Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. \n and  \n2y= 4sqrt(x) , 2y+2x=6
126378
The trick here is the comma between the two.  You have to figure out y by cancelling them out then to get sqrt after you cancel x would leave you four sqrt, 2=6, and to solve the area of the region sqrt is pretty simple from there.  Isn't there a formula that you are supposed to follow?  My Algebra teacher taught me to cancel out similar integers with respect to x and y here.  So if you got the answer of 3 for the x in () then you have found the area of the region.  So to check to see if this is correct you have to fill in the 3 for (x) and see if the two are equal.  This is not that hard unless you happen to be a student in Jr. HS who is up at a time when he should be asleep.
100940
OK, so first you have to graph the functions, using your graphing calculator and you will see that the function y1=-x^2+4x+2 is above the function y2=x+2. And then we must find the intersection, (either by hand or by calculator)\nby hand:\ny1=y2\n-x^2+4x+2=x+2\nand solve for x\n3x-x^2=0\nx(3-x)=0\nx=0 and x=3\nand you'll get the points of intersection, (0,2) and (3,5).\nSince the cross section is perpendicular to the x-axis, you are forced to use dx.\nOk, then we can draw the squares in the area between the curve, and the squares have the base of top-bottom (i.e y1-y2) which is 3x-x^2. So the area of the one square is (3x-x^2)^2 right?\nOK then put it all in the cross section formula:\nintegral (from a to b)of A(x) dx\nso...\nintegral (from 0 to 3) of (3x-x^2)^2 dx \nThen for question b, you have to use equilateral triangle.. ok, so the height for equilateral triangle is (a/2)*(sqr)3 if the base is a. (Phytagorean Theorem)\nso your integral is\nintegral (from 0 to 3) of ((1/2)(sqr(3)/2)*(3x-x^2)^2) dx \nWell, I hope that's right, sorry if I made mistake.
290140
Rec first  8 X 4 = 32\n32 X 16 = 512  this is Area of rectangle.\n\nCircles pi X r^2 = Area\n3.14159 X 8 X 8 = 201\n\n512 - (2 X 201) = shaded area\n109.87 = shaded area\n\nOK ?
363046
Slope of line AB=(2-0)/(0-(-4)=0.5\nSlope of curve y=sqrt(4+x)\n=dy/dx=1/(2*sqrt(x+4))\nSolving\nat x=-3, dy/dx=0.5
426284
both are easy\n\nbecause slope is -5 so\n\neqution of line is\n\ny = -5x+c\nput y = 6 and x = 3 to get\n\n6= -15+c \nor c= 21\n\nequaton of line \n\ny = - 5x + 21\nor 5x+y - 21 = 0 as you got it\n\nNow for the area\n\nit cuts y axis at x =0\nso y = 21\n\nwhen it cuts x axis at y =0 or x = 4.2\n\nit is a right triangle\nso area = 21 *4.2/2 = 21*2.1 = 44.1
413785
y = 3x²\n\nS(x, y) = (3x, 2y)\n\nSo y = 3x² →2y = 3(3x)²\n\nie y = 27x²/2
242093
let t=2x+4\n\ndt= 2dx\ndx=dt/2\nint{sec^2(2x+4)dx}=int(sec^2(t) dt/2}\n=tan(t)/2 +c= tan(2x+4) /2 +c
485189
1.) y = x/4 - 1/4\n\n2.) ((x^2)/36) + ((y-3)^2/36)
445303
if y=0\n\n2x+4(0)=-8\n2x+0=-8\nx=-4\n\n\nif x=0\n\n2(0)+4y=-8\n0+4y=-8\ny=-2\n\nIf your doing co-ordinate geometery, two points on the line are (0,-2) and (-4,0)
159959
Part a)\n\nx = width\ny = length\n2x + 2y = 1000, so y = 500 - x\n\nA = xy = x(500 - x) = 500x - x^2\n\nb)  x should be positive (and small enough to fit the plot of land that he is planting the garden in)\n\nc)  If one corner is at the coordinates (0,0), the opposite corner is at (x,y) as defined above.\n\nd)  A = 500x - x^2\n\nCalculus Solution:\n\ndA/dx = 500 - 2x\nWhen 500 - 2x = 0, the area is a maximum.  Here, x = 250 and y = 250, so the maximum area is 62,500 square yards.\n\nNon-calculus Solution:\n\nGraph the equation A = 500x - x^2.  You can see that the parabola crosses the x-axis at x = 0 and at x = 500.  By symmetry, you know that the peak of the parabola must be half-way between 0 and 500 at x = 250.  When x = 250, A = 62,500, which is the maximum area.
