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2 years ago

A quadrilateral can be inscribed in a circle, if and only if, the opposite angles in that quadrilateral are supplementary.

1 answer:

8 0

The opposite angles of the cyclic quadrilateral are equal to 180 degrees which are supplementary angles.

<h3>What is a cyclic quadrilateral?</h3>

If the quadrilateral is inscribed in a circle then the quadrilateral is known as a cyclic quadrilateral.

A quadrilateral can be inscribed in a circle, if and only if, the opposite angles in that quadrilateral are supplementary.

We know that the sum of the opposite angles of the cyclic quadrilateral is equal to 180 degrees.

Supplementary angle - Two angles are said to be supplementary angles if their sum is 180 degrees.

More about the cyclic quadrilateral link is given below.

brainly.com/question/15061291

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julia-pushkina [17]

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3 years ago

1/2 X plus X -15+1/2 X +100+ X -25 equals 540

Andrew [12]

Answer:

1/2 X + X -15 + 1/2 X + 100 + X -25 = 540

1/2x + x + 1/2x + x -15 + 100 -25 = 540

3 x + 60 = 540

3x + 60 - 60 = 540 - 60

3/3x = 480/3

x = 160

Step-by-step explanation:

1/2 X + X -15 + 1/2 X + 100 + X -25 = 540

1/2x + x + 1/2x + x -15 + 100 -25 = 540

3 x + 60 = 540

3x + 60 - 60 = 540 - 60

3/3x = 480/3

x = 160

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3 years ago

Mercury’s radius is what percent of Jupiter’s radius?

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10% is the correct answer to this question

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4 years ago

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3 years ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$

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3 years ago