + 8 < 0
(t-2)(t-4) < 0
2<t<4
So between 2nd and 4th second, the ball was above the cliff.
Length of time = 4-2 = 2s
iii) s = ut - 5t^2
ut - 5t^2 > 40
ut - 5t^2 - 40 > 0
5t^2 - ut + 40 < 0
Let's assume the two roots of this quadratic expression are α and β. Then we have:
α<t<β
β-α = 17
So how do we determine the difference of roots of a quadratic expression?
A general quadratic equation is:
ax^2 + bx + c = 0
With the roots:
x1 = [ -b + (b^2 - 4ac)^(1/2) ] / 2a
x2 = [ -b - (b^2 - 4ac)^(1/2) ] / 2a
x1 - x2 = (b^2 - 4ac)^(1/2) / a
So in our case,
β-α = [u^2 - 4(5)(40)]^(1/2) / 5
= (u^2 - 800)^(1/2) / 5 = 17
(u^2 - 800)^(1/2) = 85
u^2 - 800 = 7225
u^2 = 8025
u = 89.6
This last method may not be the best method, but I thought it was neat and no one else was answering so hope it makes sense
You got your drawing of a circle and a line with angle π/4, do you not get a right angle triangle ? If you do just use cos/sin π/4=x/|z|, where x is either the adjacent or opposite depending on the sketch, sorry I'm on mobile can't draw it for you maybe tommorow I do it
Fairly easy im just giving the logic you work the rest
musiclover gurl 937838 said:
Please try thinking about it:
It's not EVEN THOUGH, it's BECAUSE there is a Negative sign before the power that it it tends to zero. Qwertypoiu gave a detailed explanation about it hope it helps
Thank you! Got it!Please try thinking about it:
5^4 = 625
5^3 = 125
5^2 = 25
5^1 = 5
5^0 = 1
5^-1 = 1/5 = 0.2
5^-2 = 1/25 = 0.04
5^-3 = 1/125 = 0.008
5^-4 = 1/625 = 0.0016
We can conclude a few things from the pattern above:
1) if the power increases the final number also increases.
2) if the power decreases, the final number also decreases.
3) even if the power is negative, the final answer is positive.
4) switching the sign of the power is same as taking reciprocal
Anyhow e^inf is infinitely large. But for e^-inf, the power is infinitely small, so it's gonna be zero. Or you could think of it as 1/e^inf, which is 1/inf , or zero
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