By what percentage does the plate's area decrease?
114715
change in diameter=0.10xcoef.of thermal elongation(350-20)=y\nnew dim.=0.1-(change in diam.)=z\n%age change in area=1-(z/0.1)**2
706050
Ok... the first thing to realize is that you are finding the maximum area with the least amount of materials.  To do that you should put the shed at a corner of the rectange, that way two sides of the shed act as part of the fence.  Assume that the shed is not "usable area" for the shep pen.  Accordingly, you need to solve the problem twice, once with using the shed as part of one side, and once using the shed as a corner.\n\nOk, the first thing you need to do is show that the maximum area of a rectangle occurs when the lenght and width are equal.  That is the only part that needs any type of equation analysis, and it really doesn't require calculus.\n\nA=L*W\nAssuming that the width is at most as long as the L, then the equation goes to:\nA=L*(x*L), where x a percentage, ranging from 0 to 1.\nClearly, the area (A) will be maximum when x=1.  That shows that in your problem, the pen will have the maximum area when the length and width of the rectangles are equal.\n\nSo, you know that the fence is 160m., and you have 4 sides, s1, s2, s3 and s4.\nSo:\n160m=s1+s2+s3+s4\n\nNow, if you use the shed as part of one side, namely s4, then the perimeter equation for the shed is:\n180m=s1+s2+s3+s4+20m\n\nSince the maximum area occurs when the lenght and width are equal:\ns1=s2=s3=s4-20m, so substituting s2, s3 and s4 for s1 in the permiter equation, you get:\n4s1+20m=180m\nSolve for s1,\ns1=45m (call this the length)\ns4=45-20=25m\nin this case, the area will be S1*S1, since the lenght and width of the pen will be equal.  \nSo the area is 45*45=2025 sq. m.\n\nNow, put the shed in a corner of the sheep pen.\nSo in our example, we'll make the shed part of s1 and s4.\n\nso the primeter equation will be:\ns1+20m+s2+s3+s4+20m=200m\nNow, s1=s2-20, s3=s2 and s4=s2-20\nSo substituting for s1, s2 and s4, you get:\ns2-20m+s2+s3+s4-20m+40m=200m\nsolve for s2,\ns2=50m=s3 and s1=30m (s2-20m)\nSo max area is 50m*50m=2500 sq. m\nBut, now you need to reduce the area by the area of the shed (400 sq. m), so final usable area is:\n2100 sq. m which is more than using the shed as part of only 1 side.
854605
the question is not clear enough\n\ni think the lenght of cross section is the lenght of diagonal note it by d\nso let one side be l, the other side will be sqrt(d^2-l^2)\nThen the area is A(l) = l*sqrt(d^2-l^2).\nthe derivative of this is ...\nA'(l) = (d^2-2l^2)/(sqrt(d^2-L^2)) which is zero when d = sqrt(2) l, which happens only when the rectangle is a square. It remains to show that is a minimum( not a maximum). I let you do this...
701955
You could use graph paper to get a good approximation of the surface area. Using a large circle would increase the accuracy.\nCount the whole squares, then count the half squares. \n\nAlternatively would be to use graph paper, then precisely cut out the circle, then cut the graph paper. On another sheet of graph paper, try to fill squares with the incomplete pieces from the circle.\n\nModifying Mike's answer, using a sensitive microbalance, find an average mass of one graph square then measure the mass of the cut out circle.
177751
The particles in the dish and salt will expand when heated, or condense when cooled.  In this case the mass after it cooled down should be smaller
920664
area a to b is 1/2..\n\nhow? assume a side of the inner square as 1 so that its area would be also 1, by pythagorean theorem, half of the side of square b is equal to squareroot of 1/2, so the side of b is 2*squareroot(1/2), that makes the area to be equal to 2, getting the ratio of areas, that is 1:2 n_n
415440
r decreases becuase the ellipse is becoming thinner i think that would happen\n\nhope u find out if my answer is wrong GOOD LUCK!!!!!!!!
984623
It will be 1/(x^2)\n\nFor example\n\n(1/2)^2 = (1^2)/(2^2) = 1/ 4\n(1/3)^2 = (1^2)/(3^2) = 1/ 9\n(1/4)^2 = (1^2)/(4^2) = 1/16\n\nSo the result actually gets smaller even though you are squaring it
497121
Dissipation in a yield line\nIf the axes of rotation for two slab parts are not at the same depth measured from the slab surface,\nthe relative displacement discontinuity is no longer perpendicular to the yield line. The angle\nbetween the displacement discontinuity and the yield line changes with the depth from the slab\nsurface and this must be taken into account when calculating the dissipation.\n\nAnd you will find more from this book.
134473
The answer is obviously (A). The bar may shrink a little, but it can't disappear completely, as the options B, C and D imply!!\nHence, by process of elimination, the answer is (A). :-)
1135128
0.41\nor 41/100\n\n41 is prime so it cannot get smaller.
1014916
i second the a answer.  the field diameter decreases.  even if you solve this problem by process of elimination b c and d are all ridiculous answers if you give it any thought.  e not correct either, because it would occupy a larger portion since you are "zoomed in" and focusing on a smaller section of the slide or whatever so it would occupy a larger portion of the field.
404470
a) AX = 2.25 cm.\n\nb) area AXY upon area ABC = 0.5625\n\nc) area AXY upon area trapezium XYCB = 0.4375
297111
Think of the first as the area of a flat round plate sweeping out the inside of the figure you get from the rotation.  What will be its area? πr² where the r is the y value of √(9 + X²).  That changing area will sweep out the volume from whatever the limits are (you don't say), so it's the integral of π(√(9 + X²))², or π(9 + X²).\n\nIn the second case, the thing sweeping out the volume is the surface area of a cylinder as its radius grows.  Thats the circumfrence of the base times the height.  The circumfrence is 2πr where r is the X value and the height is the difference between Y = x and Y = X/2.  So it's the integral from 0 to 2 of:\n\n   2πX(X-X/2).\n\nThinking of an area sweeping thru the volume helps.
460116
It depends what it looks like.\nWithout seeing the actual picture I cannot find the area.
826403
hmmm taking me way back here.  when you break the area under a curve into rectangles of differential width,  either the left top corner will lie on the curve or the right top corner will.  \n\nas you decrease the differential width the area will become more exact until bada bing you are doing calculus.  i think.
798812
Yes its the area...say of the parking lot.  This is the Rational Method.  It looks alot like darcy's law.  It is very simple and easy to use.  However, for undeveloped areas....forested, grass...try the SCS Method.  It is a little more complicated but produces better results.