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Romashka-Z-Leto [24]
3 years ago
6

4

Mathematics
1 answer:
o-na [289]3 years ago
4 0

Answer:

The measure of exterior angle is 6 is 122

You might be interested in
There is a manhole cover with a radius of 1 foot. What is the manhole cover's diameter?
zhuklara [117]

Step-by-step explanation:

This seems like you just want to figure out the circumference of the manhole cover. The formula for the circumference of a circle is pi (3.14) multiplied by the diameter (d) of the circle so, circumference=πd. (π is the symbol for pi and approx. equals 3.14)

Circumference = πd

= 3.14(d)

= 3.14(3)

= 9.42 ft.

The length of the brass grip-strip will be 9.42 ft.

If the problem was stated in terms of the radius of the manhole cover then the formula would be circumference = 2πr which is the radius multiplied by 2 then multiplied by pi.

The radius of a circle is the distance from the center to the edge and the diameter is the distance from one edge of the circle to the other passing through the center of the circle.

Well, if the grip strip were of no width and could be straightened out to a line (which a piece of rubber cut in a circle couldn't be), then the length of the grip would correspond to the circumference of the manhole cover.

Circumference = 2*PI*radius = PI*diameter so your answer is 3*PI feet long.

6 0
2 years ago
Find the slope between the points (0,8) and (4,4)
Oxana [17]

Answer:

Slope=-1

Step-by-step explanation:

Slope=y1-y2/x1-x2

Where (X1,y1)(0,8) and (X2,y2) (4,4)

Slope=8-4/0-4

=4/-4

=-1

So slope is -1

5 0
2 years ago
What is .333 written in its simplest fraction from
gayaneshka [121]
Hi.
your answer is 1/3, or 2/6, or 3/9, or 4/12, etc.
hope this helps!!!


6 0
3 years ago
Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering p
tankabanditka [31]

Answer:

1.778 times more or 16/9 times more

Step-by-step explanation:

Given:

- Mirror 1: D_1 = 8''

- Mirror 2: D_2 = 6"

Find:

Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering power?

Solution:

- The light gathering power of a mirror (LGP) is proportional to the Area of the objects:

                                           LGP ∝ A

- Whereas, Area is proportional to the squared of the diameter i.e an area of a circle:

                                           A ∝ D^2

- Hence,                              LGP ∝ D^2

- Now compare the two diameters given:

                                           LGP_1 ∝ (D_1)^2

                                           LGP ∝ (D_2)^2

- Take a ratio of both:

                           LGP_1/LGP_2 ∝ (D_1)^2 / (D_2)^2

- Plug in the values:

                               LGP_1/LGP_2 ∝ (8)^2 / (6)^2

- Compute:             LGP_1/LGP_2 ∝ 16/9 ≅ 1.778 times more

6 0
3 years ago
Questions 16-17 | Math 1 - 0 points Solve the graph Help needed !!
iris [78.8K]

Answer:

16) The area of the circle is 25.1 units²

17) JKLM is a parallelogram but not a rectangle

Step-by-step explanation:

16) Lets talk about the area of the circle

- To find the area of the circle you must find the length of the radius

- In the problem you have the center of the circle and a point on

 the circle, so you can find the length of the radius by using the

 distance rule

* Lets solve the problem

∵ The center of the circle is (1 , 3)

∵ The point on the circle is (3 , 5)

- Using the rule of the distance between two points

* Lets revise it

- The distance between the two points (x1 , y1) and (x2 , y2) is:

 Distance = √[(x2 - x1)² + (y2 - y1)²]

∴ r = √[(3 - 1)² + (5 - 3)²] = √[4 + 4] = √8 = 2√2 units

∵ The area of the circle = πr²

∴ The area of the circle = π (2√2)² = 8π = 25.1 units²

* The area of the circle is 25.1 units²

17) To prove a quadrilateral is a parallelogram, prove that every

     to sides are parallel or equal or the two diagonal bisect

     each other

* The parallelogram can be rectangle if two adjacent sides are

  perpendicular to each other (measure of angle between them is 90°)

 or its diagonals are equal in length

- The parallel lines have equal slopes, then to prove the

  quadrilateral is a parallelogram, we will find the slopes of

  each opposite sides

* Lets find from the graph the vertices of the quadrilateral

∵ J = (0 ,2) , K (2 , 5) , L (5 , 0) , M (3 , -3)

- The opposite sides are JK , ML and JM , KL

- The slope of any line passing through point (x1 , y1) and (x2 , y2) is

 m = (y2 - y1)/(x2 - x1)

∵ The slop of JK = (5 - 2)/(2 - 0) = 3/2 ⇒ (1)

∵ The slope of LM = (-3 - 0)/(3 - 5)= -3/-2 = 3/2 ⇒ (2)

- From (1) and (2)

∴ JK // LM

∵ The slope of KL = (0 - 5)/(5 - 2) = -5/3 ⇒ (3)

∵ The slope of JM = (-3 - 2)/(3 - 0)= -5/3 ⇒ (4)

- From (3) and (4)

∴ KL // JM

∵ Each two opposite sides are parallel in the quadrilateral JKLM

∴ It is a parallelogram

- The product of the slopes of the perpendicular line is -1

* lets check the slopes of two adjacent sides in the JKLM

∵ The slope of JK = 3/2 and the slope of KL = -5/3

∵ 3/2 × -5/3 = -5/2 ≠ -1

∴ JKLM is a parallelogram but not a rectangle

4 0
3 years ago
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