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Sophie [7]
3 years ago
10

Consider the function arctan(x/6)arctan⁡(x/6). write a partial sum for the power series which represents this function consistin

g of the first 5 nonzero terms. for example, if the series were ∑∞n=03nx2n∑n=0∞3nx2n, you would write 1+3x2+32x4+33x6+34x81+3x2+32x4+33x6+34x8. also indicate the radius of convergence.
Mathematics
1 answer:
Natali [406]3 years ago
3 0
<span>(x^2)/36 - (x^4)/1944 + (23x^6)/2099520 - (11x^8)/44089920 + (563x^10)/95234227200 radius of convergence = DNE (the series does not converge)</span>
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A park in a subdivision has a triangular shape. Two adjacent sides of the park are 533 feet and 525 feet. The angle between the
pishuonlain [190]
<span>Two adjacent sides of the park say x and y are,
</span><span>x=533feet
y=525feet
A=53º
area=1/2*b*c*sin(A)
111739 feet^2</span>
3 0
3 years ago
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Step-by-step explanation:

8 0
2 years ago
Which of the following is a solution of y &gt; |x| - 5?
Strike441 [17]
B is the answer!!!!!!!!!
5 0
2 years ago
37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the
Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy

and I assume <em>C</em> has a positive orientation in both cases

(a) It looks like the region has the curves <em>y</em> = <em>x</em> and <em>y</em> = <em>x</em> ² as its boundary***, so that the interior of <em>C</em> is the set <em>D</em> given by

D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}

• Compute the line integral directly by splitting up <em>C</em> into two component curves,

<em>C₁ </em>: <em>x</em> = <em>t</em> and <em>y</em> = <em>t</em> ² with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}

(b) <em>C</em> is the boundary of the region

D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}

• Compute the line integral directly, splitting up <em>C</em> into 3 components,

<em>C₁</em> : <em>x</em> = <em>t</em> and <em>y</em> = 0 with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = <em>t</em> with 0 ≤ <em>t</em> ≤ 1

<em>C₃</em> : <em>x</em> = 0 and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}

• Using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}

4 0
3 years ago
Simplify the expression and show work 4(3x-2)-7(x+5)
Gennadij [26K]

Answer:

5x - 43

Explanation:

Given the below expression;

4\left(3x-2\right)-7\left(x+5\right)

To simplify the above, we'll go ahead and use the numbers outside the parentheses to multiply the ones inside to clear the parentheses as seen below;

\begin{gathered} 12x-8-7x-35 \\ =12x-7x-8-35 \\ =5x-43 \end{gathered}

So the answer is 5x - 43

7 0
9 months ago
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