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Heptagon
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How many diagonals does a nine sided polygon?
For an n-sided polygon, the number of diagonals is 1+2+3+....+(n-4)+2(n-3). For a nine-sided polygon, there are 1+2+3+4+5+2(6)=27 diagonals.
What is the vertex for the parabola y equals x squared plus 4x plus 5?
The vertex of a parabola is the minimum or maximum value of the parabola. To find the maximum/minimum of a parabola complete the square: x² + 4x + 5 = x² + 4x + 4 - 4 + 5 = (x² + 4x + 4) + (-4 + 5) = (x + 2)² + 1 As (x + 2)² is greater than or equal to 0, the minimum value (vertex) occurs when this is zero, ie (x + 2)² = 0 → x + 2 = 0 → x = -2 As (x + 2)² = 0, the minimum value is 0 + 1 = 1. Thus the vertex of the parabola is at (-2, 1).
What are the coordinates of the vertex of y x2 - 4x - 5?
Y = X2 - 4X - 5set to zeroX2 - 4X - 5 = 0X2 - 4X = 5halve the linear term ( - 4 ) then square it and add that result to both sidesX2 - 4X + 4 = 5 + 4factor on the left and gather terms together on the right(X - 2)2 = 9(X - 2)2 - 9 = 0==============vertex form(2, - 9)======vertex
What is the y-coordinate of the vertex of a parabola with the following equation y equals x2 - 8x plus 18?
You can work this out by taking the derivative of the equation, and solving for zero: y = x2 - 8x + 18 y' = 2x - 8 0 = 2x - 8 x = 4 So the vertex occurs where x is equal to 4. You can then plug that back into the original equation to get the y-coordinate: y = 42 - 8(4) + 18 y = 16 - 32 + 18 y = 2 So the vertex of the parabola occurs at the point (4, 2), leaving 2 as the answer to your question.
Identify the vertex of the parabola y equals 4x2 - 24x plus 38?
There are two ways to solve this problem; one that involves basic calculus skills and one that can be done with algebraic methods. I will do both, and you can choose which one you understand based on your knowledge of mathematics. I will start with the algebraic method. The end goal of this is to get the equation in vertex form, where you will be able to simply read the vertex off of the equation. To do this, we will employ a method called "completing the square". Your original function is: y=4x2-24x+38 This is a parabola, but unfortunately it is not in vertex form. A parabola in vertex form will read as y=(x-a)2+c where "a" and "c" are real numbers. To get the component (x-a)2, we must have a trinomial that can be factored to be a product of the same term squared, that is it must be a trinomial that results from multiplying (x-a)*(x-a). In its current form, the trinomial in your equation is not a perfect square. You cannot express 4x2-24x+38 as product of two of the same binomial terms. So, we must make it a trinomial that is a perfect square. We will use the "completing the square" method. First, however, we must make it so no coefficient exists in front of the x2 term. This will make "completing the square" much easier. To do this, we will have to divide everything by 4. Remember what you do to one side of an equation you must do to both sides. y/4=4x2/4-24x/4+38/4 Simplify this to get: y/4=x2-6x+(19/2) To "complete the square", take the coefficent of the single-x term and halve it, then square it. For this trinomial, the single-x term is -6x: -6/2=3 32=9 This is the number that must be at the end of your trinomial to make it a perfect square. If you can manipulate the equation to get a trinomial of x2-6x+9, this will be a perfect square. Currently, you have x2-6x+(19/2). 19/2 is the same as 9.5. You could make 19/2 into the 9 that you need if you subtracted 1/2. You can do this as long as you do it to both sides of the equation. So, you get: y/4-(1/2)=x2-6x+(19/2)-(1/2) y/4-(1/2)=x2-6x+9 The trinomial you now have on the right side of the equation can be rewritten as (x-3)2, so you get: y/4-(1/2)=(x-3)2 By isolating y, you get an equation of a parabola in vertex form: y/4=(x-3)2+(1/2) y=4((x-3)3+(1/2)) y=4(x-3)3+2 This is vertex form, which reads as y=a(x-b)2+c. In this case, a=4, b=3, and c=2. You can get the vertex of a parabola in this form to be simply (b,c). So, the vertex for this