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Sladkaya [172]
3 years ago
6

two sides of a polygon are parallel line segments. what is the least number of sides the polygon could have?

Mathematics
2 answers:
Darya [45]3 years ago
8 0
4 is the least the amount of sides the polygon can have
atroni [7]3 years ago
5 0
I belive the aneser is five you are welcome for your help

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I need the answers for these pls help
DanielleElmas [232]

Answer:

1)

A. m arc QRS = 100°

B. m arc QRT = 255°

C. m arc UTS = 180°

D. m arc UTR = 205°

2)

1] The major arc CA = 255°

2] The minor arc AB = 90°

Step-by-step explanation:

In a circle:

  • The measure of an central angle is equal to the measure of its subtended arc
  • The measure of a circle is 360°
  • Any chord divides a circle into two arcs, a minor arc its measure is < 180° and a major arc its measure is > 180°, the sum of the measures of the minor and major arcs is 360°
  • If the measures of the minor and major arcs are equal, then each arc represents a semi-circle

1)

In circle O

A.

∵ m∠QOR = 75°

∵ ∠QOR subtended by arc QR

- By using the 1st note above

∴ m of arc QR = m∠QOR

∴ m of arc QR = 75°

∵ m∠ROS = 25°

∵ ∠ROS subtended by arc RS

∴ m of arc RS = m∠ROS

∴ m of arc RS = 25°

The measure of arc QRS is the sum of the measures of arcs QR and RS

∴ m arc QRS = 75° + 25° = 100°

B.

∵ m∠SOT = 155°

∵ ∠SOT subtended by arc ST

∴ m of arc ST = m∠SOT

∴ m of arc ST = 155°

The measure of arc QRT is the sum of the measures of arcs QR, RS and ST

∴ m arc QRT = 75° + 25° + 155 °= 255°

C.

∵ m∠UOT = 25°

∵ ∠UOT subtended by arc UT

∴ m of arc UT = m∠UOT

∴ m of arc RUT = 25°

The measure of arc UTS is the sum of the measures of arcs UT and TS

∴ m arc UTS = 25° + 155° = 180°

D.

The measure of arc UTR is the sum of the measures of arcs UT, TS and SR

∴ m arc UTR = 25° + 155° + 25 = 205°

2)

1]

∵ The sum of the measures of the minor and major arcs is 360°

∵ m of minor arc AC = 105°

- Subtract 105 from 360° to find the measure of the major arc CA

∴ m of major arc CA = 360° - 105°

∴ m of major arc CA = 255°

2]

∵ m of major arc AB = 270°

- Subtract 270° from 360° to find the measure of the minor arc AB

∴ m of minor arc AB = 360° - 270°

∴ m of minor arc AB = 90°

8 0
3 years ago
Given sec(theta)= 5... find cot(90-theta)
Mumz [18]
\sec(\theta)=\frac{1}{\cos(\theta)}&#10;\\&#10;\\\cos(\theta)=\frac{1}{\sec(\theta)}=\frac{1}{5}&#10;\\&#10;\\ \sin^2\theta+\cos^2\theta=1&#10;\\&#10;\\\sin\theta= \sqrt{1-\cos^2\theta}= \sqrt{1-( \frac{1}{5})^2 }  = \sqrt{1- \frac{1}{25} } = \sqrt{ \frac{25}{25} -\frac{1}{25} }=\frac { \sqrt{24}}{5} &#10;\\&#10;\\ \cot(90^o-\theta)=\tan{\theta}= \frac{\sin\theta}{\cos\theta} = \frac{\frac { \sqrt{24}}{5} }{ \frac{1}{5} } = \sqrt{24} = \sqrt{4\times6}=2 \sqrt{6}  &#10;\\&#10;
3 0
3 years ago
Need help with number 60
vampirchik [111]
You could write 42/20 for #60.
5 0
3 years ago
Write an equation in standard form for the line that passes<br> through (-1,-2) and (3.0)
Agata [3.3K]

Answer:

y+2=1/2(x+1)

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(0-(-2))/(3-(-1))

m=(0+2)/(3+1)

m=2/4

m=1/2

y-y1=m(x-x1)

y-(-2)=1/2(x-(-1))

y+2=1/2(x+1)

3 0
3 years ago
Find the length of the diameter of circle 0
r-ruslan [8.4K]

Answer:

Diameter = 19.33

Step-by-step explanation:

Imaging a radius line from O to the endpoints of the 7. call this line R.

Label the part of the vertical line from O to the 90 degree intersection y.

Now you have a right triangle.

Using the Pythagorean theorem:

R² = 7² + y²

also

y = R - 3

substitute for y:

R² = 49 + (R-3)²

R² = 49 + R² - 6R + 9

simplify:

0 = 58 - 6R

6R = 58

R = 9.6667

Diameter = 2(9.6667) = 19.33

5 0
2 years ago
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