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kaheart [24]
2 years ago
13

How to solve x + y = -4

Mathematics
1 answer:
Sav [38]2 years ago
6 0
<span><span>x+y</span>+<span>−y</span></span>=<span><span>−4</span>+<span>−<span>y
Answer: </span></span></span>x=<span><span>−y</span>−<span>4</span></span>
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Determine if each of the following sets is a subspace of Pn, for an appropriate value of n.
snow_tiger [21]

Answer:

1) W₁ is a subspace of Pₙ (R)

2) W₂ is not a subspace of Pₙ (R)

4) W₃ is a subspace of Pₙ (R)

Step-by-step explanation:

Given that;

1.Let W₁ be the set of all polynomials of the form p(t) = at², where a is in R

2.Let W₂ be the set of all polynomials of the form p(t) = t² + a, where a is in R

3.Let W₃ be the set of all polynomials of the form p(t) = at² + at, where a is in R

so

1)

let W₁ = { at² ║ a∈ R }

let ∝ = a₁t² and β = a₂t²  ∈W₁

let c₁, c₂ be two scalars

c₁∝ + c₂β = c₁(a₁t²) + c₂(a₂t²)

= c₁a₁t² + c²a₂t²

= (c₁a₁ + c²a₂)t² ∈ W₁

Therefore c₁∝ + c₂β ∈ W₁ for all ∝, β ∈ W₁  and scalars c₁, c₂

Thus, W₁ is a subspace of Pₙ (R)

2)

let W₂ = { t² + a ║ a∈ R }

the zero polynomial 0t² + 0 ∉ W₂

because the coefficient of t² is 0 but not 1

Thus W₂ is not a subspace of Pₙ (R)

3)

let W₃ = { at² + a ║ a∈ R }

let ∝ = a₁t² +a₁t  and β = a₂t² + a₂t ∈ W₃

let c₁, c₂ be two scalars

c₁∝ + c₂β = c₁(a₁t² +a₁t) + c₂(a₂t² + a₂t)

= c₁a₁t² +c₁a₁t + c₂a₂t² + c₂a₂t

= (c₁a₁ +c₂a₂)t² + (c₁a₁t + c₂a₂)t ∈ W₃

Therefore c₁∝ + c₂β ∈ W₃ for all ∝, β ∈ W₃ and scalars c₁, c₂

Thus, W₃ is a subspace of Pₙ (R)

8 0
3 years ago
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