All categories

2 years ago

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°

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

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

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

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

Then B and D can be circumscribed by a circle

You might be interested in

at a local fitness center, members paya a $6 membership fee and $5 for each aerobics class. nonmembers pay $6 for each aerob

Anna007 [38]

The members 6 classes will be the same price of 11 nonmember classes

7 0

3 years ago

MATH HELP PLZ!!!

RoseWind [281]

Answer:

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

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

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

cos 330° = 1- 2 sin² (165°)

Step-by-step explanation:

Step(i):-

By using trigonometry formulas

cos2∝ = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ = 2 cos² ∝/2

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

cos2∝ = 1- 2 sin² ∝

cos∝ = 1- 2 sin² ∝/2

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

Step(i):-

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

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

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

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

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

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

cos∝ = 1- 2 sin² ∝/2

put ∝ = 330°

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

cos 330° = 1- 2 sin² (165°)

3 0

3 years ago

Find the inverse of this matrix. 1 -1 -1 -1 2 3 1 1 4

zheka24 [161]

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>Solution:</h3>

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

Last year only 1,200 students enrolled.

To find: percent increase in the enrollment

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.

The percent increase is given as:

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

Here initial value (last year) = 1200 and final value(this year) = 1272

Substituting the values in above formula,

$\begin{aligned}&\text { percentage increase }=\frac{1272-1200}{1200} \times 100\\\\&\text { percentage increase }=\frac{72}{1200} \times 100=6\end{aligned}$

Thus percent increase is 6 %

4 0

3 years ago