What is the area for y=x^3-x  and   y=3x?
342990
Take the derivative of each equation. Subtract one from the other. That'll get you the area enclosed between the two.
174683
The simple way to find the area of any shape is to break it up in to trangles and squares.  \n\n\nOr use integration... but i do not think that is what you are looking for
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
164954
x^3 – 3x^2 + 4x – 12\n= x^2(x – 3) + 4(x – 3)\n= (x-3)(x^2 +4)
831999
y = (x+3)(x-1)(x-2) = x^3 - 7x + 6
1015745
here goes again .. seems like im gona do the whole hw for u ;) to find the limit of the area that ur gona revolve around the x=-2 axis u have to draw the two equations again and the area enclosed is basicly the shape of a circle (not a perfect circle though) and its located right above the x-axis touching it from below and the center of it is on the y-axis .. now the limit is the intersection of these two euations y=x^2 and y=2-x^2 that is x^2= 2- x^2 ==> 2*x^2 = 2 ==> x^2=1 ==> | x | = 1 ==> x = +1 or x = -1 .. and thats ur interval limits
1117529
y = 3x^2 + 12x - 7\nWhat do the directions say?
1017799
V = area of base times height\nheight = 14\nbase = 3*3*pi = 9pi\nV = 14*9pi\nV = 126pi\nV = 395.8
921006
Area = πab where a and b are the semi-major and semi-minor axis length
900288
A=3*sqrt3 *a^2\nhere a =4\nso A= 3*sqrt3*16\n= 48*sqrt3\nor= 16*2.59808\n= 41.56928 sq.m
389991
height times the base area. divided by three.
698387
y= -(2/3)x^2 + (2/3)x + 4\n\n\n(P): y = ax^2 + bx + c\n\nPoints (0;4)  (-2;0)  (3;0) are element of (P)\n-->  4 = 0a + 0b + c --> c = 4\n0 = 4a - 2b + c\n0 = 9a + 3b + c\n\n--> 5a + 5b = 0 --> b = -a\n--> 4a + 2a + 4 = 0 --> a = -2/3,  b=2/3
1148527
You need calculus. You can find that the places where the two curves intersect are at x=pi/3 and at x=-pi/3. So using the formula \nArea = integral (r^2/2), the area is:\n4(pi/3)n^2 - 4*(3^(1/2))*n^2 - 16n^2 (pi/3 + (3^(1/2))/2)
1066777
x-3=3\nx=3+3\nx=6\n\n\n15-(x)=7\n15-7=x\n8=x
182640
I'm not sure if this is what you're asking, but if...\nd = length\nd-3 = width\n\nthen the area is d(d-3), a.k.a. d^2 - 3d .\n\nHope this helps!
409697
As x becomes very large, 3^(-x) becomes very small compared to 3^x, and can thus be neglected. Therefore, you have (3^x)/(3^x) = 1
843398
6x(x-3y)-3(6x^2-xy) = 6x^2-18xy-18x^2+3xy=-12x^2-15xy ans.