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nataly862011 [7]
3 years ago
6

4 times the sum of x and 0.5 *

Mathematics
2 answers:
horsena [70]3 years ago
5 0

Answer:

4(x+0.5)

Step-by-step explanation:

the sum of x and 0.5 is x+0.5

it says 4 times (x+0.5)

so the equation is

Fynjy0 [20]3 years ago
4 0
4(x+0.5) is the answer if your trying to simplify it
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​ Quadrilateral ABCD ​ is inscribed in this circle.
Sunny_sXe [5.5K]

Answer: m\angle A=116\°

Step-by-step explanation:

The missing figure is attached.

For this exercise it is important to remember that, by definition, the opposite interior angles of an inscribed quadrilateral are supplementary, which means that their sum is 180 degrees.

Based on this, you can identify that the angle D and the angle B are opposite and, therefore, supplementary.

Knowing that, you can write the following equation:

x+28\°=180\°

Now you must solve for "x" in order to find its value. This is:

x=180\°-28\°\\\\x=152\°

Then:

m\angle D=152\°

You know that:

m\angle A=(x-36)\°

Therefore, since you know the value of "x", you can substitute it into   m\angle A=(x-36)\° and then you must evaluate, in order to find the measure of the angle A. This is:

 m\angle A=152\°-36\°\\\\m\angle A=116\°

4 0
3 years ago
If your car goes 50 kilometers in 2 hours,what is its average speed
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Answer:

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8 0
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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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