Which of the following subsets of P2P2 are subspaces of P2P2? A. {p(t) | p′(t){p(t) | p′(t) is constant }} B. {p(t) | p(−t)=p(t){p(t) | p(−t)=p(t) for all t}t} C. {p(t) | p′(6)=p(7)}{p(t) | p′(6)=p(7)} D. {p(t) | ∫10p(t)dt=0}{p(t) | ∫01p(t)dt=0} E. {p(t) | p′(t)+2p(t)+8=0}{p(t) | p′(t)+2p(t)+8=0} F. {p(t) | p(5)=6}{p(t) | p(5)=6}

The correct options are;

E. {p(t) | p′(t)+2p(t)+8=0}{p(t) | p′(t)+2p(t)+8=0}

F. {p(t) | p(5)=6}{p(t) | p(5)=6}

Step-by-step explanation:

A. {P(t)| P'(t)} is constant is given by

P(t)=a+bt+ct^2

P'(t) = b + 2ct which is not a constant, therefore

{P(t)| P'(t)} is not a subspace of P2

B. {p(t) | p(−t)=p(t){p(t) | p(−t)=p(t) for all t}t}

Here, we have

P(t)=a+bt+ct^2 and

P(-t)=a-bt+ct^2

P(t) is not equal to P(-t)

Therefore {p(t) | p(−t)=p(t){p(t) | p(−t)=p(t) for all t}t} is not a subspace of P2

C. {p(t) | p′(6)=p(7)}{p(t) | p′(6)=p(7)}

p'(t) = b+ct and p′(6) = b+6c

p(7) = a+7b+49c

Therefore

p′(6) is not equal to p(7) and

{p(t) | p′(6)=p(7)}{p(t) | p′(6)=p(7)} is not a subspace of P2

D. {p(t) | ∫10p(t)dt=0}{p(t) | ∫01p(t)dt=0}

∫10p(t)dt

= at+0.5×bt^2+(ct^3)/3

= a + 0.5b+cr/3 which is not equal to 0

Therefore

{p(t) | ∫10p(t)dt=0}{p(t) | ∫01p(t)dt=0} is not a subspace of P2

E. {p(t) | p′(t)+2p(t)+8=0}{p(t) | p′(t)+2p(t)+8=0}

Here, we have

p′(t)+2p(t)+8=0 given by

b+2c+2 (a+bt+ct^2)+8=0...(1)

When t=2, we have

a = -(16+5b+4c)/2

Substituting the value of a into the equation (1), and simplying we have

(2t-4)×b+(2t-4)×c+ 2t^2-8=0

Therefore when t=2 the above equation =0

Hence

{p(t) | p′(t)+2p(t)+8=0}{p(t) | p′(t)+2p(t)+8=0} is a subspace of P2

F. {p(t) | p(5)=6}{p(t) | p(5)=6}

Here we have

p(t) = a+bt+ct^2

p(5) = a+5b+25c = 6

a=6-5b-25c

Substitution gives

6-5b-25c+bt+ct^2

Which gives on factorization

(t-5)b+(t^2-25)c+6 which is equal to 6 when t=5

Therefore

{p(t) | p(5)=6}{p(t) | p(5)=6} is a subspace of P2

proofs involving isosceles triangles often require special consideration because an isosceles triangle has several distinct properties that do not apply to normal triangles.(more about triangle types) therefore, when you are trying to prove that two triangles are congruent, and one or both triangles, are isosceles you have a few theorems that you can use to make your life easier.

step-by-step explanation:

an isosceles triangle has two congruent sides and two congruent angles. the congruent angles are called the base angles and the other angle is known as the vertex angle. ∠bac and ∠bca are the base angles of the triangle picture on the left. the vertex angle is ∠abc

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.