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2 years ago

Find the value of x

Mathematics

1 answer:

givi [52]2 years ago

3 0

Answer:

110°

Step-by-step explanation:

In triangle AOB

OA = OB (radii of same circle)

$m\angle OAB = m\angle OBA =20\degree$

$m\angle AOB = 180\degree - (20\degree +20\degree)$

$m\angle AOB = 180\degree - 40\degree$

$m\angle AOB = 140\degree$

$m\widehat {ACB} =m\angle AOB = 140\degree$

(measure of minor arc is equal to corresponding central angle)

$x\degree = \frac{1}{2} (360\degree - 140\degree)$

(inscribed angle theorem)

$x\degree = \frac{1}{2} \times 220\degree$

$\huge\purple {x = 110}$

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30 + 6 = 36

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42 + 6 = 48 <-- the answer is 48!

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3 years ago

If the measure 3= (2y+20) degrees, and the measure of 5= (210-4y) degrees, what is the measure of 4?

Sidana [21]

Answer:

110°

Step-by-step explanation:

y = 25

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Angle 4 equals 110°

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2 years ago

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irakobra [83]

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Step-by-step explanation:

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3 years ago

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3 years ago

Let R be the region in the first quadrant of the xy-plane bounded by the hyperbolas xyequals1, xyequals9, and the lines yequa

Tema [17]

Answer:

The area can be written as

$\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = 0.2274$

And the value of it is approximately 1.8117

Step-by-step explanation:

x = u/v

y = uv

Lets analyze the lines bordering R replacing x and y by their respective expressions with u and v.

x*y = u/v * uv = u², therefore, x*y = 1 when u² = 1. Also x*y = 9 if and only if u² = 9

x=y only if u/v = uv, And that only holds if u = 0 or 1/v = v, and 1/v = v if and only if v² = 1. Similarly y = 4x if and only if 4u/v = uv if and only if v² = 4

Therefore, u² should range between 1 and 9 and v² ranges between 1 and 4. This means that u is between 1 and 3 and v is between 1 and 2 (we are not taking negative values).

Lets compute the partial derivates of x and y over u and v

$x_u = 1/v$

$x_v = u*ln(v)$

$y_u = v$

$y_v = u$

Therefore, the Jacobian matrix is

$\left[\begin{array}{ccc}\frac{1}{v}&u \, ln(v)\\v&u\end{array}\right]$

and its determinant is u/v - uv * ln(v) = u * (1/v - v ln(v))

In order to compute the integral, we can find primitives for u and (1/v-v ln(v)) (which can be separated in 1/v and -vln(v) ). For u it is u²/2. For 1/v it is ln(v), and for -vln(v) , we can solve it by using integration by parts:

$\int -v \, ln(v) \, dv = - (\frac{v^2 \, ln(v)}{2} - \int \frac{v^2}{2v} \, dv) = \frac{v^2}{4} - \frac{v^2 \, ln(v)}{2}$

Therefore,

$\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = \int\limits_1^2 (\frac{1}{v} - v \, ln(v) ) (\frac{u^2}{2}\, |_{u=1}^{u=3}) \, dv= \\4* \int\limits_1^2 (\frac{1}{v} - v\,ln(v)) \, dv = 4*(ln(v) + \frac{v^2}{4} - \frac{v^2\,ln(v)}{2} \, |_{v=1}^{v=2}) = 0.2274$

4 0

3 years ago

Find the value of x ​

Find the value of x