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3 years ago

What is the sum of the exterior angles of a 14-gon?

Mathematics

2 answers:

Zanzabum3 years ago

5 0

Answer:

360 degrees.

Step-by-step explanation:

The sum of exterior angles of every polygon is 360 degrees so the What is the sum of the exterior angles of a 14-gon is also equal to 360 degrees.

ad-work [718]3 years ago

4 0

Answer:

360 degrees

Step-by-step explanation:

The sum of all exterior angles in any convex polygon is 360 degrees.

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Two column proof:

gulaghasi [49]

Answer:

Given : BRDG is a kite that is inscribed in a circle,

With BR = RD and BG = DG

To prove : RG is a diameter

Proof:

Since, RG is the major diagonal of the kite BRDG,

By the property of kite,

∠ RBG = ∠ RDG

Also, BRDG is a cyclic quadrilateral,

Therefore, By the property of cyclic quadrilateral,

∠ RBG + ∠ RDG = 180°

⇒ ∠ RBG + ∠ RBG = 180°

⇒ 2∠ RBG = 180°

⇒ ∠ RBG = 90°

⇒ ∠ RDG = 90°

Since, Angle subtended by a diameter or semicircle on any point of circle is right angle.

⇒ RG is the diameter of the circle.

Hence, proved.

8 0

3 years ago

Helpppppppppppppppppppppppp

Nonamiya [84]

Answer:

the answer is - 133

Step-by-step explanation:

6 0

2 years ago

Olivia flips a two-sided counter and then spins a spinner with six equal-size sections labeled 1 through 6. One side of the coun

Novay_Z [31]

Answer:

Step-by-step explanation:

A 2 - sided counter ; (red, yellow)

A spinner (1,2,3,4,5,6)

Number of trials = 80

P(red and number > 3) :

P(red) = 1/2 ;

P(number >3) : numbers greater Than 3 = (4, 5, 6)

Hence, P(number <3) = 3 /6 = 1/2

Theoretical probability = 1/2 *1/2 = 1/4

Expected number of outcomes :

1/4 * number of trials

1/4 * 80 = 20

Experimental outcome :

Relative frequency = number of outcomes / number of trials

Relative frequency = 2/5

Hence,

2/5 = number of outcomes / 80

Cross multiply :

160 = number of outcomes * 5

Number of outcomes = 160 /5 = 32

Actual outcomes = 32

Difference between actual and expected :

32 - 20 = 12

7 0

3 years ago

Find the smallest positive $n$ such that \begin{align*} n &\equiv 3 \pmod{4}, \\ n &\equiv 2 \pmod{5}, \\ n &\equiv

Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number $x$ such that taking it mod 4, 5, and 7 leaves the desired remainders:

$x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6$

Taken mod 4, the last two terms vanish and we have

$x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4$

so we multiply the first term by 3.

Taken mod 5, the first and last terms vanish and we have

$x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5$

so we multiply the second term by 2.

Taken mod 7, the first two terms vanish and we have

$x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7$

so we multiply the last term by 7.

Now,

$x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147$

By the CRT, the system of congruences has a general solution

$n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}$

or all integers $27+140k$ , $k\in\mathbb Z$ , the least (and positive) of which is 27.

3 0

3 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago