(pi*r2)/2 + 4000
The surface area of a whole sphere, of radius r, is 4*pi*r2.The area of the curved surface of a hemisphere is, therefore, 2*pi*r2.The area of the flat surface is that of a circle of radius r, that is = pi*r2.So the total surface area of a hemisphere is 2*pi*r2 + pi*r2 = 3*pi*r2.
Lateral Surface Area = π(r1 + r2)s = π(r1 + r2)√((r1 - r2)2 + h2)Top Surface Area = πr12Base Surface Area = πr22Total Surface Area = π(r12 + r22 + (r1 * r2) * s) = π[ r12 + r22 + (r1 * r2) * √((r1 - r2)2 + h2) ]
If the radius of the circle is R, and the length of the flat region is c, then the area of the circle containing a flat is: A = R2 [Pi - Arcsin(c / 2R)] + [(c / 2) (R2 - (c / 2)2)1/2] After some calculus and algebra
Surface area = 2*pi*r2+ 2*pi*r*h = 1105.8 sq units.=Surface area = 2*pi*r2+ 2*pi*r*h = 1105.8 sq units.=Surface area = 2*pi*r2+ 2*pi*r*h = 1105.8 sq units.=Surface area = 2*pi*r2+ 2*pi*r*h = 1105.8 sq units.=
Release 2
(R1 * R2) / (R1 + R2) = 2 Parallel R1 + R2 = 9 Series Treating the two as simultaneous equations, and substituting for R1: ((9-R2) * R2) / (9 - R2 + R2) = 2 R2^2 - 9R2 + 18 = 0 Solving the quadratic, we get: R2 = 6 ohm R1 = 3 ohm Which you can check by substituting back into the original equations.
assume the following configuration: battery connected to 2 parallel resistors with an ammeter in series with the battery... observe the current measurement ... remove one of the resistors .... observe the current again, it will have decreased: if the resistors were of equal value, the current will decrease to half of its original value when one of the resistors is removed. Mathematics: V=IR (V- voltage, I - current, R - resistance in a parallel circuit, R=(R1*R2)/(R1+R2) where R1 and R2 are the values of resistance of the resistors. Before removal- Ib=V*(R1+R2)/(R1*R2) After removal (assume R2 is removed)- Ia=V/R1 so Ia/Ib=(R1*R2)/(R1*(R1+R2)) or Ia=Ib*(R2/(R1+R2) if R1=R2 then Ia=Ib*R2/(2*R2) or Ia=Ib/2 so the current after is 1/2 of that before.
x2 - (-b/a)x + (c/a) = 0 or x2 - (sum of the roots)x + (product of the roots) = 0 Let the roots be r1 and r2. So we have: r1 + r2 = 5 (r1)2 + (r2)2 = 15 r1 = 5 - r2 (express r1 in term of r2) (5 - r2)2 + (r2)2 = 15 25 - 10r2 + (r2)2 + (r2)2 = 15 2(r2)2 - 10r + 25 = 15 (subtract 15 to both sides) 2(r2)2 - 10r + 10 = 0 (divide by 2 to both sides) (r2)2 - 5r + 5 = 0 (use the quadratic formula) r2 = [-b + &- sq root of (b - 4ac)]/2a r2 = {-(-5) + &- sq root of [(-5)2 - 4(1)(5)]}/2(1) = [5 + &- sq root of (25 - 20)]/2 = (5 + &- sq root of 5)/2 r1 = 5 - r2 r1 = 5 - (5 + &- sq root of 5)/2 Thus, when r2 = (5 + sq.root of 5)/2, r1 = (5 - sq.root of 5)/2 or vice versa. Since the given equation is x2 + bx + c = 0, a = 1, then c equals to the product of roots. So that, c = (r1)(r2) = [(5 - sq.root of 5)/2][(5 + sq.root of 5)/2] = [52 - (sq.root of 5)2]/4 = 5
F=GM1M2/R2 dF=GM1dM2/R2 You will find that this is a linear relationship between F and M2. The slope of the line is GM1/R2 the y intercept is 0
r1,r2,l1,r2,left,down,right,up
yes press R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2, then press select to complete the entire game
2003 Ohms. R = (R1 x R2)/( R1 + R2) Where R = 667 and R1 = 1000 then R2 = 2003
2 r2
R2, r2, r1, r2, l1, r2, left, down, right, up, left down, right, up
look up cheatcc on google and r2 r2 x down is not a cheat!
It could be lots of things. One answer can be: r2 + 9 = r2 + 32.