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Bingel [31]
2 years ago
6

Simplify this expression for 19 point! -0.5p+1/2p-2.75+2/3p

Mathematics
1 answer:
defon2 years ago
7 0

Answer:

you mean 10?

-2.75+\frac{2p}{3}

Step-by-step explanation:

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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
2 years ago
Please helpppp<br><br> Extra points and brainleist
vladimir1956 [14]

Answer:

B

Step-by-step explanation:

x² - 4x - 7 = 0

(x - 2)² = 11

5 0
3 years ago
A sleep time of 15.9 hours per day for a newborn baby is at the 10h percentile of the distribution of sleep times for all newbor
Nuetrik [128]

Answer:

option (D) 16.5

Step-by-step explanation:

Data provided:

Sleep time of , X = 15.9 is at the 10th percentile

Standard deviation, σ = 0.5  hour

also,

10th percentile of normal distribution is

Z = F⁻¹(0.10)

or

Z = - 1.28  (Using Standard Normal distribution)

Now,

\\Z=\frac{X-\mu}{\sigma}

or

X = μ + ( σ × Z )

or

μ = X - ( σ × Z )

on substituting the respective values, we get

μ = 15.9 - 0.5 × (-1.28)

or

μ = 16.5

Hence, the correct answer is option (D) 16.5

8 0
2 years ago
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