Multiply, (2x^4-3y^2)(x^2+5y^4)

Mathematics

2 answers:

Ad libitum [116K]3 years ago

7 0

10x^4y^4+2x^6-15y^6-3x^2y^2

aleksandr82 [10.1K]3 years ago

4 0

To answer this question, I use FOIL (First, Outside, Inside, Last)

First: 2x^4 x x^2 = 2x^6
Outside: 2x^4 x 5y^4 = (10xy)^8
Inside: -3y^2 x x^2 = (-3yx)^4
Last: -3y^2 x 5y^4 = -15y^6

Now you just add all the answers up :)

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a park covers a total area of 644 acres there are 482 more acres of woods than grassy areas in the park how many acres are there

maw [93]

Answer:

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Step-by-step explanation:

3 0

3 years ago

Consider the line integral Z C (sin x dx + cos y dy), where C consists of the top part of the circle x 2 + y 2 = 1 from (1, 0) t

pashok25 [27]

Direct computation:

Parameterize the top part of the circle $x^2+y^2=1$ by

$\vec r(t)=(x(t),y(t))=(\cos t,\sin t)$

with $0\le t\le\pi$ , and the line segment by

$\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)$

with $0\le t\le1$ . Then

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)$

$=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt$

$=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}$

Using the fundamental theorem of calculus:

The integral can be written as

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}$

If there happens to be a scalar function $f$ such that $\vec F=\nabla f$ , then $\vec F$ is conservative and the integral is path-independent, so we only need to worry about the value of $f$ at the path's endpoints.