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.
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.
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
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?
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
To rationalize the denominator in the expression 1/(1 + √3 - √5), we need to eliminate the square roots from the denominator.
To do this, we'll use a method called rationalizing the denominator. The idea is to multiply both the numerator and the denominator by the conjugate of the denominator.
The conjugate of the denominator 1 + √3 - √5 is 1 + √3 + √5. Multiplying both the numerator and denominator by this conjugate will eliminate the square roots.
Here's the step-by-step process:
1. Multiply the numerator and denominator by the conjugate of the denominator:
(1/(1 + √3 - √5)) * ((1 + √3 + √5)/(1 + √3 + √5))
2. Simplify the expression:
= (1 * (1 + √3 + √5)) / ((1 + √3 - √5) * (1 + √3 + √5))
3. Use the FOIL method to simplify the denominator:
= (1 + √3 + √5) / (1 + √3 - √5 + √3 + 3 - √15 + √5 + √15 - 5)
4. Combine like terms in the denominator:
= (1 + √3 + √5) / (4 + 2√3 - 2√5)
Now, the denominator is rationalized, containing no square roots.
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Moving on to the second part of your question:
You're given the formula A = 2πr^2 + 2πrh.
To solve for positive r, we need to isolate the r term on one side of the equation. Here's how:
1. Start with the equation: A = 2πr^2 + 2πrh.
2. Subtract 2πrh from both sides to move it to the right side of the equation:
A - 2πrh = 2πr^2.
3. Divide both sides by 2π to isolate r^2:
(A - 2πrh) / (2π) = r^2.
4. Take the square root of both sides:
√((A - 2πrh) / (2π)) = r.
However, since you mentioned solving for positive r, it's important to note that if A - 2πrh is negative, there won't be a positive real solution for r, as the square root of a negative number is not a real number.