1/[1+ sqrt3- sqrt5]
First multiply both top and bottom by
1 + sqrt 3 + sqrt5
and you get
[1 + sqrt 3 + sqrt5]/(1 + 2sqrt3 + 3 - 5)
=[1 + sqrt 3 + sqrt5]/(2sqrt 3 -1)
Now multiply top and bottom by (2sqrt3 +1) to get rid of the radical in the denominator. The new denominator will be (2sqrt3 +1)(2sqrt3 -1)= 11.
Rationalize the denminator in the expression:
1/(1+(sqrt of 3)-(sqrt of 5))
and thenn...
A=2PIr^2 + 2PIrh
*solve for positive r
**PI is equal to 3.14etc.
5 answers
A = 2 pi(r^2 + rh)
A/(2 pi)= r^2 + rh
Treat this as a quadratic equation in r, with c = -A/(2 pi)
r^2 + rh + c = 0
r = [-h +/-sqrt (h^2-4ch)]/2
The - sign gives a negative r, so ignore it.
r = [-h + sqrt(h^2+2Ah/pi)]/2
A/(2 pi)= r^2 + rh
Treat this as a quadratic equation in r, with c = -A/(2 pi)
r^2 + rh + c = 0
r = [-h +/-sqrt (h^2-4ch)]/2
The - sign gives a negative r, so ignore it.
r = [-h + sqrt(h^2+2Ah/pi)]/2
I was following your work until I noticed that the answer you got is not the answer i'm suppose to get. Our teacher gave us the problem and the answer and we just need to come up with the work inbetween. The answer is (-pih+sqrt(pi^2h^2+2piA))/(2pi)
Is there another way to solve it other than what you did before?
Is there another way to solve it other than what you did before?
Well, theoretically speaking all other methods of solving this equation to derive r positive would come from this formula
whqt is the answer for i am a number .go round the corners of the given figure times.when you my value with the number of corners you have crossed you get 46
1 answer