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In a certain regular polygon, the measure of each interior angle is $2$ times the measure of each exterior angle.  Find the number of sides in this regular polygon.     

The sum of the interior angle and the exterior angle at a vertex is 180^o.  

Since the value of the interior angle is twice that of the exterior angle, the exterior angle must equal 60^o.  

The sum of all the exterior angles of a regular polygon is 360^o irrespective of the number of vertexes.  

Since one exterior angle is 60^o there must be six of them to add up to the total of 360^o. 

Since there are the same number of sides that there are vertexes, the polygon has six sides.  

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Let's solve this problem step-by-step:

1. Define variables:

Let L be the length, W be the width, and H be the height of the rectangular prism.

Let D be the length of the diagonal joining one corner of the prism to the opposite corner.

2. Use surface area information:

The total surface area of the prism is the sum of the areas of its six faces. Using the formula for rectangular face area, we can write:

2 * L * W + 2 * W * H + 2 * L * H = 56

3. Use edge length information:

The sum of all edge lengths is equal to the perimeter of the base plus the perimeter of the lateral faces. We can write:

2 * (L + W) + 4 * H = 64

4. Simplify equations:

From equation 2, we can isolate H: H = 30 - (L + W).

Substitute this expression for H in equation 1: 2 * L * W + 2 * W * (30 - (L + W)) + 2 * L * (30 - (L + W)) = 56.

Expand and simplify: 60L - 2W^2 - 2LW = 56.

Rewrite as a quadratic equation: -2W^2 - LW + 4L - 56 = 0.

5. Solve for L and W:

We can use factoring or the quadratic formula to solve for L and W. One possible solution is L = 8 and W = 2.

6. Calculate diagonal length:

Using the Pythagorean theorem in the right triangle formed by the diagonal and the sides of the prism, we can find D: D^2 = L^2 + W^2 + H^2.

Substituting the values: D^2 = 8^2 + 2^2 + (30 - 10)^2 = 960.

Taking the square root: D = √960

Therefore, the length of the diagonal joining one corner of the prism to the opposite corner is sqrt(960) units.

Here's how to find the value of n for the equilateral triangle, square, and regular n-gon:

Analyze the pentagons and decagon:

Two pentagons and a decagon can surround a point because the angles at each vertex of the decagon can be paired with two non-consecutive angles of a pentagon, forming a straight line through the central point.

Apply similar logic to the triangle and square:

Consider the equilateral triangle. For this shape to surround a point with the square and n-gon, each vertex of the triangle must coincide with a corner of the square and a vertex of the n-gon.

Similarly, for the square, each vertex must coincide with a corner of the equilateral triangle and a vertex of the n-gon.

Connect angles:

Since each vertex of the triangle coincides with a corner of the square, the three angles at that vertex (all 60° in the triangle) must add up to 360°, the angle at the corresponding vertex of the n-gon.

Therefore, the internal angle at each vertex of the n-gon is 360° / 3 = 120°.

Identify n-gon:

A regular polygon with an internal angle of 120° is an equilateral triangle or a hexagon. However, an equilateral triangle already fills one of the spaces, so the n-gon must be a regular hexagon.

Conclusion:

Therefore, the value of n is 6. A regular hexagon, along with an equilateral triangle and a square, with all sides the same length, can perfectly surround a point, similar to the two pentagons and a decagon.

This solution demonstrates how understanding the geometric relationships between different shapes and their angles can help solve spatial reasoning problems.

In a regular polygon with n sides, we can draw nC2 diagonals (excluding the n sides themselves). Therefore, we have the equation:

nC2 = 2n

To solve for n, we can rewrite the binomial coefficient using factorials:

n(n-1) / 2 = 2n

Simplifying the equation:

n(n-1) = 4n

n^2 - n - 4n = 0

n^2 - 5n = 0

Factoring the equation:

n(n-5) = 0

Therefore, the possible values of n are 0 and 5. However, a regular polygon cannot have 0 sides. So, the only valid solution is n = 5.

Therefore, the regular polygon has 5 sides. This corresponds to a pentagon, where the number of diagonals (10) is indeed twice the number of sides (5).

