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The first thing required is to find out what dy/dx is in terms of x. For this, we need to differentiate y with respect to x which be can so to each term of the polynomial. All you need to do is mutiply th...

∫(x³+x²+6)dx= ∫x³dx +∫x²dx+∫6dx =x⁴/4 +x³/3 +6x+C.

We can divide by (cosx)^2 across the identity (sinx)^2 + (cosx)^2 = 1 (which can be derived from properties of the unit circle and a bit of Pythagoras’ theorem) to achieve

[(sinx)^2 / (cosx)^2] + [...

P(J>2) = P(J=0)+P(J=1)     [split it up]

P(X=t)= (V^t)/t!*e^V       where V=4 in this case  [use the formula]

P(J>2) = 4^0/0!*e^4 + 4^1/1!*e^4

          =1/e^4 + 4/e^4  =  5e^-4...

P- (3,0) y=ln(x/3)     u=x/3    y=ln(u) ​​​​​​            du = 1/3  dy = 1/u = 3            dx       du dy= du x dy dx dx  du   = 1/3 x 3 = 1 gradient at normal = -1 equ...