is it possible to to draw a geometric figure that veries the formula (a+b)^4?
351583
how about instead of (a+b)^4 you plot (x+y)^4?\n\nit's quite a funny 3D plot.\n\nLooks like a conehead balancing on his head.\n\nSeriously... draw a cone rounded off on the top and then flip the paper upside down and you'll get an approximation of (a+b)^4.
624769
You can use sin and cosine.
882816
You can't do that on a calculator. What you can do is use the calculator to get the graph drawn then use your algebra to figure out where to shade. \nLooking at the graph, you know that it's a parabola with the vertex (4, -3). test the point (0,0) outside of the parabola.....\nsince you get\n0 is greater than or equal to 0 - 0 + 13\nsince that is a false mathematical statement, you know that you shade inside of the parabola....\nlets test a point inside/on the parabola and see, like (4, -3)\n- 3 is greater than or equal to 16 - 32 + 13\n-3 is greater than or equal to - 3\nthat is true so you now know to shade inside your parabola\nthat's how you do problems like that w/a calculator\nsince your calculator can do all of the operations, it's really easy
780311
You are on the right track.\nThis a geometrical problem that requires drawings.\nCheck this link http://departments.kings.edu/chemlab/chemlab_v2/packgeo.html\n\nI'll get back to you later \n\nso later it is\n\n3--------R=r(1+2/3^.5)\n7--------R=3r\n19------R=5r\n37------R=7r\n61------R=9r
720161
Yes, I can:\n4! + 4 \n=\n(4 x 3 x 2 x 1) + 4 = 28\n\n(BTW, 4squared + 4squared = 32)
743164
Remember the definitons of tangent, cotangent, etc. as they relate to right triangles.  Draw a triangle for both α and β.  Use the pythagorean theorem to finr the hypotneuses.  You will find the sides of the α triangle to be 3, 4, 5.  so:\n\nsin α = 4/5\ncos α = 3/5\n\nDo the same for β and the triangle is 5, 12, 13 with:\n\nsin β = 12/13\ncos β = 5/13\n\nFinaly, use the identity:\n\n sin(α-β) = sin α  cos β - sin β cos α \n\nSubstituting:\n\nsin(α-β) = (4/5)(5/13) - (12/13)(3/5) = (20 - 36)/65 = - 16/65
618490
Just draw a right triangle with short sides of 10 and 24.\n\nThe hypotenuse can be calculated by the Pythagorean theorem as the square root of 24^2 + 10^2.  That works out to 576 + 100 or 676.  The square root of 676 is 26.\n\nThe direction is roughly ENE (East North-East).
175327
Try this, maybe?\n\nhttp://mathforum.org/library/drmath/drmath.elem.html
328129
just draw a triangle has right angle which dimensions 1,sqrt(3),and hypothenus 2. if you want it, i can send to you
650879
Use string, paper, and a ruler.  The string can represent the perimeter, and you can draw a grid on the circle to represent the units of area.  Just don't get too in depth on the origins of pi.
101215
Obviously n=2 is a sphere, and n=1 is plane at 45degrees, and n=0 is a defective case.\n\nYou're talking about some geometric generalization of a sphere, with a non-Cartesian distance function (X^n + Y^n + Z ^n).\n\nFor general integer n>=3, or for non-integer n, I can't see a nice geometric interpretation, other than this:\nIf you set n~2 like say n = 2.1 or 1.9 you get a distorted sphere. For n>2, the sphere will bulge inwards in the 8 regions where X,Y,Z are not close to 0, for n<2 it bulges outwards.\n\nTry plotting it in Matlab or Octave and see.\n\nThings that this is *not*:\n- an n-dimensional hypersphere with n=3 (x^2 + y^2 + z^2 + ...x_n^2 = r^2), since the exponent is always 2\n- a spheroid\n\n\n\nThis is _remotely_ similar to visualizing Diophantine equations or Fermat's Last Theorem, but allowing X,Y,Z to be reals, not just integers, i.e. (X,Y,Z) not restricted to lattice points (also Fermat only had X,Y,R-terms, no Z-term, but similar)?\nVery interesting question, not aware if people have got anywhere.\n\nGood luck and let us know what you find.
712502
You know that the diagonal forms a right triangle with the length and the width. So you can use the Pythagorean Theorem.\nx=width=a\nx+1=length=b\n4=diagonal=c\n(a^2)+(b^2)=(c^2)\n(x^2)+(x+1)^2=(4^2)\n(x^2)+(x^2)+2x+1=16\n2(x^2)+2x+1=16\n2(x^2)+2x-16=0\nIf you use the quadratic formula thing (-b plus or minus, etc.)\nthen you get:\nx=2.2838\nor\nx=-7.5\nBut you know that a dimension cannot be negative, so it must be the first one.\nSo if \nx=2.2828\nthat means that that's the width. If you add one to that, you'll get the length.\nSo, the dimensions are:\n2.2838 x 3.2838
863713
The only thing I remember about geometry is...a2 + b2 = c2.\n\nSomehow, I doubt that helps you.....sorry.
1024676
For a simple wheel, you would use the formula.. \n\n(sin2 x)^2+(cos x)^2 =1\n\nsimilarly, every arc or every line drawn is formed of lines whose distance depends on the adjacent line or your screen size. (/2 or /4or some ration of it for position)
99265
I asked that same question to my teacher last time. She told me that I have to Seperated the shape into different common shapes. (for example like lets say I wanted to seperate a "square" into some common shapes I could turn it into two seperate triangles or four different triangles. Well Just try to use the formulas you know for the areas of the different disected figures you get.
187653
in geometric sequence \nUn = U1[r^(n-1)]  \nwhere,\nU1 = first term\nr = common ratio\nn = number of terms\n\nthen we can say that,\nU4 = U1[r^(4-1)]\n24 = U1[r^3]    -> let this be equation (1)\nand\nU7 = U1[r^(7-1)]\n192 = U1[r^6]   -> let this be equation (2)\n\nif you divide equation (1) by equation (2), you'll get\n\n(24/192) = U1[r^3] / U1[r^6]\n(1/8) = 1/r^3\ntherefore \nr = 2\nif r=2 then you can get U1 by substituting r in equation (1)\n\n24 = U1[r^3]\nU1 = 24/8\nU1 = 3\n\ntherefore,\n\nUn = 3[2^(n-1)]
891091
Yes, It is possible if we use x,y,z plane in plotting graphs. When using this method we should be very precise.\n\n3D model gives very spectacular graphs. There is algebraic expression for many 3D shapes e.g sphere, cones