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3 years ago

Find the measure of an interior and an exterior angle of a regular 46-gon

1 answer:

7 0

For any polygon, the formula of the interior angle is 180 - 360/n where n is the number of sides of the polygon. The exterior angle is the supplementary angle of the interior angle. Hence when n is equal to 46, then interior angle is equal to 172.17 degrees and the exterior is equal to 7.83 degrees.

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There is 1/4 ounce of yeast in every 2 1/4 teaspoon of yeast. A recipe calls for 2 teaspoons of yeast. How many ounces of yeast

MrRissso [65]

1/4 ounce of yeast ... 2 1/4 teaspoons of yeast
x ounce of yeast = ? ... 2 teaspoons of yeast

If you would like to know how many ounces of yeast need to be in the recipe, you can calculate this using the following steps:

1/4 * 2 = x * 2 1/4
1/2 = x * 9/4 /*4/9
x = 1/2 * 4/9
x = 2/9

Result: 2/9 ounce of yeast needs to be in this recipe.

7 0

4 years ago

Gary wants to buy a video game with a selling price of 48 on sale for 50% off the sales tax in his state is 4.5% how much will G

NeTakaya

Answer:

$25.08

Step-by-step explanation:

48/2= 24(50% of 100% simplifies to 1/2)

24(0.045)=1.08

24 + 1.08 = 25.08

8 0

3 years ago

ABC is a straight line where BC = 3AB. OA = a, AB = b Express OC in terms of a and b.

Aleks [24]

That would be 3AB because of the expression you have

7 0

2 years ago

11 ÷ 924 in long division

Goryan [66]

Answer:

231 your welcome

Step-by-step explanation:

4 0

3 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago