The shape of a supporting arch can be modeled by h(x)+-0.03x^2+3x, where h(x) represents the height of the arc
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your equation is a little weird - i believe you mean:\nh(x)=-0.03x^2+3x\nin any case, finding the "maximum" or "minimum" of anything with a given equation is simply finding the derivative of the equation, finding "x" when that equals 0, and finding the height or whatever when x=that number.  for example, the equation above is a quadratic that looks like an arc if you graph it.  finding the derivative:\n\nh(x)=-0.03x^2+3x\nh'(x)=-(2)0.03x+3\nh'(x)=-0.06x+3\n\nnow let h'(x)=0.  this means, by looking at the graph, when the height of the arc is no longer increasing or decreasing, or has reached its max or min.  find x when h'(x)=0:\n\nh'(x)=-0.06x+3=0\n-0.06x+3=0\n-0.06x=-3\nx=3/(0.06)\nx=50\n\ntherefore the maximum height of the arch occurs when x=50.  to find the height at this point, just plug it into the given equation:\n\nh(50)=-0.03(50)^2+3(50)\nh(50)=-75+150\nh(50)=75\n\ntherefore the maximum height of the arch is 75 meters.  hope this helped!
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V=1/2(b1+b2)*h*w  where b1 and b2 are the two parallel sides of the trapezoid and h is the height and w is the width.
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Cylinder: (pi)*(radius^2)*(height)\nPrism: (area of the base)*(height)
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height times the base area. divided by three.
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find the area of the base and multiply it by the height\n\n(1/2)*b*h1*h2  for a triangular prism h1 is height of the triangle, h2 is the height of the prism\n\npi*(r^2)*h for cylinder, h is height, r is radius of the base
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since this starts as a square.  when we place the diagonal line from the bottom right corner to the upper left corner we end up with two right triangles.  This is the perfect set up to use the Pythagorean theorem, which states "the square of the hypotenuese is qual to the sum of the squares of the base and height. you may have seen it as \n       h^2 = x^2 + y^2\nwe started with a square so we know the base (x) and height (y) are each 7 sq rt 2.\nsquare the values then add them together.  this will give you that diagonal's square. ****remember to take the sqrt of the answer to find the length of the diagonal.
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go to your graphing calculator and hit the Y= button, then input your problem, it looks like an arch, that goes extremly high.. it starts on -3
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Volume of cylinder = Area of the base X height = πr2h.
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c^2=a^2+b^2 applies for right trianges.  a and b are the two legs of the triange; therefore, c, the hypothenuse = the square root of (10^2+10^2) which equals 14.14
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Cones Volume = (1/3) x (volume of cylinder) = (1/3) x (base x height) = (1/3) x (pi x r^2 x h)
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h = -16x² + 48x + 8\n\nNow clearly it is projected from an elevated position and that position is 8 ft above the horizontal ( when x = 0 h = 8)\n\nSo the NEXT time it reaches 8 ft will be the second time it is there (on the way down)\n\n8 = -16x² + 48x + 8\nie 0 = -16x² + 48x\n= 16x(3 - x)\nSo x = 0 or 3\n\nThus the next time it reaches 8 ft is in 3 second
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Let h = height, then \nb = (1/2)h\nA = (1/2)bh\n64 = (1/2)(1/2)h*h\n64*4 = h^2\n256 = h^2\nh = 16
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Let R and H be the radius and altitude of the CONE\nLet r and h be the radius and altitude of the CYLINDER\nVc=Vol of Cylinder\n\nThen Vc = pi*r^2*h\n\nIf the radius of the cylinder is zero (r=0), its altitude is the height of the cone (h=H).\n\nIf the radius of the cylinder is the same than the radius of the cone (r=R), its altitude is zero (h=0).\n\nFrom the above conclude:\n\nh=H when r=0   and   h=0 when r=R\n\nTreat h as a dependent variable (y) and r as an independent variable (x). Then, the slope of the cone (from the line above) is:\n\n(H-0)/(0-R) = -H/R\n\nTherefore the altitude of the cylinder "h" (the dependent variable) as a function its radius "r" (the independent variable) is:\n\nh=H-(H/R)r\n\nPlug this "h" variable in the formula of Vc\n\nVc = pi*r^2(H-(H/R)r)\n\nCompute the derivative of this function with respect to r (the independent variable. Remember H and are unknown but constants) and equate to zero. Next solve for r.\n\nr=0 (would yield a minimum)\n\nr=(2/3)R is the maximum\n\nPlug this value in the formula of "h" given above to obtain:\n\nh=H/3\n\nPlug this two values of r and h in the original formula Vc and obtain YOUR ANSWER:\n\n(4/27)*pi*H*R^2\n\nCompare to the volume of the cone, a little less than half of it.
