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otez555 [7]
3 years ago
12

What is the smallest fredholl number that is prime

Mathematics
2 answers:
kaheart [24]3 years ago
7 0
Smallest Fredholl no. that's prime is 13
mylen [45]3 years ago
5 0
13 because:
1-9 aren't fredholl numbers.
<span>10 and 12 aren't prime.
11 is not a fredholl number.
</span>So the smallest fredholl number, which is prime is 13.
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Two times the sum of a number and 7
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2(x+7)

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The range of the function f(x) = x + 5 is {7, 9}. What is the function’s domain?
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Answer:

Step-by-step explanation:

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3 years ago
Find a power series representation for the function. (Give your power series representation centered at x = 0.) f(x) = x/6x^2 +
timama [110]

Looks like your function is

f(x)=\dfrac x{6x^2+1}

Rewrite it as

f(x)=\dfrac x{1-(-6x^2)}

Recall that for |x|, we have

\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n

If we replace x with -6x^2, we get

f(x)=\displaystyle x\sum_{n=0}^\infty\frac(-6x^2)^n=\sum_{n=0}^\infty (-6)^n x^{2n+1}

By the ratio test, the series converges if

\displaystyle\lim_{n\to\infty}\left|\frac{(-6)^{n+1} x^{2(n+1)+1}}{(-6)^n x^{2n+1}}\right|=6|x^2|\lim_{n\to\infty}1=6|x|^2

Solving for x gives the interval of convergence,

|x|^2

We can confirm that the interval is open by checking for convergence at the endpoints; we'd find that the resulting series diverge.

3 0
3 years ago
Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw
vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

<em>x</em> ² - 10<em>x</em> + <em>y</em> ² = 0

<em>x</em> ² - 10<em>x</em> + 25 + <em>y </em>² = 25

(<em>x</em> - 5)² + <em>y</em> ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

<em>x</em> = <em>r</em> cos(<em>θ</em>)

<em>y</em> = <em>r</em> sin(<em>θ</em>)

Then

<em>x</em> ² + <em>y</em> ² = 100   →   <em>r </em>² = 100   →   <em>r</em> = 10

<em>x</em> ² - 10<em>x</em> + <em>y</em> ² = 0   →   <em>r </em>² - 10 <em>r</em> cos(<em>θ</em>) = 0   →   <em>r</em> = 10 cos(<em>θ</em>)

<em>y</em> = 5   →   <em>r</em> sin(<em>θ</em>) = 5   →   <em>r</em> = 5 csc(<em>θ</em>)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to <em>y</em> = 5 at the point (0, 5).

Split up the region at 3 angles <em>θ</em>₁, <em>θ</em>₂, and <em>θ</em>₃, which denote the angles <em>θ</em> at which the curves intersect. They are

<em>θ</em>₁ = 0 … … … by solving 10 = 10 cos(<em>θ</em>)

<em>θ</em>₂ = <em>π</em>/6 … … by solving 10 = 5 csc(<em>θ</em>)

<em>θ</em>₃ = 5<em>π</em>/6  … the second solution to 10 = 5 csc(<em>θ</em>)

Then the area of the region is given by a sum of integrals:

\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)

=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta

To compute the integrals, use the following identities:

sin²(<em>θ</em>) = (1 - cos(2<em>θ</em>)) / 2

cos²(<em>θ</em>) = (1 + cos(2<em>θ</em>)) / 2

and recall that

d(cot(<em>θ</em>))/d<em>θ</em> = -csc²(<em>θ</em>)

You should end up with an area of

=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta

=\boxed{25\sqrt3+\dfrac{125\pi}3}

We can verify this geometrically:

• the area of the larger circle is 100<em>π</em>

• the area of the smaller circle is 25<em>π</em>

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line <em>y</em> = 5, has area 100<em>π</em>/3 - 25√3

Hence the area of the region of interest is

100<em>π</em> - 25<em>π</em> - (100<em>π</em>/3 - 25√3) = 125<em>π</em>/3 + 25√3

as expected.

3 0
3 years ago
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