Use the total differential to estimate the error in the calculated area of the triangle.?
816108
dA = ½ (b.dh + h.db)\nSo δA ≈ ½ (b.δh + h.δb)\n= ½ (100 x ±1 + 10 x ±2)\n= ½  * ±120\n= ±60ft²\n\nCheck:\nMin area = ½ * 98 * 9 = 441ft²\nMax area = ½ * 102 * 11 = 561ft²\n\nNow 500±60 ft² = (440 to 560)ft²\n\nSo estimate is pretty close!!!
481212
the area under the curve f(x)!
754846
the arc length differential is da^2=dx^2+dy^2. divide RHS by dx^2 and square root both sides to get\n\nds=sqrt(1+(dy/dx)^2)\n\nso y(x)=x^4 => dy/dx=4x^3\n\nso to get arc length we integrate ds with respect to x with limits x= 0 to x= 1\n\nintegral from 0 to 1 of sqrt(1+16x^6) with respect to x\n\nsince a rounded answer is required i assume you must use a numerical scheme such as Simpson's rule to obtain answer but again the integrand is sqrt(1+16x^6) limits x= 0 to 1
1019108
Do you have limits between which to calculate the area? If not you can solve for the values where f(x) = g(x)\nyou should get two values since g(x) is a second degree function.\nThe 2 values are your limits for integration.\nIntegrate each function with respect to dx between the limits and take the difference of the two. Since area is always positive the value should be the absolute value of the difference.\n\nI think that is enough hint. I won't do it for you. But here is what you should start with:\n\nsolve for x where f(x) = g(x), the two values for x are the two limits for your integration, for example x1 and x2.\nthen integrate\n  x1\n| ∫( f(x) - g(x) ) dx |\n  x2
716347
Of three points, A,B,C\n Check the slope of A-B,  B-C, A-C.  If any of the slopes are the same, you have parallel lines and thus no triangle.  So, to do (2), do (4) first.\n  If you know the distance between A-B, B-C, A-C, you can calculate the area with the formula, so (5) also depends on (4).\n\n You write your own code.  yA isn't a site to do your homework, since if you don't do it yourself, you don't pass the class!
184797
find the angle between two sides.
577829
There are two ways to measure the area of a shaded figure. The third is by integral calculus.\n\nMethod 1. \n\nStep 1. Draw the figure on a graph paper by choosing suitable scales in x and y axes.\nStep 2. Count the small squares that are shaded. If a square is not whole, then count it as one if it is greater than half a square. Skip it if it is less than half. And of course, two halves make a whole.\nStep 3. Convert it to area by multiplying the number of squares with area of 1 square.\n\nMethod 2.\nCase 1:\nIf the figure is a polygon, draw diagonals from from corner points and break it down to triangles.\nNow, the area of triangles can be calculated mainly by two formulae.\n\nArea of a triangle = (1/2) base * altitude\n= square root of [s(s-a)(s-b)(s-c)] \n\nThe second formula is very useful and practised in surveying. Here, a,b, and c are the three sides of a triangle; and s is the semi-perimeter i.e. s=(1/2)(a+b+c)\n\nCase 2:\nIf the boundary of the figure is curved, the the matter is a bit harder. If it is a circle or a part of a circle, use formula: area= pi*r squared; pi is approximately 3.1416..  Or else, you have to replace the curve by a straight line such that the area remains the same by eye.\n\nBy integral calculus, area can be found by using table of formula.
73727
Angle of deviation, d = i1 + i2 - A\nA= r1 + r2\nIn minimum deviation, µ = {sin(A+D)/2}/(sinA/2)
152408
Use excel to give delta y/delta x on the curve. Look for a point where this value (the derivative) shoots up in value. That's your point.
863802
area=(1/2)d1*d2\n=(1/2)*28*13\n=14*13\n=182 sft
1136463
It's been a while since I had calculus, but I think it's something like this:\n\nYou'd have to take the integral from 0 to k (treat k as a constant). When you have that, take that value (which is equivalent to the area) and set it equal to 3. Then, algebraically, solve for k.
