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steposvetlana [31]
2 years ago
13

The two triangles above are similar.

Mathematics
1 answer:
kondaur [170]2 years ago
4 0

Answer:

Part a) x=15\ cm

Part b) x=15\ cm

Part c) The answers are the same because the triangles are similar, therefore the  ratio of their corresponding sides are equal

Step-by-step explanation:

we know that

If two triangles are similar, then the ratio of their corresponding sides are equal  and is called the scale factor

Part a) Find x using the ratio of the sides 12 cm and 16 cm

so

\frac{12}{16}=\frac{x}{20}

\frac{3}{4}=\frac{x}{20}

The ratio of their corresponding sides is called the scale factor and in this problem is equal to \frac{3}{4}

solve for x

x=20*3/4\\x=15\ cm

Part b) Find x using the ratio of the sides 6 cm and 8 cm

so

\frac{6}{8}=\frac{x}{20}

\frac{3}{4}=\frac{x}{20}

Observe that the scale factor is equal to \frac{3}{4}

solve for x

x=20*3/4\\x=15\ cm

Part c) The answers are the same because the triangles are similar, therefore the  ratio of their corresponding sides are equal

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Answer:

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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

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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

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2)

let W₂ = { t² + a ║ a∈ R }

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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)

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