318266
The previous two answerers found the volume of the region under that surface and above R. That is not what is being asked for.\n\nThe quantity you have to integrate is \nsqrt(1+(dz/dx)^2 +(dz/dy)^2), where the derivatives are partial derivatives.\n\nIn this case, dz/dx=-1 and dz/dy=-2y, so we have to integate\nsqrt(2+4y^2) over the region R.\nThis gives\nint(0,2)(0,x) sqrt(2+4y^2)dydx.\nThis is unpleasant, so we try to interchange thew order of integration:\nint(0,2)(y,2) sqrt(2+4y^2)dxdy\n=int(0,2) (2-y)sqrt(2+4y^2) dy.\nThis is still rather messy, but can be broken into two integrals, the first of which is done by trig substitution and the second by dsimple substitution u=2+4y^2.\n\nI'll let you do the rest.
424467
Easiest way is to write them as two equations equal to zero.  Then equate them...\n\nStarting with your first example:\nx² + y² - 4 = 0\nx² + (y-6)² - 25 = 0\n\nNow equate these:\nx² + y² - 4 = x² + (y-6)² - 25\n\nExpand out the (y-6)²\nx² + y² - 4 = x² + (y²-12y+36) - 25\n\nNow cancel like terms (like x², y², etc.) which are on both sides:\n-4 = -12y + 36 - 25\n\nNow simplify:\n12y = 36 - 25 + 4\n12y = 15\ny = 15/12\ny = 5/4\n\nNow use the value you found (y=5/4) and solve for the other variable using one of the original equations:\nx² + (5/4)² = 4\nx² + (25/16) = 4\nx² = 64/16 - 25/16\nx² = 39/16\nHere you need to take the sqrt and you'll have two answers:\nx = sqrt(39/16)\nx = +/- sqrt(39)/4\n\nSo your two points of intersection for the first pair of equations are:\n(sqrt(39)/4, 5/4) and (-sqrt(39)/4, 5/4)\n\nThe same method works for the other problems...
518789
Area = pi r^2\nd Area/ dr = 2 pi r\n\nd Area = 2 pi r dr\n\nGiven d Area = 2 dr,  so substitute to get 2 dr = 2 pi r dr\n\n1 = pi r\n\n1/pi = r
196843
((x^2-2x)/4y)/((x^2-4)/6)\n=(6x(x-2))/(4y*(x-2)(x+2))\n=3x/(2y(x+2))
278104
Would be happy to help, but some information is missing.  I am assuming that you are asking about the region bounded by the following four equations:\n{2x + y < 25}  {x + y < ?}  {x < 12}  {x > 0}\n\nNeed to know what ? is in order to give an exact answer.
392716
4x+2y=8\n\nSubtract the 4x to the other side to get\n\n2y= -4x + 8\n\nDivide the 2 on all sides to get:\n\n2y(/2)= -4x(/2) + 8 ( /2)\n\nYour final answer will be:\n\ny= -2x + 4\n\n4= y intercept and -2 equals the slope.
478048
the area of a rectangle is:  a = b*h\n\nSo, the area with the path is 400 (given) = (30-2x)(20-2x)\n400=600-100x+4x^2\n0 = 200-100x+4x^2\n0 = 50 - 25x +x^2\nand then solve from there using quad eq.\n\naddendum:  using the quadratic equation, the answer is about 2.1922.\n\n2 ft is not correct:  the area you are trying to get is 400.  if the path were 2 feet, then the area would be b * h = (30-4)(20-4) [the path goes all the way around so 2 ft from each side].  So 26*16=416.