406687
Here's the basic idea:  We usually estimate area under the curve using Riemann Sums, ie rectangles.  The height of these rectangles is most often based off the right endpoint, left endpoint, or the midpoint (though we could use any point - and we do when trying to prove the Fundamental Theorem of Calculus).\n\nI often ask students, "Given a Riemann Sum, how could we make it more accurate?"  Well, the answer I'm generally looking for is: use more rectangles... But, another common (and worthwhile) answer is: Average the Right Riemann sum and the Left Riemann Sum.  \n\nIn fact, this simple suggestion results in the trapezoid rule.  Try drawing some figures and convincing yourself that the average of the right and left triangles can be drawn as a trapezoid...\n\nSo, suppose we have n rectangles and an interval from a to b.  I will use the following notation:\n(delta)x = (b-a)/n\nx_0 = a\nx_1 = x_0 + (delta)x\n...\nx_n = b\n\nThen, we can write the Right Riemann Sum as follows:\nR = (delta)x*f(x_1) + (delta)x*f(x_2) + ... + (delta)x*f(x_n)\n=[f(x_1) + f(x_2) + ... + f(x_n)]*(delta)x\nWe can also write the Left Riemann Sum as follows: \nL = (delta)x*f(x_0) + (delta)x*f(x_1) + ... + (delta)x*f(x_n-1)\n=[f(x_0) + f(x_1) + ... + f(x_n-1)]*(delta)x\n\nTo get the trapezoid rule, we average them: T = (R+L)/2\n\nNotice that f(x_0) and f(x_n) are each used in only one formula each, but all the others are used in both sums.  So, basically, we get:\nT =(1/2)*[f(x_0) + 2*f(x_1) + 2*f(x_2)... + 2*f(x_n-1)+ f(x_n)]*(delta)x\n\nor, just to write it different we could say:\nT =[(1/2)*f(x_0) + f(x_1) + f(x_2)... + f(x_n-1)+ (1/2)*f(x_n)]*(delta)x
1068661
Perimeter of the triangle multiplied by the height of the prism.
666057
*\n1/2 base X altitude + area (triangle)\n\n\n45 square centimeters
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
450833
join the 2 pointz frm da arc vit da radius.measure out da value of da angle forned here.then substitute da values 4 this formula:  2*pie*radius/da angle u measured\n\nhappy solvin!
771519
The triangle formed by the three centres has area\nsqrt(3)/4  (Standard result for equilateral triangle).\n\nEach of the little bits between that triangle and the arcs that surround it is\n(area of sector with angle pi/3) - area of triangle\n= pi/6 -sqrt(3)/4, and so the total area of that inside bit (a sort of curved triangle formed by the three centres and the arcs joining them) is\n3*(pi/6 -sqrt(3)/4)  + sqrt(3)/4\n=pi/2 - sqrt(3)/2   [I'm sure there must be a slick trick to avoid all this!]\n\nNow look at the region common to any two of the circles -- I think it's called a "lune".  If you join the points of intersection of the circles, so as to halve the lune, each half has area\n=(area of sector with angle 2*pi/3) - (1/2)*1*sin(2*pi/3)\n= 2*pi/3 - sqrt*(3)/4\n\nDouble this to get the area of the "lune":\n4*pi/3 - sqrt(3)/2\n\nNotice that if we add the areas of two circles, we've included the area of a "lune" twice, so to get the total area of two circles we'd have to subtract the area of a "lune".  So:\n\nNow add the areas of the three circles, subtract the areas of three "lunes" because each of them is included twice.  However the bit which is common to the three lunes has then been subtracted three times, so we should add twice its area.  i.e.\n\n3 circles (3*pi) - 3 lunes (3*(4*pi/3 - sqrt(3)/2) + 2*middle bit  (2*(pi/2) - sqrt(3)/2)\n\nYou can finish the calculation, but I bet I've made a mistake somewhere.  Haven't time to check.  Sorry.