2x2 = x2 - 2
Subtract x2 from both sides: x2 = -2
Take square roots: x = + or - i*sqrt(2) where i is the imaginary sqrt of -1.
(x3 + 3x2 - 2x + 7)/(x + 1) = x2 + 2x - 4 + 11/(x + 1)(multiply x + 1 by x2, and subtract the product from the dividend)1. x2(x + 1) = x3 + x22. (x3 + 3x2 - 2x + 7) - (x3 + x2) = x3 + 3x2 - 2x + 7 - x3 - x2 = 2x2 - 2x + 7(multiply x + 1 by 2x, and subtract the product from 2x2 - 2x + 7)1. 2x(x + 1) = 2x2 + 2x2. (2x2 - 2x + 7) - (2x2 + 2x) = 2x2 - 2x + 7 - 2x2 - 2x = -4x + 7(multiply x + 1 by -4, and subtract the product from -4x + 7)1. -4(x + 1) = -4x - 42. -4x + 7 - (-4x - 4) = -4x + 7 + 4x + 4 = 11(remainder)
The square of 22 is 484- 22 X22 ----- 44 44X ------ 484
v38
726 square feet
I tried to find f by integrating the partial derivatives, but since 1/r is multiplying the whole vector, I just took it out, I'm not sure if I can do that. Like this: ∂f∂x(x,y,z)=x ∂f∂y(x,y,z)=y ∂f∂z(x,y,z)=z thus f(x,y,z)=x22+g(y,z) f(x,y,z)=y22+h(x,z) f(x,y,z)=z22+k(x,y) for some functions g, h, and k, so if g=y22+z22, h=x22+z22 and k=x22+y22, the function f is: f(x,y,z)=1r(x22+y22+z22)=12r⋅r2=r2 Am I correct? If not, how can I solve this correctly, should I integrate x/r, y/r and z/r instead?
there is x18 x19 and x22 they are like sport bikes but smaller the x22 is 27in seat height
(x3 + 3x2 - 2x + 7)/(x + 1) = x2 + 2x - 4 + 11/(x + 1)(multiply x + 1 by x2, and subtract the product from the dividend)1. x2(x + 1) = x3 + x22. (x3 + 3x2 - 2x + 7) - (x3 + x2) = x3 + 3x2 - 2x + 7 - x3 - x2 = 2x2 - 2x + 7(multiply x + 1 by 2x, and subtract the product from 2x2 - 2x + 7)1. 2x(x + 1) = 2x2 + 2x2. (2x2 - 2x + 7) - (2x2 + 2x) = 2x2 - 2x + 7 - 2x2 - 2x = -4x + 7(multiply x + 1 by -4, and subtract the product from -4x + 7)1. -4(x + 1) = -4x - 42. -4x + 7 - (-4x - 4) = -4x + 7 + 4x + 4 = 11(remainder)
v38
33"x22"
=current price of gold X22/24
The square of 22 is 484- 22 X22 ----- 44 44X ------ 484
36 can never be a prime. You can find its prime factorization. 32 x22 =36
v38
726 square feet
I tried to find f by integrating the partial derivatives, but since 1/r is multiplying the whole vector, I just took it out, I'm not sure if I can do that. Like this: ∂f∂x(x,y,z)=x ∂f∂y(x,y,z)=y ∂f∂z(x,y,z)=z thus f(x,y,z)=x22+g(y,z) f(x,y,z)=y22+h(x,z) f(x,y,z)=z22+k(x,y) for some functions g, h, and k, so if g=y22+z22, h=x22+z22 and k=x22+y22, the function f is: f(x,y,z)=1r(x22+y22+z22)=12r⋅r2=r2 Am I correct? If not, how can I solve this correctly, should I integrate x/r, y/r and z/r instead?
16 ft x 22 ft = 352 ft2
You can solve this to the accuracy of your liking by using Newton's method: xn+1 = xn - f(xn) / f'(xn) In this case, we'll say f(x) = x2 - cos(x) f'(x) would then be 2x + sin(x) Let's take a rough guess, and start with x0 = 0.5 x1 = 0.5 - (0.52 - cos(0.5)) / (2(0.5) + sin(0.5)) = 0.92420692729319751536 x2 = x1 - (x12 - cos(x1)) / (2x1 + sin(x1)) = 0.82910575599741780916 x3 = x2 - (x22 - cos(x2)) / (2x2 + sin(x2)) = 0.82414613172819520712 x4 = x3 - (x32 - cos(x3)) / (2x3 + sin(x3)) = 0.8241323124099124229 x5 = x4 - (x42 - cos(x4)) / (2x4 + sin(x4)) = 0.82413231230252242297 x6 = x5 - (x52 - cos(x5)) / (2x5 + sin(x5)) = 0.82413231230252242296 Now we can test our answer: 0.824132312302522422962 = 0.67919406818110235182 cos(0.82413231230252242296) = 0.67919406818110235183 So we're accurate to the nearest ten quintillionth.