parabola is (3,2). As a side note: the "a" value of a parabola in vertex form (4 for your equation) does not affect its vertex, it simply affects how "fat" or "skinny" the parabola will be. Second side note: If the squared term had happened to have been (x+3)2, the x coordinate of the vertex would be -3, not 3. Since vertex form reads as (x-a)2, "a" must be negative for an equation have (x+a)2 Now, for the calculus approach. This method is much simpler, but you must understand the basic of calculus to do it. I will not explain all of that, because it would take far too long. For this example, I will assume you know the meanings and applications of function maxima and minima are and how to derive a function. Since the function is a parabola, it will only have one critical point, its vertex, and since it is a parabola that opens upward, this critical point will be a minimum. So, the absolute minimum of the parabola is its vertex. By finding the absolute minimum, we find the vertex. To do this, we must first find the first derivative of the function. y=4x2-24x+38 y'=8x-24 By finding where the first derivative equals zero, we find the critical points to investigate on this function to determine if they are minima or maxima. y'=0 8x-24=0 8x=24 x=3 Since there is only one critical point, and since the function is a parabola, we know that this one point is the vertex. There is no need to verify that it is a minimum, although you can do that if you want to. We have found the x-value for the vertex of the function, so now we must find its corresponding y-value for the original function. By plugging in 3 for x in the original function, we get: y=4(3)2-24(3)+38 y=4*9-72=38 y=36-72+38 y=2 So, for x=3, y=2, meaning our vertex is (3,2).
How many diagonals can you draw from one vertex in a polygon with 4 sides?
Only one.
How many diagonals can be drawn from one vertex of a 7-sided polygon?
7-3 = 4
How many diagonals does a 8 sided polygon have?
An eight-sided polygon has 8 vertices. Consider1st vertex = 7 diagonals to other vertices2nd vertex = 6 diagonals to other vertices (since one has now been used)3rd vertex = 54th vertex = 45th vertex = 36th vertex = 27th vertex = 18th vertex = 0, so7+6+5+4+3+2+1 = 28Improved Answer:-Formula for finding diagonals of a polygon 0.5*(n2-3n) when n is the number of sides0.5*(82-24) = 20 diagonals
How many diagonals can be drawn from one vertex of a seven sided polygon?
There can be 14 lines in a seven side shape * * * * * That is the total number of diagonals from ALL vertices. Not what the question asked, though. From one vertex, there can be 4. One to every other vertex except for itself and one each on either side.
How many diagonals can you draw from one vertex of a hexagon?
four (4)
A hexagon can be divided into how many triangles by drawing all of the diagonals from one vertex?
A hexagon (six-sided polygon) can be divided into 4 triangles by drawing all of the diagonals from one vertex (only three lines can be drawn in this case, since each vertex already connects to two others on the edges of the form). If you instead drew lines from the center to each vertex, you would get 6 triangles.
A hexagon can be divided into how many triangles by drawing diagonals from one vertex?
4
How many triangles can you form in a hexagon if you draw all the diagonals from only one vertex?
Either 12,14,16,or18 I think 18. * * * * * The correct answers is 4. You can draw only 3 diagonals from one vertex.
How many triangles can you form in a hexagon if you draw all of the diagonals from only one vertex?
4.
How many diagonals are in a 5 sided polygon?
4
Why can't you draw a diagonal in a triangle?
Because it has only 3 sides and a polygon must have 4 or more sides to have diagonals. A diagonal of a polygon is a line segment which connects two non-adjacent vertices. Since from any vertex of a triangle, you are adjacent to both of the other vertex, there are not any non-adjacent vertices.
How many diagonals does a nine sided polygon?
For an n-sided polygon, the number of diagonals is 1+2+3+....+(n-4)+2(n-3). For a nine-sided polygon, there are 1+2+3+4+5+2(6)=27 diagonals.
