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3 years ago

What are the solutions to the equation x2 - 8x = 10

Mathematics

1 answer:

mojhsa [17]3 years ago

5 0

Answer:

x = 4 ±√26

Step-by-step explanation:

Please write this as x^2 - 8x = 10, or x^2 - 8x - 10 = 0. " ^ " indicates exponentiation.

Let's complete the square:

x^2 - 8 x can be rewritten as

x^2 - 8x + 16 - 16, and so

x^2 - 8x = 10 becomes x^2 - 8x + 16 - 16 - 10 = 0, or

(x - 4)^2 = 26

Taking the square root of both sides, we get:

x - 4 = ±√26, or

x = 4 ±√26

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3x^2+15x+25=−3x−2 whats the discrimate?

Zepler [3.9K]

Answer:

The discriminate is 0

Step-by-step explanation:

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8 0

3 years ago

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Identify the polygon that has vertices A(−10,−1), P(−7,3), E(−3,0), and X(−6,−4), and then find the perimeter and area of the po

Bogdan [553]

Answer:

square; perimeter 20 units; area 25 square units.

Step-by-step explanation:

As the attachment shows, each side of the polygon is the hypotenuse of a 3-4-5 right triangle, so has length 5 units. The perimeter is the sum of those lengths, 4×5 = 20; the area is the product of the lengths of adjacent sides, 5×5 = 25.

The figure is a square of side length 5 units.

The perimeter is 20 units; the area is 25 square units.

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3 years ago

What equation represents a parabola that opens UP with the vertex at 2,4

lubasha [3.4K]

Answer:

y = (x -2)^2 +4

Step-by-step explanation:

The vertex form of the equation of a parabola is ...

y = a(x -h)^2 +k

for vertex (h, k) and vertical scale factor "a". When a > 0, the parabola opens upward.

One equation for your parabola could be ...

y = (x -2)^2 +4

In standard form this one is ...

y = x^2 -4x +8

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3 years ago

Is the number rational or irrational? Explain. A. -38 B. 400 C. 2 3 27

Alik [6]

A. No, because -38 isn't a fraction.

2, No because 400 is a real number.

3. Yes because fractions, decimals, and percents are rational numbers.

hope that helped

6 0

3 years ago

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

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)

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)

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