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3 years ago

How to use order of operation in 20 divide by 4 multiply 5 = 1

Mathematics

1 answer:

I am Lyosha [343]3 years ago

6 0

Use PEMDAS. 20/(4*5)=1
Multiplication before dividion

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Let vector F = (6 x^2 y + 2 y^3 + 4 e^x) i + (7 e^{y^2} + 54 x) j . Consider the line integral of vector F around the circle of

balu736 [363]

Denote the circle of radius $a$ by $C$ . $C$ is simple and closed, so by Green's theorem the line integral reduces to a double integral over the interior of $C$ (call it $D$ ):

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_C(6x^2y+2y^3+4e^x)\,\mathrm dx+(7e^{y^2}+54x)\,\mathrm dy$

$=\displaystyle\iint_D\left(\frac{\partial(7e^{y^2}+54x)}{\partial x}-\frac{\partial(6x^2y+2y^3+4e^x)}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy$

$D$ is a circle of radius $a$ , so we can write the double integral in polar coordinates as

$\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^a(54-6r^2)r\,\mathrm dr\,\mathrm d\theta$

a. For $a=1$ , we have

$\displaystyle\int_0^{2\pi}\int_0^1(54-6r^2)r\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^1(54r-6r^3)\,\mathrm dr=\boxed{51\pi}$

b. Let $I(a)$ denote the integral with unknown parameter $a$ ,

$I(a)=12\pi\int_0^a(9r-r^3)\,\mathrm dr\,\mathrm d\theta$

By the fundamental theorem of calculus,

$I'(a)=12\pi(9a-a^3)$

$I(a)$ has critical points when

$12\pi(9a-a^3)=12\pi a(9-a^2)=0\implies a=0,a=\pm3$

If $a=0$ , then line integral is 0, so we ignore that critical point. For the other two, we would find $I(\pm3)=243\pi$ .

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3 years ago

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Rina8888 [55]

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