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3 years ago

(01.06 LC) Expand and simplify: (2x + 5y) (x - 3y) (5 points)

Mathematics

1 answer:

laiz [17]3 years ago

4 0

Answer:

2x² - xy - 15y²

Step-by-step explanation:

(2x + 5y)(x - 3y)

Each term in the second factor is multiplied by each term in the first factor, that is

2x(x - 3y) + 5y(x - 3y) ← distribute parenthesis

= 2x² - 6xy + 5xy - 15y² ← collect like terms

= 2x² - xy - 15y²

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Alinara [238K]

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The circulation of the field f(x) over curve C is Zero

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The function $f(x)=(x^{2},4x,z^{2})$ and curve C is ellipse of equation

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Theory: Stokes Theorem is given by:

$I= \int \int\limits {{Curl f\cdot \hat{N }} \, dx$

Where, Curl f(x) = $\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\F1&F2&F3\end{array}\right]$

Also, f(x) = (F1,F2,F3)

$\hat{N} = grad(g(x))$

Using Stokes Theorem,

Surface is given by g(x) = $16x^{2} + 4y^{2} - 3$

Therefore, tex]\hat{N} = grad(g(x))[/tex]

$\hat{N} = grad(16x^{2} + 4y^{2} - 3)$

$\hat{N} = (32x,8y,0)$

Now, $f(x)=(x^{2},4x,z^{2})$

Curl f(x) = $\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\F1&F2&F3\end{array}\right]$

Curl f(x) = $\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\x^{2}&4x&z^{2}\end{array}\right]$

Curl f(x) = (0,0,4)

Putting all values in Stokes Theorem,

$I= \int \int\limits {Curl f\cdot \hat{N} } \, dx$

$I= \int \int\limits {(0,0,4)\cdot(32x,8y,0)} \, dx$

I=0

Thus, The circulation of the field f(x) over curve C is Zero

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3 years ago