G5. (5 points) next, let’s turn our attention to the vector space p2, which is the set of polynomial with degree at most 2, together with polynomial addition and scalar multiplication. let p1 = 1−x+2x2 p2 = 3+x p3 =5−x+4x2 p4 =−2−2x+2x2. is the span{p1,p2,p3,p4}={k1p1 +k2p2 +k3p3 +k4p4 : k1,k2,k3,k4 ∈r}=p2? that is, can every polynomial of degree at most two be written as a linear com- bination of p1, p2, p3 and p4? explain.

No, those vectors do not span P₂

Step-by-step explanation:

At first, notice we have 4 vectors from a vector space of dimension 3 ({1,x,x²} is the standard base of P₂) so we are sure they are linearly dependent (because there are more than 3 vectors) i.e. it´s possible to write one of them in terms of other in the same set:

p4=p1-p2

Now, we conclude the span {p1,p2,p3,p4} = {p1,p2,p3}

If we want to know the subspace that this span gets, we work with the matrix of the vectors made as follows:

The row-reduction of this matrix show us the basis of the same row-subspace even in the case that some of them are linearly dependent:

This shows that the span {p1,p2,p3}={(1+0.5x²),(x-1.5x²)}

Looking in the previous equality, the span {p1,p2,p3} is generated by two linearly independent vectors, but P₂ needs 3 linearly independent vectors to be generated.

The answer is that {p1,p2,p3,p4}⊂P₂.

muyb ajo

step-by-step explanation:

jegosu

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