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2 years ago

Find the domain of - y/(y-6) +15/(y+6) - y-10/y^2+3

Mathematics

1 answer:

ArbitrLikvidat [17]2 years ago

5 0

Answer:

Step-by-step explanation:

Domain of a function represents the set of x-values (input values) and y-values (output values) of the function represent the Range of the function.

Given function is,

If Domain (input values) of this function is,

{-12, -6, 3, 15}

Table for the input-output values of this function,

x -6 3 15 -12

y 11 5 -3 15

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alekssr [168]

Answer:

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Step-by-step explanation:

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2 years ago

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jeka57 [31]

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2 years ago

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arlik [135]

Answer:

$x=-3, d=2, e=3$

Step-by-step explanation:

$\frac{9a^3b^5}{-3ab^2} =\frac{9}{-3} \cdot\frac{a^3}{a} \cdot \frac{b^5}{b^2}$

Recall how to divide exponents. For instance, $a^3/a^1=a^{3-1}=a^2$ .

$=-3a^2b^3$

Thus, $x=-3, d=2, e=3$

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2 years ago

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Alja [10]

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3 years ago

Find the counterclockwise circulation and outward flux of the field F=7xyi+5y^2j around and over the boundary of the region C en

dezoksy [38]

Split up the boundary of C (which I denote ∂C throughout) into the parabolic segment from (1, 1) to (0, 0) (the part corresponding to y = x ²), and the line segment from (1, 1) to (0, 0) (the part of ∂C on the line y = x).

Parameterize these pieces respectively by

r(t) = x(t) i + y(t) j = t i + t ² j

and

s(t) = x(t) i + y(t) j = (1 - t ) i + (1 - t ) j

both with 0 ≤ t ≤ 1.

The circulation of F around ∂C is given by the line integral with respect to arc length,

$\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T \,\mathrm ds$

where T denotes the tangent vector to ∂C. Split up the integral over each piece of ∂C :

• on the parabolic segment, we have

T = dr/dt = i + 2t j

• on the line segment,

T = ds/dt = -i - j

Then the circulation is

$\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(\mathbf i+2t\,\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i-\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (7t^3+10t^5)\,\mathrm dt - 12 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{-\frac7{12}}$

Alternatively, we can use Green's theorem to compute the circulation, as

$\displaystyle\int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \iint_C\frac{\partial(5y^2)}{\partial x} - \frac{\partial(7xy)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = -7\int_0^1\int_{x^2}^x x\,\mathrm dx \\\\ = -7\int_0^1 xy\bigg|_{y=x^2}^{y=x}\,\mathrm dx \\\\ =-7\int_0^1(x^2-x^3)\,\mathrm dx = -\frac7{12}$

The flux of F across ∂C is

$\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N \,\mathrm ds$

where N is the normal vector to ∂C. While T = x'(t) i + y'(t) j, the normal vector is N = y'(t) i - x'(t) j.

• on the parabolic segment,

N = 2t i - j

• on the line segment,

N = - i + j

So the flux is

$\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(2t\,\mathbf i-\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i+\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (14t^4-5t^4)\,\mathrm dt - 2 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{\frac{17}{15}}$

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3 years ago