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3 years ago

A regular decagon has 10 sides. How many reflectional symmetries does a regular decagon have?

2 answers:

lina2011 [118]3 years ago

7 0

The rule for regular polygons are very easy. The number of reflectional symmetries is same as the number of sides. Regular polygons have all sides the same length and all angles same. Reflection symmetry means that you can fold the shape along that line and it will match up.

For this question, we want the number of reflectional symmetries of a regular decagon. Decagon is a 10 sided figure. Hence, the number of reflectional symmetries is 10.

There are 5 symmetry lines from one vertex to opposite vertex and 5 more symmetry lines form midpoint of one side to midpoint of opposite side.

ANSWER: 10

qaws [65]3 years ago

4 0

Answer:

10,10,36

Step-by-step explanation:

You might be interested in

X^2 + Y^2 +ax + 2y + 3 = 0 What number is this equation the equation of a circle?

Sav [38]

Answer:

Eq: (x+a/2)²+(y+1)²=(a²-8)/4

Center: O(-a/2, -1)

Radius: r=0.5×sqrt(a²-8)

Mandatory: a>2×sqrt(2)

Step-by-step explanation:

The circle with center in O(xo,yo) and radius r has the equation:

(x-xo)²+(y-yo)²=r²

We have:

x²+y²+ax+2y+3=0

But: x²+ax=x²+2(a/2)x+a²/4-a²/4= (x+a/2)²-a²/4

And

y²+2y+3=y²+2y+1+2=(y+1)²+2

Replacing, we get:

(x+a/2)²-a²/4+(y+1)²+2=0

(x+a/2)²+(y+1)²=a²/4-2=(a²-8)/4

By visual inspection we note that:

- center of circle: O(-a/2, -1)

- radius: r=sqrt((a²-8)/4)=0.5×sqrt(a²-8). This means a²>8 or a>2×sqrt(2)

6 0

3 years ago

Will give BRAINLEST :)

DiKsa [7]

Answer:

Step-by-step explanation:

This explanation mostly depends on what you're learning right now. The first way would be to convert this matrix to a system of equations like this.

g + t + k = 90

g + 2t - k = 55

-g - t + 3k = 30

Then you solve using normal methods of substitution or elimination. It seems to me that elimination is the quickest method.

g + t + k = 90

-g - t + 3k = 30

____________

0 + 0 + 4k = 120

4k = 120

k = 30

No you can plug this into the first two equations

g + t + (30) = 90

g + t = 60

and

g + 2t - (30) = 55

g + 2t = 85

now use elimination again by multiplying the first equation by -1

g + 2t = 85

-g - t = -60

_________

0 + t = 25

t = 25

Now plug those both back into one of the equations. I'll just do the first one.

g + (25) + (30) = 90

g = 35

Therefore, we know that Ted spent the least amount of time on the computer.

The second method is using matrix reduction and getting the matrix in the row echelon form, therefore solving using the gauss jordan method. If you would like me to go through this instead, please leave a comment.

3 0

2 years ago

Work out and simplify where possible a) 2/5+1/3 b) 5/9-1/2

Vlad1618 [11]

The answer is
a) 11/15
b) 1/18

5 0

3 years ago

An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach

ollegr [7]

Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

velocity, a is the acceleration of gravity (32 feet/second²) and x

is the time

18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

of the motion

is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

- When the rocket hits the ground h(x) = 0

∵ h(x) = -16x² + 128x + 144

∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16

∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x

∵ x² - 8x - 9 = 0

∴ (x - 9)( x + 1) = 0

∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

* The time for the object to hit the ground is 9 seconds

8 0

2 years ago

A woman begins driving her car 25 km north of calgary. Some time later, the woman and her car are 100 km north of calgary.

VARVARA [1.3K]

Well the displacement of milage from the 2 places if 75km however, the women and her car are in the same place so there is no displacement. But 75km is the correct mathematical answer.

4 0

3 years ago