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tatuchka [14]
3 years ago
11

#1) * Find the value of x. Write your answer in simplest form.​

Mathematics
1 answer:
Usimov [2.4K]3 years ago
5 0
Answer is 4
in the 45 45 90 triangle rule, the longest length is x sq root 2
so since the longest length is 4 sq root 2,
the two others are just 4
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Answers for the 2 boxes please :)​
monitta

Answer:

x=1

y=2

Step-by-step explanation:

Multiply the second equation by 2

and then solve!

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Which expression is equivalent to 6^-3?
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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
2 years ago
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