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liq [111]
3 years ago
6

-50 Points- If segment AC and segment BC are tangent to circle O, find the value of x.

Mathematics
2 answers:
Shkiper50 [21]3 years ago
4 0

Answer:

x = 170 deg

Step-by-step explanation:

A tangent to a circle is perpendicular to the radius of the circle at the point of tangency. That means that in quadrilateral AOBC, angles A and B are right angles and measure 90 deg.

The sum of the measures of the angles of an n-sided polygon is

(n - 2)180 deg

In this case, n = 4, so

(4 - 2)180 deg = 360 deg

m<A + m<O + m<B + m<C = 360

90 + x + 90 + 10 = 360

x + 190 = 360

x = 170

Answer: 170 deg

erik [133]3 years ago
3 0

AOB ( x)  + ACB = 180

AOB (x) + 10 = 180

x = 180-10

x = 170

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3 years ago
Quadrilateral ABCD ​ is inscribed in a circle.
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When a quadrilateral is inscribed in a circle, the opposite angles are supplementary.

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Solve the Differential equation (x^2 + y^2) dx + (x^2 - xy) dy = 0
natita [175]

Answer:

\frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C

Step-by-step explanation:

Given differential equation,

(x^2 + y^2) dx + (x^2 - xy) dy = 0

\implies \frac{dy}{dx}=-\frac{x^2 + y^2}{x^2 - xy}----(1)

Let y = vx

Differentiating with respect to x,

\frac{dy}{dx}=v+x\frac{dv}{dx}

From equation (1),

v+x\frac{dv}{dx}=-\frac{x^2 + (vx)^2}{x^2 - x(vx)}

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x\frac{dv}{dx}=\frac{1 + v^2}{v-1}-v

x\frac{dv}{dx}=\frac{1 + v^2-v^2+v}{v-1}

x\frac{dv}{dx}=\frac{v+1}{v-1}

\frac{v-1}{v+1}dv=\frac{1}{x}dx

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v-2ln(v+1)=lnx+C

Now, y = vx ⇒ v = y/x

\implies \frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C

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