I Urgently Need Help With Geometry Questions. Help.

I need help with a geometry question. I don't need the answer, but a walk through oh how to find it?

Ok so the first thing you should do is set up an equation stating the two quantities that are equal. In this case, it would be:(Volume of small pipe) + (Volume of big pipe) = (Volume of new pipe)To find the volumes of the original pipes, the radius is needed first:1. Diameter/2 = 6/2 = 3 = radius2. Diameter/2 = 8/2 = 4 = radiusNow, the formula the volume of a cylinder [(pi)(r^2)(h)] can be put into the equation, and because they all have the same height, the height will be written as h:(pi)(3^2)(h) + (pi)(4^2)(h) = (pi)(r^2)(h)Now because there is a pi and an h in every term, both sudes can be divided by (pi)(h) to guve the following equation:3^2 + 4^2 = r^2Simplification can then give the answer:9 + 16 = r^225 = r^25 = rNow, the question asks for the diameter of the pipe, so all you have to do is multiply 5 by 2 as 2r = d, so the diameter is 10 inches. Please excuse any typos.

Where can I get help with a geometry question?

In most graphics libraries these days, you have the concept of a “current transformation matrix”. It might be called something else, but I’ll refer to it as the CTM.This matrix establishes a coordinate system for the subsequent graphics drawing operations. As drawing proceeds, the matrix is frequently modified, for example when drawing takes place in a particular window, the CTM is changed to establish coordinates local to that window.Every coordinate system has an origin - a place where the coordinate value 0,0 can be found. Graphics are drawn relative to that.So in your figure, you want to make C2 the origin, apply the rotation, then put the origin back where it was (if necessary). You should find within your graphics library simple functions that modify the CTM, such as translating the origin by a certain distance in x and y, or rotating it by a certain number of degrees, or scaling by a given factor in x and y. These operations can be combined within the single CTM, which is hugely efficient, because it means that just the one matrix can be used for all coordinate calculations. But critically, matrix operations are not commutative - that is, they have to be applied in a specific order. If you rotate then translate, you’ll get a different result than if you translate then rotate.In your case, you’d probably do:Translate the CTM by -c2.x, -c2.yRotate the CTM by 45° (or whatever)Translate the CTM again by c2.x, c2.yDraw the figure.You should find that the rotation is about the point C2.Note: the above translations might seem to be back-to-front, but due to the mathematics of matrices, they’re not. Effectively, the ‘steps’ that you are baking into the matrix (which end up just modifying numbers that are applied once), run from last to first.

Can you help me with this solid geometry question?

The total (one-sided) area of the plastic sheet wrapping is[math](2\pi{+}18)r\times{L}[/math]where [math]r[/math] is the radius of the circle and [math]L[/math] is the length of the cylinder.ExplanationThe area of the sheet is the length of one of those cylinders multiplied by the length of the triangle-shaped line surrounding the circular faces in the diagram, so most of the difficulty would be expended in to calculating the length of this line.Consider that a perfectly-stacked pile of cylinders will have the centre of each of the circular faces forming equilateral triangles. We can extend the sides of the triangle to make a mirrored upside-down triangle:The angle spanned by [math]a[/math] is, of course, [math]60\deg[/math]. The sides of the triangles are parallel with the lines formed by the plastic wrapping. However, the intersection between the triangle and the circles are not where the wrapping first touches the top cylinder, which is actually [math]30\deg[/math] more on each side of the triangle:One side of the triangle line extension shown. The same reasoning applies to the other triangle side. Right angle drawn for clarification.The shortest distance between two parallel lines, one touching the surface of a circle and the other crossing the centre of the circle, is the length of a perpendicular line segment which runs between both of them.Extending the angle [math]a=60\deg[/math] on both sides by [math]30\deg[/math] results in the angle [math]b[/math][math]=60\deg{+}2\times{30}\deg=[/math][math]120\deg[/math].Now [math]120\deg[/math] is the angle spanned by the pie segment of a third of a circle. So, the distance of the line wrapped around one of the corner circles (shown by the orange outline below) is [math]1/3[/math] of the circle’s circumference.There are three of these corner circles, so the total distance wrapped around the corner circles is[math]3\text{(corner circles)}\times{1/3}\times{2\pi{r}}=2\pi{r}[/math]The rest of the wrapping spans 3 circle diameters for each side of the wrapped bunch, so the total distance required to wrap around the stacked bunch of circles is[math]2\pi{r}+3\text{(for the three sides of the wrapped bunch)}\times{3}\text{(circle diameters)}\times{2r}=(2\pi{+}18)r[/math]The total (one-sided) area of the plastic sheet wrapping is then[math](2\pi{+}18)r\times{L}[/math]

Can you help me with these analytic geometry questions?

For the first you need to find a circle touching the x-axis and passing through the points (3,1) and (10,8).Circles have the equation [math](x-x_0)^2 + (y-y_0)^2 = r^2[/math]. As its touching the x-axis the radius r is the same as the y-coordinated of the center y_0 so the equation simplifies to[math] (x-x_0)^2 + (y-y_0)^2 = y_0^2[/math].Expand to get [math]x^2 - 2 x x_0 + x_0^2 + y^2 - 2 y y_0 + y_0^2 = y_0^2 [/math]and eliminate the last term[math]x^2 - 2 x x_0 + x_0^2 + y^2 - 2 y y_0 = 0[/math]Now put the two known sets of know values[math]9 - 6 x_0 + x_0^2 + 1 - 2 y_0 = 0[/math]     (A)[math]100 - 20 x_0 + x_0^2 + 64 - 16 y_0 = 0 [/math]   (B)to eliminate[math] y_0[/math] use 8 A - B [math]-28 + 28 x_0 + 7 x_0^2 -56 = 0[/math]divide by 7 and rearrange[math]x_0^2 + 4 x_0 -12 = 0[/math]Which can be factorised as[math](x_0 - 6) (x_0 +2) = 0[/math]Hence [math]x_0 = 6[/math]. Put this value into A, [math]9 - 36 + 36 + 1 - 2 y_0 = 0[/math]Solving gives y_0 =5 and the equation is [math](x-6)^2 + (y-5)^2 = 25[/math].As this looks a lot like homework I will not do any of the others. If there is one you have a problem, at least take the trouble to type it in so people do not need to look sideways at a screen shot.For Q3: The sides of a triangle are on the lines 3x - y - 5 = 0, x + 3y - 1= 0, and x - 3y + 7 = 0. Find the equation of the circle inscribed in the triangle.We want to find the Incircle of a triangle the way you would do this geometrically would be to find the bisectors of the three angles and the point is intersection of these three lines is the center of the incircle. This is not the way I would do it, instead I would use the  distance from a point to a line for a point [math](h,k)[/math] and a line [math]a x + b y + c = 0[/math], then the distance from the point to the line is [math]r=\frac{|a h + b k + c|}{\sqrt{a^2+b^2}}[/math]its a little easier to use the squared distance[math]r^2=\frac{(a h + b k + c)^2}{a^2+b^2}[/math]to solve you have three equations one for each line and they must all have the same distance. Find the equation of the parabola with axis x=2 and passing through points (-2,3) and (10,-3)Parabola are given by the equation[math] y = a (x - x_0)^2 + b[/math]. Here the axis are [math]x=2[/math] so the equation is [math]y = (x-2)^2 + b[/math]. Putting the two known values in [math]3 = a (-2 - 2)^2 + b = 16 a + b[/math][math]-3 = a (10-2)^2 +b = 64 a + b[/math]you can then solve the simultaneous equations to find a and b.