Question

P(t)=2+t+2t²-(t)³. P=0;P=1;P=2

Answer 1

Answer:

p(0)=2

p(1)=4

p(2)=4

Step-by-step solution:

p(0) = 2+0+2(0)2–(0)3=2

p(1) = 2+1+2(1)2–(1)3=2+1+2–1=4

p(2) = 2+2+2(2)2–(2)3=2+2+8–8=4

Answer 2

Given,

t = 0

t = 1

t = 2

To Find,

P(t) $=2+t+2t^{2} -(t)^{3}$ when t=0, t=1, t=2

Solution,

We can solve the given problem by using the following process.

If t=0,

                                P(0) $= 2+0+2(0)^{2}-(0)^{3}$

                                        $=2+0\\\\=2$

If t=1,

                                P(1)$=2+1+2(1)^{2} -(1)^{3}$

                                       $=3+2-1\\\\=4$

If t=2,

                                 P(2)$=2+2+2(2)^{2} -(2)^{3}$

                                       $=4+8-8\\\\=4$

Hence, At t=0, P(t)=2,

            at t=1, P(t)=4,

            at t=2, P(t)=4.