Here's how to find the side length of the hexagon:

Formula for area of a regular hexagon: The area of a regular hexagon can be calculated using the formula:

A = (3√3 / 4) * s^2

where A is the area and s is the side length.

Substituting known value and solving for s: We are given that A = 3/2 square units. Plugging this into the formula:

3/2 = (3√3 / 4) * s^2

3/2 * 4/3√3 = s^2

2/√3 = s^2

Taking the square root of both sides (remembering both positive and negative solutions):

s = ± √(2/√3)

Simplifying the radical:

s = ± √(2√3 / 3)

Therefore, the possible side lengths of the hexagon are:

s1 = + √(2√3 / 3) ≈ 1.051 units (approximately)

s2 = - √(2√3 / 3) ≈ -1.051 units (negative value not applicable)

Since the side length of a polygon cannot be negative, the valid side length of the hexagon is approximately 1.051 units.

Note: This answer provides the solution in both exact and approximate forms. You can use whichever form is more appropriate for your specific context.

In a regular polygon, the sum of the interior angles is (n-2)*180 degrees, where n is the number of sides. Each interior angle of the given polygon measures 2x degrees, where x is the measure of each exterior angle.

Therefore, we have the equation:

(n-2)*180 = n * 2x

Simplifying this equation:

180n - 360 = 2nx

2x = 180 - 360/n

x = (180 - 360/n) / 2

Now, we know that the measure of each interior angle is 2x:

2x = 2 * (180 - 360/n) / 2

x = 180 - 360/n

We are given that the measure of each interior angle is twice the measure of each exterior angle, so:

2x = 2 * (180 - 360/n) = 360/n

Substituting the expression for x from before:

2 * (180 - 360/n) = 360/n

360 - 720/n = 360/n

720/n = 0

Since n cannot be zero, the only solution is if n = 720.

Therefore, the regular polygon has 720 sides.

A sector of a circle is shown below.  The sector has an area of $60 \pi.$  What is the radius of the circle?  

I'm not sure what \pi means.  Does the backslant mean divided by pi, or is the backslant just another one of those useless symbols like all those dollar signs in every problem I look at.  I don't know about you, but I would use a forward slant to mean divided by.  I'm going to treat the backslant as meaningless.  

The shaded area is a fourth of the circle, indicated by the right angle in the center. 

                                                          A_circle  

                                                        –––––––  =  60 π  

                                                             4  

                                                           A  =  240 π  

                                                           A  =  π r²   

therefore                                       240 π  =  π r²     

                                                            r²  =  240  

                                                            r  =  sqrt(240)  

                                                            r  =  4 • sqrt(15)  

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We can solve this problem similarly to the previous one by analyzing the discriminant of the quadratic:

Discriminant and no real solutions: Recall that for the quadratic ax^2 + bx + c = 0 to have no real solutions, the discriminant b^2 - 4ac must be negative. In this case, a = 1, b = y, and c = 1.

Therefore, the condition for no real solutions becomes:

y^2 - 4 * 1 * 1 < 0

Simplifying the inequality:

y^2 - 4 < 0

Adding 4 to both sides:

y^2 < 4

Taking the square root of both sides (remembering that we need both positive and negative solutions):

-2 < y < 2

Integer solutions: Now, we need to count the number of integer values of y within this range. Since -2 and 2 are not integers, we consider the closed interval [-1, 1]. However, we need to be careful because the original inequality involved x^2 instead of just x.

Special cases: The original inequality x^2 + yx + 1 < 0 becomes equal to zero for x = -1 and x = 1 when y = -1 and y = 1, respectively. These values of y cannot satisfy the condition of no real solutions because they introduce real roots through x = -1 and x = 1.

Therefore, we exclude -1 and 1 from the possible values of y.

Final count: After excluding -1 and 1, we are left with the closed interval [-1, 0) and (0, 1]. Both intervals contain 1 integer each: 0 for the first and 1 for the second.

Therefore, there are 1 + 1 = 2 possible values of y for which the inequality has no real solutions (excluding y = -1 and y = 1).

Note: This case is different from the previous one because the original inequality involves x^2, which can produce double roots at x = 0. This necessitates excluding y = -1 and y = 1, which would otherwise satisfy the condition but introduce double roots that violate the requirement of no real solutions.