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Well, it is somewhere between

**R**and L, with speakers varying between...not individual letters. So you can't write**R**along, but you can write syllables that start with**R**...8 Answers · Society & Culture · 27/05/2008

**r**= ± 3 - √3, ± 1 - √3 so**r**IS irrational therefore it must be irrational s = ±√[**r**^2 + (√3)**r**.... Solve the resulting quartic equation:**r**^4 - 4r^2s^2 - 6r^2 + 4s^4 - 12s^2 + 9 = 0 for either "**r**" or...1 Answers · Science & Mathematics · 24/01/2010

Let

**r**= (x(1), x(2), ..., x(n)). So, |**r**...x(2))^2 + ... + (x(n))^2]. Hence,**r**(hat) = (x(1)/√[(x(1))^2 + (x(2))^2...x(1))^2 + ... + (x(n))^2)^(-1/2) = (n-1)/|**r**|. In the case where n = 3, this ...1 Answers · Science & Mathematics · 14/03/2013

this can be written as

**r**^2 + 3r - 2 = 0 Now to solve this you can use the discriminat method (delta = b^2...the method of completing the square.**r**^2 + 3r + (3/2)^2 - (3.2)^2 -2 =0 (**r**+ 3/2)^2 = 9/4 + 2 = 17/4**r**+3/2 = +or- sqrt(17/4)**r**= + sqrt(17/4) - 3/2 or...3 Answers · Science & Mathematics · 15/09/2010

(

**r**− 3)(**r**+ 9)=-9 =>**r**(**r**+ 9) −3(**r**+ 9)= -9 =>**r**^2 + 9r −3r − 27 = -9 =>**r**^2 + 6r − 18 = 0 now...4 Answers · Science & Mathematics · 15/12/2013

y =

**r***sin(T) We know that**r**= 5*sin(T) which gives sin(T) =**r**/5 And**r**^2 = x^2...3 Answers · Science & Mathematics · 03/05/2010

p(

**r**) gives you the charge per unit volume of the region. To find the charge... charge in the region = charge per unit volume x volume So dq = p(**r**) x dV where dV is the volume of the thin shell. (Note...2 Answers · Science & Mathematics · 26/05/2011

**r**^2 - 5r - 6r + 30 = 0**r**(**r**- 5) - 6(**r**- 5) = 0 (**r**- 6)(**r**- 5) = 0**r**- 6 = 0 or**r**- 5 = 0**r**= 6 or**r**= 53 Answers · Science & Mathematics · 06/02/2011

The DVD-

**R**format was developed by Pioneer in the autumn of 1997. ... system of tracking and speed control used by DVD+**R**being less susceptible to interference and error than the LPP system used...3 Answers · Computers & Internet · 03/10/2007

Since

**r**^2 = (-**r**)^2, we can assume without loss of generality that**r**≥ 0. For each fixed**r**≥ 0, A(**r**) is a circle centered at (0, 0) with radius**r**. (When**r**= 0, then A(0) is simply {(0, 0)}.) Hence...1 Answers · Science & Mathematics · 27/03/2012

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