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pogonyaev
2 years ago
10

ken purchased a plot of land shaped like the figure shown. how can he find the length of the side labeled x if he knows the area

A of this lot ?
Mathematics
1 answer:
OLga [1]2 years ago
3 0
What figure?  anyways if it is a square you simply square root the area. EX: A=81 S=9
it it is a rectangle, you must know the other side before dividing
and so on
sorry I couldn't see the figure
You might be interested in
لیست داده و
choli [55]

Given parameters:

Cost price of the article = Nu.28.30

Selling price of the article  = Nu.29.30

Unknown:

Gain percentage = ?

The gain percentage is the same as the percentage profit on a trade.

The formula is given as:

       Gain percentage  = \frac{Profit}{Cost price}  x 100

         Profit = Selling price  - Cost price

                   = Nu.29.30 - Nu.28.30  

                   = Nu. 1

Now input the parameters and solve;

            Gain percentage  = \frac{1}{28.3} x 100

                                          = 3.5%

The gain percent is  3.5%

3 0
3 years ago
Multiply. 1 1/2 x 3 2/3
sasho [114]

Answer:

5.5

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Please help me thank you!
Romashka [77]

Answer:

the answer is 1¢26 if not try h5

7 0
2 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
Find the surface area of the right triangular prism shown ​
Inessa05 [86]

The surface area of the triangular prism is 144 cm²

<u>Explanation:</u>

Given:

Base of the triangle = 4

Height of the triangle = 3

Hypotenuse of the triangle = 5

Length of the prism = 11

Surface area of the rectangular prism = ?

Surface area, A = ah+bh+ch+\frac{1}{2} \sqrt{-a^4 + 2(ab)^2 + 2(ac)^2 - b^4 + 2(bc)^2 - c^4}

On substituting the value, we get:

A = 4 X 11+ 3X11+5X11+\frac{1}{2} \sqrt{-(4)^2 + 2(4X3)^2 + 2(4X5)^2 - 3^4 + 2(3X5)^2-5^4} \\\\A = 144 cm^2

Therefore, the surface area of the triangular prism is 144 cm²

6 0
3 years ago
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