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Chord length = R (see dimensions of rectangle ie flat part of slice)\nSo the size of the angle (θ) subtended at centre of circle is 60° as all sides of the triangle so formed = R.\nSo θ = π/3\n\nSo length of side of rectangle forming curved surface = length of Arc\n= Rθ\n= πR/3\n\nAlso area of segment at either end = ½R²(θ - sinθ)\n= ½R²(π/3 - sinπ/3)\n= ½R²(π/3 - √3/2)\n\nSo total surface area = Area of curved part + area of flat rectangle + 2 * area of end segment\n= πR/3 * 4R + R*4R + 2 * ½R²(π/3 - √3/2)\n= R²(5π/3 + 4 - √3/2)\n\nVolume = Area of segment * height\n= ½R²(π/3 - √3/2) * 4R\n= 2R³(π/3 - √3/2)
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A cone and pyramid have one thing in common, the base area is proportional\nto the square of the height.  If you keep the angles and overall shape the\nsame and just increase the size (like a slight zoom of a camera lens), the\nbase area is some constant times the square of the height:  A=cH^2.  The "c"\nis a constant that will eventually divide out.  Also, the volume is\nproportional to the cube of the height:  V=kH^3.  We want to find out how k\nmust relate to c.\n\nConsider two cones with identical shapes: one has height H, one has height\nH+h.  "h" is extremely small.  If H were 10 meters, h would be a millimeter.\nThe base area and volume for the first are A1=cH^2 and V1=kH^3.  The other's\narea and volume are A2=c(H+h)^2 and V2=k(H+h)^3.\n\nIf we subtract the larger smaller volume from the larger volume, we get the\nvery thin sheet at the bottom.  If h is small enough, the thin sheet will\nhave area A1=cH^2 and height h.  The thin sheet will have a volume (cH^2)h.\nIf you don't say that h is very small, then the sheet will have a more\ncomplex volume formula.  We now have V2-V1=(cH^2)h.  Let's do the algebra:\nk(H+h)^3- kH^3=(cH^2)h\nkH^3+(3kH^2)h+(3kH)h^2+kh^3  -  kh^3  = (cH^2)h\n(3kH^2)h + (3kH)h^2 + kh^3 =(cH^2)h\n\nNow, the limits:\n\n"h" is tiny, so the second and third terms are extremely small compared to\nthe first.  They are like adding a penny to a million dollars.  They really\ndon't count enough to keep.  Limit theory lets us drop them under the\ncircumstances.  This yields:  (3kH^2)c=(cH^2)h.  Divide out common terms and\nget:  3k=c, or k=(1/3)c.  Put this in the original formula for volume:\nV=kH^3=(1/3)cH^3=(1/3)(cH^2)H=(1/3)AH.  Volume equals (1/3)Base area times
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It sounds to me like you are talking about a trapezoid. Go to the website I have listed under my sources and see if what you are describing looks like the trapezoid on there. If it is a trapezoid, the area formula is  Area = 1/2h(b1 + b2). h=height, b1=base 1, b2=base 2
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volume = (1/3) b h = 1/3 pi r 2 h \nSurface area = pr2 + prs\ns= the height of the outside of the cone, while h = the height of the cone at its center