answersLogoWhite

0

yes you can re new it but you need to scrape half the top of the card of then put it in the machine so it gets re printed make sure you still have your bottom half, half way from the rank up.....

User Avatar

Wiki User

15y ago

What else can I help you with?

Related Questions

Derivative of 5ex plus 2?

5ex+2?d/dx(u+v)=du/dx+dv/dxd/dx(5ex+2)=d/dx(5ex)+d/dx(2)-The derivative of 5ex is:d/dx(cu)=c*du/dx where c is a constant.d/dx(5ex)=5*d/dx(ex)-The derivative of 2 is 0 because it is a constant.d/dx(5ex+2)=(5*d/dx(ex))+(0)d/dx(5ex+2)=5*d/dx(ex)-The derivative of ex is:d/dx(eu)=eu*d/dx(u)d/dx(ex)=ex*d/dx(x)d/dx(5ex+2)=5*(ex*d/dx(x))-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(5ex+2)=5*(ex*1)d/dx(5ex+2)=5*(ex)d/dx(5ex+2)=5ex5ex+2?d/dx(cu)=c*du/dx where c is a constant.d/dx(5ex+2)=5*d/dx(ex+2)-The derivative of ex+2 is:d/dx(eu)=eu*d/dx(u)d/dx(ex+2)=ex+2*d/dx(x+2)d/dx(5ex+2)=5*(ex+2*d/dx(x+2))-The derivative of x+2 is:d/dx(u+v)=du/dx+dv/dxd/dx(x+2)=d/dx(x)+d/dx(2)d/dx(5ex+2)=5*[ex+2*(d/dx(x)+d/dx(2))]-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1-The derivative of 2 is 0 because it is a constant.d/dx(5ex+2)=5*[ex+2*(1+0)]d/dx(5ex+2)=5*[ex+2*(1)]d/dx(5ex+2)=5*[ex+2]d/dx(5ex+2)=5ex+2


How do you find the definition of derivative of f of x equals 3x plus 2?

The definition of the derivative, at a point X = x is the limit, as dx tends to 0, of [f(x+dx)-f(x)]/dx. In this case, therefore, it is lim[3*(x+dx)+2 - (3*x+2)]/dx = lim[3x + 3*dx +2 - 3x - 2]/dx = lim[3*dx/dx] = lim[3] = 3.


What is the derivative of the square root of 1-sinx?

√(1-sinx)=(1-sinx)1/2Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(1-sinx)1/2=(1/2)(1-sinx)1/2-1*d/dx(1-sinx)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*d/dx(1-sinx)-The derivative of 1-sinx is:d/dx(u-v)=du/dx-dv/dxd/dx(1-sinx)=d/dx(1)-d/dx(sinx)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[d/dx(1)-d/dx(sinx)]-The derivative of 1 is 0 because it is a constant.-The derivative of sinx is:d/dx(sinu)=cos(u)*d/dx(u)d/dx(sinx)=cos(x)*d/dx(x)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*d/dx(x))]-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*1)]d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x))]d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[-cos(x)]d/dx(1-sinx)1/2=(-cosx)/[2√(1-sinx)]


What is the derivative of ktansqrt4x?

k*tan[√(4x)]=k*tan[(4x)1/2], where k is a constant:Multiplying by a constant, multiply the derivative of u by the constant c: d/dx d/dx(cu)=c*du/dxd/dx(k*tan[(4x)1/2])=k*d/dx(tan[(4x)1/2])-The derivative of tan[(4x)1/2] is:d/dx(tan u)=sec2(u)*d/dx(u)d/dx(tan[(4x)1/2])=sec2([(4x)1/2])*d/dx([(4x)1/2])d/dx(tan[(4x)1/2])=sec2(2√x)*d/dx([(4x)1/2])d/dx(k*tan[(4x)1/2])=k*{sec2(2√x)*d/dx([(4x)1/2])}-The derivative of (4x)1/2 is:Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(4x)1/2=(1/2)*(4x)1/2-1*d/dx(4x)d/dx(4x)1/2=(1/2)*(4x)-1/2*4d/dx(4x)1/2=4/[2(4x)1/2]d/dx(4x)1/2=4/[2(2√x)]d/dx(4x)1/2=4/[4√x]d/dx(4x)1/2=1/(√x)d/dx(k*tan[(4x)1/2])=k*sec2(2√x)*(1/√x)d/dx(k*tan[(4x)1/2])=[k*sec2(2√x)]/√x


What is the derivative of 2 to the power of 5x?

25x?d/dx(au)=au*ln(a)*d/dx(u)d/dx(25x)=25x*ln(2)*d/dx(5x)-The derivative of 5x is:d/dx(cu)=c*du/dx where c is a constantd/dx(5x)=5*d/dx(x)d/dx(25x)=95x*ln(2)*(5*d/dx(x))-The derivative of x is:d/dx(x)=1x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(25x)=25x*ln(2)*(5*1)d/dx(25x)=25x*ln(2)*(5)-25x can simplify to (25)x, which equals 32x.d/dx(95x)=32x*ln(2)*(5)


How to differentiate x plus 2?

d/dx(x + 2) = d/dx(x) + d/dx(2) = 1 + 0 = 1


How would you evaluate the indefinite integral -2xcos3xdx?

Integrate by parts: ∫ uv dx = u ∫ v dx - ∫ (u' ∫ v dx) dx Let u = -2x Let v = cos 3x → u' = d/dx -2x = -2 → ∫ -2x cos 3x dx = -2x ∫ cos 3x dx - ∫ (-2 ∫ cos 3x dx) dx = -2x/3 sin 3x - ∫ -2/3 sin 3x dx = -2x/3 sin 3x - 2/9 cos 3x + c


Integral of 3x-2 dx?

5


Integral of sin square root x?

For ∫ sin(√x) dx let y = √x = x1/2 → dy = 1/2 x-1/2 dx → 2x1/2 dy = dx → 2y dy = dx → ∫ sin(x1/2) dx = ∫(sin y) 2y dy Now: ∫ uv dx = u∫v dx - ∫(u'∫v dx) dx → ∫(sin y) 2y dy = ∫2y sin y dy = 2y ∫sin y dy - ∫(2 ∫sin y dy) dy = -2y cos y + 2 sin y + C = 2 sin y - 2y cos y + C → ∫ sin(√x) dx = 2 sin(√x) - 2(√x) cos(√x) + C


Integrate square root of 4 xdivide by x?

∫[√(4x) / x] dx = ∫(2 / √x)dx = 2∫(x-1/2) dx = 2(2x1/2 + C) = 4√x + C


What is the Integral of 23x e2x dx?

This browser is pathetic for mathematical answers but here's the best that I can do:Let u = 23x therefore du/dx = 23let dv/dx = e^(2x) therefore v = 1/2*e^(2x)then, integrating by parts,I = I(u*dv/dx) dx = u*v - I(du/dx*v) dx= 23x*(1/2)*e^(2x) - I(23*(1/2)*e^(2x) dx= 23/2*xe^(2x) - 23/2*I(e^(2x)) dx= 23/2*xe^(2x) - 23/2*(1/2)*e^(2x)= 23/2*xe^(2x) - 23/4*e^(2x)or 23/4*e^(2x)*(2x - 1)


How do you find dy by dx of y equals e square 5x?

You have : y = e^(5x)^2 and de^u/dx = [ e^u ] [ du/dx ]dy/dx = [ e^(5x)^2 ] [ 10x ]