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15 days ago

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Determine whether each of the following functions is a solution of Laplace's equation uxx + uyy = 0. (Select all that apply.) u

= e^(−x) cos(y) − e^(−y) cos(x) u = sin(x) cosh(y) + cos(x) sinh(y)

1 answer:

PIT_PIT [3.9K]15 days ago

4 0

Answer with Step-by-step explanation:

We are given Laplace's equation:

$u_{xx}+u_{yy}=0$

We need to verify if the presented functions satisfy this equation.

A function solves Laplace's equation only if the sum of its second partial derivatives with respect to x and y equals zero.

For the first function: $u=e^{-x}cosy-e^{-y}cosx$

Compute the first derivative with respect to x:

$u_x=-e^{-x}cosy+e^{-y}sinx$

Then find the second derivative with respect to x:

$u_{xx}=e^{-x}cosy+e^{-y}cosx$

Next, derive with respect to y:

$u_y=-e^{-x}siny+e^{-y}cosx$

Then the second derivative with respect to y:

$u_{yy}=-e^{-x}cosy-e^{-y}cosx$

Substitute into Laplace's equation:

$e^{-x}cosy+e^{-y}cosx-e^{-x}cosy-e^{-y}cosx=0$

Therefore, the function satisfies Laplace's equation.

For the second function: $u=sinx coshy+cosx sinhy$

First partial derivative w.r.t x:

$u_x=cosx coshy-sinx sinhy$

Second derivative w.r.t x:

$u_{xx}=-sin x coshy-cosxsinhy$

First derivative w.r.t y:

$u_y=sinx sinhy+cosx coshy$

Second derivative w.r.t y:

$u_{yy}=sinx coshy+cosx sinhy$

After substituting, we have:

$-sinx coshy-cosxsinhy+sinxcoshy+cosx sinhy=0$

Hence, this function also solves Laplace's equation.

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The proof diagram to complete the question state the missing reason in the proof for the letter given

tester [3916]

ANSWER

Initially, it was established that line p is parallel to line r.

To begin with the proof:

As stated in the prompt:

∠1 ≈∠5

∠1 and ∠5 are corresponding angles.

Utilizing the property of corresponding angles,

if two lines are intersected by a transversal such that the corresponding angles are

congruent, then those lines must be parallel.

In the diagram, q acts as the transversal.

Thus, based on this characteristic,

line p is parallel to line r.

Proof of 1(a)

REASON

Vertically opposite angle

When two lines intersect, the angles formed are referred to as vertically opposite angles.

Therefore,

∠4 and ∠1 are vertically opposite angles,

hence,

∠4 ≈∠1

Proof of 2(b)

REASON

Alternate interior angle

The angles situated on either side of the transversal, within the two lines, are termed alternate interior angles. When two parallel lines are crossed by a transversal, the alternate interior angles produced will be congruent.

Since line p is parallel to line r (as proven above)

and q is the transversal,

then

∠4 ≈∠5

Thus, it is established.

Proof of 3 (c)

Given that ∠4 ≈ ∠5 (as demonstrated above)

REASON

If two lines are intersected by a transversal such that the alternate interior angles are congruent, then those lines are parallel.

Thus, based on the previously mentioned property,

line p is parallel to line r.

Therefore, it is proven.

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Answer:

Error made by Andrew: He identified incorrect factors based on the roots.

Step-by-step explanation:

The roots of the polynomial consist of: 3, 2 + 2i, 2 - 2i. By the factor theorem, if a is a root of the polynomial P(x), then (x - a) must be a factor of P(x). According to this premise:

(x - 3), (x - (2 + 2i)), (x - (2 - 2i)) represent the factors of the polynomial.

<pBy simplification, we obtain:

(x - 3), (x - 2 - 2i), (x - 2 + 2i) as the respective factors.

This is where Andrew's mistake occurred. Factors should always be in the form (x - a), not (x + a). Andrew expressed the complex factors incorrectly, resulting in an erroneous conclusion.

Thus, the polynomial can be expressed as:

$(x - 3)(x - 2 - 2i)(x - 2+2i)=0\\\\ (x-3)(x^{2}-2x+2xi-2x+4-4i-2xi+4i-4i^{2})=0\\\\ (x-3)(x^{2}-4x+4+4)=0\\\\ (x-3)(x^{2}-4x+8)=0\\\\ x^{3}-4x^{2}+8x-3x^{2}+12x-24=0\\\\ x^{3}-7x^{2}+20x-24=0$

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Answer:

A. C. E. F.

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Step-by-step explanation:

Within geometry, exterior angles are defined as those angles formed between any side of a polygon and the extension of an adjacent side, as demonstrated in the diagram.

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Step-by-step explanation:

Consider the expression

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Recall that

$cot^2(64\°)=\frac{cos^2(64\°)}{sin^2(64\°)}$

$sec^{2}(26\°)=\frac{1}{cos^2(26\°)}$

For two complementary angles A and B (where A+B=90°),

the identity is

cos(A) = sin(B)

Here, 26° and 64° are complementary angles, so

$\frac{1}{cos^2(26\°)}=\frac{1}{sin^2(64\°)}$

Substituting values,

$\frac{1}{sin^2(64\°)}-\frac{cos^2(64\°)}{sin^2(64\°)}$

$\frac{1-cos^2(64\°)}{sin^2(64\°)}$

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By substitution,

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