The taylor polynomial of degree 100 for the function f about x=3 is given by
\[p(x)= (x-3)^2 - ((x-3)^4)/2! + + [(-1)^n+1] [(x-3)^n2]/n! + - ((x-3)^100)/50! ]/
what is the value of f^30 (3)?

answer: it's d

step-by-step explanation:

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.