All categories

3 years ago

Find x and y i will give brainliest

Mathematics

2 answers:

Serga [27]3 years ago

6 0

Answer: N= 135, M=45.

Step-by-step explanation:The formula to find the degree inside of a shape is: (n-2)180. This is an octagon, so it has 8 sides, plug this in for n.

8-2 * 180

1080 divide by 8 to get the value of n

n=135

The formula for an angle outside of the shape is:

360/n n is still eight, so plug this in for n

360/8

m=45

sashaice [31]3 years ago

5 0

Answer:

n = 135 degrees ( ° ) ,

m = 45 degrees ( ° )

Step-by-step explanation:

* Sorry I didn't answer this before *

1. In this figure we can see that this visual is a representation of an octagon, provided it's 8 sides.

2. Knowing this, let us split this figure into triangle 180 degrees each ⇒ compute the total interior angles in degrees.

3. If you were to draw lines from one vertex to any other non-adjacent vertices, it would be that 6 triangles are formed.

4. One triangle ⇒ 180 degrees ( ° ) so that this octagon ⇒ 180° * number of triangles, or ⇒ 180 * 6 = 1080°

5. Knowing that this shape is a regular polygon, all angles are congruent such that if x represents one interior angle ⇒ 8x = 1080, ⇒ x = 135°. As n acts as one of the interior angles in this figure ⇒ n = 135 degrees ( ° )

6. Now through supplementary angles, the interior angle adjacent to angle m would form a linear pair, adding to 180°.

7. This would mean that m + interior angle = 180, and as all interior angles were found to be 135 degrees ⇒ m + 132 = 180 ⇒ m = 45 degrees ( ° )

You might be interested in

The price of a car has been reduced from $15,500 to $10,695. What is the percentage decrease of the price of the car?

IceJOKER [234]

(10,695/15,500 -1) * 100% = -31%

The price was decreased 31%.

5 0

4 years ago

Determine whether Point A lies on the circle whose center is Point C and whichcontains the Point P. Justify your answer mathemat

Arada [10]

The Solution:

Given:

Center = (0,0)

Point A = (-5,2) being a point on the circle.

We are required to check if point P = (2,-5) is on the circle.

Solving the given problem graphically, we have:

From the above graph, it is clear that point P(2,-5) is a point on the circle.

8 0

1 year ago

Im 4 and i need help with my math the question is 11+10

Arada [10]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Can someone answer this ill give 20 brainly points

antiseptic1488 [7]

Answer:

The answer you're looking for is B. 27.55 m

6 0

2 years ago

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago