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Sindrei [870]
2 years ago
12

Which figures can be circumscribed by a circle? Select all that apply.

Mathematics
2 answers:
Darina [25.2K]2 years ago
7 0

Answer:

A, B, C, D

Step-by-step explanation:

They All add up to 360

White raven [17]2 years ago
6 0

Answer:

B and D

Step-by-step explanation:

figures which can be circumscribed by a circle are called cyclic quadrilaterals.

the opposite angles of a cyclic quadrilateral sum to 180°

A

∠ B + ∠ D = 82° + 56° = 138° ≠ 180° ← not cyclic

B

∠ B + ∠ D = 103° + 77° = 180° and ∠ A + ∠ C = 90° + 90° = 180° ← cyclic

C

∠ B + ∠ D = 112° + 90° = 202° ≠ 180° ← not cyclic

D

∠ B + ∠ D = 90° + 90° = 180° and ∠ A + ∠ C = 90° + 90° = 180°← cyclic

Then B and D can be circumscribed by a circle

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MATH HELP PLZ!!!
RoseWind [281]

Answer:

a)    tan (157.5) = \frac{1-cos 315}{sin315}

b)

            sin (165) =\sqrt{ \frac{1-cos (330) }{2}}

c)

      sin^{2} (157.5) = \frac{1-cos (315) }{2}

d)

  cos 330° = 1- 2 sin² (165°)

       

         

Step-by-step explanation:

<u><em>Step(i):-</em></u>

By using trigonometry formulas

a)

cos2∝  = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ =  2 cos² ∝/2

cos^{2} (\frac{\alpha }{2}) = \frac{1+cos\alpha }{2}

b)

cos2∝  = 1- 2 sin² ∝

cos∝  = 1- 2 sin² ∝/2

sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

<u><em>Step(i):-</em></u>

Given

              tan\alpha = \frac{sin\alpha }{cos\alpha }

          we know that trigonometry formulas

        sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )

         1- cos∝ =  2 sin² ∝/2

      Given

         tan(\frac{\alpha }{2} ) = \frac{sin(\frac{\alpha }{2} )}{cos(\frac{\alpha }{2}) }

put ∝ = 315

      tan(\frac{315}{2} ) = \frac{sin(\frac{315 }{2} )}{cos(\frac{315 }{2}) }

     multiply with ' 2 sin (∝/2) both numerator and denominator

        tan (\frac{315}{2} )= \frac{2sin^{2}(\frac{315)}{2}  }{2sin(\frac{315}{2} cos(\frac{315}{2}) }

Apply formulas

 sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )

  1- cos∝ =  2 sin² ∝/2

now we get

 tan (157.5) = \frac{1-cos 315}{sin315}

       

b)

          sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

               put ∝ = 330° above formula

             sin^{2} (\frac{330 }{2}) = \frac{1-cos (330) }{2}

            sin^{2} (165) = \frac{1-cos (330) }{2}

            sin (165) =\sqrt{ \frac{1-cos (330) }{2}}

c )

         sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

               put ∝ = 315° above formula

             sin^{2} (\frac{315 }{2}) = \frac{1-cos (315) }{2}

            sin^{2} (157.5) = \frac{1-cos (315) }{2}

           

d)

     cos∝  = 1- 2 sin² ∝/2

   put      ∝ = 330°

       cos 330 = 1 - 2sin^{2} (\frac{330}{2} )

      cos 330° = 1- 2 sin² (165°)

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Let's use Gaussian elimination. Consider the augmented matrix,

\left[\begin{array}{ccc|ccc}1 & -1 & -1 & 1 & 0 & 0\\-1 & 2 & 3 & 0 & 1 & 0\\1 & 1 & 4 & 0 & 0 & 1\end{array}\right]

• Add row 1 to row 2, and add -1 (row 1) to row 3:

\left[\begin{array}{ccc|ccc}1 & -1 & -1 & 1 & 0 & 0\\0 & 1 & 2 & 1 & 1 & 0\\0 & 2 & 5 & -1 & 0 & 1\end{array}\right]

• Add -2 (row 2) to row 3:

\left[\begin{array}{ccc|ccc}1 & -1 & -1 & 1 & 0 & 0\\0 & 1 & 2 & 1 & 1 & 0\\0 & 0 & 1 & -3 & -2 & 1\end{array}\right]

• Add -2 (row 3) to row 2:

\left[\begin{array}{ccc|ccc}1 & -1 & -1 & 1 & 0 & 0\\0 & 1 & 0 & 7 & 5 & -2\\0 & 0 & 1 & -3 & -2 & 1\end{array}\right]

• Add row 2 and row 3 to row 1:

\left[\begin{array}{ccc|ccc}1 & 0 & 0 & 5 & 3 & -1\\0 & 1 & 0 & 7 & 5 & -2\\0 & 0 & 1 & -3 & -2 & 1\end{array}\right]

So the inverse is

\begin{bmatrix}1&-1&-1\\-1&2&3\\1&1&4\end{bmatrix}^{-1} = \boxed{\begin{bmatrix}5&3&-1\\7&5&-2\\-3&-2&1\end{bmatrix}}

3 0
3 years ago
Find the value of X.
Doss [256]

Answer:

x=13

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
This year, 1,272 students enrolled in night courses a a local college. Last year only 1,200 students enrolled. What was the perc
Ahat [919]

The percent increase in enrollment is 6 %

The operation used in first step is finding the difference between final value and initial value

<h3><u>Solution:</u></h3>

Given that This year, 1,272 students enrolled in night courses a a local college

Last year only 1,200 students enrolled.

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The percent increase between two values is the difference between a final value and an initial value, expressed as a percentage of the initial value.

<em><u>The percent increase is given as:</u></em>

\text { percentage increase }=\frac{\text { final value - initial value}}{\text { initial value }} \times 100

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Thus percent increase is 6 %

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