Determine which properties of a vector space are met by the set Q2 of polynomials with real coefficients that have degree equal to 2, together with the usual definition of addition and scalar multiplication for polynomials. (Select all that apply.) Property 1: If vi and v2 are in V, then so is v v2
Property 2: If c is a real scalar and v is in V, then so is c .
Property 3: There exists a zero vector 0 in V such that 0 + v = v for all v in V.
Property 4: Property 3 holds and for each v in V there exists an additive inverse vector-win v such that v + (-v) = 0 for all v in v.
Property 5(a): For all v1 and v2 in V, we have v1 + v2-v2 + v1- 1x1
Property 5(b): For all vi, v2, and v3 in V, we have (vi + V2) + v3 = vi + (v2+ v3)
Property 5(c): For all vi and v2 in V and real scalars c1, wehave cv+ v2) -c1vv2 ld
Property 5(d): For all v1 in Vand real scalars c1 and c2, we have (c1 + c2)V1 = c1 v1 + c2y.
Property 5(e): For all v1 in V and real scalars c1 and c2, we have (c1c2)v1 = c1(c2v1)
Property 5(f): For all VI in V, we have 1-vi-vi. none of these Submit Answer Save Progress

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.

describing distributions is one of the key skills you’ll need to earn a high score on the ap statistics exam. if you need proof of this, just flip through some past exam questions, which can be found at the collegeboard website. you’ll notice that the first free response question is almost always a question requiring you to look at a graph and describe it.

step-by-step explanation: just trying to get points sorry : (

i think its a and d

step-by-step explanation: