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2 years ago

Quadrilateral ABCD is inscribed in a circle.

Mathematics

1 answer:

vesna_86 [32]2 years ago

3 0

Answer:

135°

Step-by-step explanation:

this is a cyclic quadrilateral, so the sum of opposite sides are 180
hence,

3x+6 + x+2 = 180
4x = 180 - (6+2)
4x = 172
x = 172/4
x = 43

therefore, m∠A is :

3(43) + 6
= 129 + 6
= 135°

hope it helps

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Solve using quadratic formula 1) y=3x^2-x-6 2) y=-5x^2+7x+3

Paul [167]

1) Solving $3x^2-x-6$ we get $x=\frac{1+\sqrt{73}}{6}\,\,and\,\,x=\frac{1-\sqrt{73}}{6}$

2) solving $-5x^2+7x+3$ we get $x=\frac{-7+\sqrt{109}}{-10}\,\,and\,\,x=\frac{-7-\sqrt{109}}{-10}$

Step-by-step explanation:

We need to solve the equations using quadratic formula.

The first equation is:

1) $y=3x^2-x-6$

The quadratic formula is:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

where a =3, b = -1 and c= -6

Putting values:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-1)\pm\sqrt{(-1)^2-4(3)(-6)}}{2(3)}\\x=\frac{1\pm\sqrt{1+72}}{6}\\x=\frac{1\pm\sqrt{73}}{6}\\so, x=\frac{1+\sqrt{73}}{6}\,\,and\,\,x=\frac{1-\sqrt{73}}{6}$

The second equation is

2) $y=-5x^2+7x+3$

The quadratic formula is:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

where a =-5, b = 7 and c= 3

Putting values:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(7)\pm\sqrt{(7)^2-4(-5)(3)}}{2(-5)}\\x=\frac{-7\pm\sqrt{49+60}}{-10}\\x=\frac{-7\pm\sqrt{109}}{-10}\\So, x=\frac{-7+\sqrt{109}}{-10}\,\,and\,\,x=\frac{-7-\sqrt{109}}{-10}$

Keywords: Solve Using Quadratic Formula

Learn more about Solve Using Quadratic Formula at:

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Quadrilateral ABCD​ is inscribed in a circle.

Quadrilateral ABCD is inscribed in a circle.