Find the area of the regions between the cure abd the horizontal axis?
957759
a. 0 = 7-x^2, x^2=7, x = +/-sqrt(7)\nArea = Integral[-sqrt(7), sqrt(7)] (7-x^2)dx = 7x - (1/3)x^3 |-sqrt(7), sqrt(7) = 2[7sqrt(7) - (1/3)sqrt(7)^3] ~= 24.69 u^2\n\nb. 0 = x^4-8, x^4 = 8, x = +/-8^(1/4)\nArea = (1/5)x^5-8x |-8^(1/4), 8^(1/4) = 2[(1/5)8^(5/4)-8(8)^(1/4)] ~absv= 21.53 u^2
572318
1) Let x:the length , y:the width of the picture.Note:x,y>0\nThe first equation we have is : x=y+4 (1)\nThe combined area has the length is x+4,the width is y+4       \nso the second equation we have is: (x+4)*(y+4)=140 (2)\nReplace (1) in (2)-->(x+4)x=140--> x^2+4x-140=0 -->x=10 or x=-14 .since x>0 then x=10(cm)-->y=6(cm).\n2)  Let x: the length,y:the width.Note:x,y>0\n    The first equation we have is x=y+3 (1)\n    The second is: the area=x*y=550 (2)\n    Replace (1) in (2)--> (y+3)y=550--> y^2+3Y-550=0 --> y=22 or y=-25.Since y>0 then y=22(cm)-->x=25(cm)\n3)  Let x,y:two positive integers.Supposed that x>y>0 \n    we have two equation:x*y=240(1) and x=y+8(2)\n    Replace (2) in (1)-->y(y+8)=240--> y^2+8y-240=0 -->y=12 or y=-20.Since y>0 then y=12-->x=20.
992822
Let the breadth of each red rectangle be x\n\nSo, area of red portion = 4x + x(3-x)\n\nSo, area of white portion is (4-x)(3-x)\n\nSo, 4x+x(3-x) = (4-x)(3-x)\nor 4x + 3x -3x^2 = x^2 -7x+12\nor 4x^2-14x+12 = 0\nor 2x^2-7x+6 = 0\nor 2x^2-6x-x+6=0\nor(2x-1)(x-3)=0\n\nx=1/2 or 0.5 foot is the reasonable answer.
380933
Surface areas and lateral areas are always fun to find.  Surface areas for right prisms and cylinders is always the sum of the lateral area and two times a base.  (Lateral area is the sum of all the surfaces that aren't bases.)\n\nIn the case of a cylinder, the lateral area is really a rectangle - imagine cutting open a roll of toilet paper.  The length is really the circumference of the top (base) and the height is, well, the height of the roll.  So the lateral area is 2*pi*r*h.\n\nThe surface area is the lateral area plus two bases.  A base in this case is given by pi*r^2.  So the total surface area is 2*pi*r^2 + 2*pi*r*h.\n\nThe idea behind lateral and surface areas will carry over when you start working on right prisms (if you're not already doing those).  Lateral area is, in the case of a prism, the perimeter of the base (similar to cylinders, except we used "circumference" in place of "perimeter") times the height of the prism.  And the surface area is the lateral area plus two bases - in this case, the bases vary (you may have squares, triangles, trapezoids, etc.), so your formula for the total surface area will vary as well.\n\nThis is probably more than you wanted to know, but it's always good to relate things you know to things that you're learning.
657713
The area of a parallelogram is b × h, where b is the length of the base of the parallelogram, and h is the corresponding height.   \n\nHence, as per measurements given area of the parallelogram  would be 6 X 8 = 48 inches  and dividing it by two (to get half) gives us 24 Therefore the area of half of the parallelogram would be 24 sq.in.
587075
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.
332600
to find areas of parallelograms use the formula: \n\nA = base×height.
191332
Find the dementions and multiply.
957793
This is a huge question........here is the easiest way. Find the area between the line and the x-axis between the limits of integration. Areas above the axis are positive and below the axis are negative.
856302
Surface Area = (2 •  p   • r²) + (2 • p  • r • height)\n\nWhere (2 • p • r² ) is the surface area of the "ends" and\n(2 • p • r • height) is the lateral area (the area of the "side").\n\nVolume   =   p • r² • height   =   ¼ • p • d² • height\n\np = Pi
85251
The area under any function can be got by the integration from an initial point to a final point\n\nfor example y(x)=x^2 and we are intrested to solve the area between x=1 to x=3\n\nthe integration = x^3/3 from 1 to 3\nthe area = (3^3/3) -(1^3/3) = 9 - 1/3 = 8.666667
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
359961
Find the areas of all 6 faces of the rectangular solid by muliplying each one's width by length.  Then add them together for the total surface area.  To find volume, multiply length by width by height.  The ratio will be the total surface area divided by the volume (or expressed as a fraction the total surface area over the volume).
729233
limit is not specified so we asume whole area\n\nx^2+y^2 = a^2\n\nthis is a circle of radius a\n\nso area = pi*a^2
599733
I believe it would be one-half times the length of each diagonal.\n\nIn other words, if the diagonals are 4 & 6 the area would be:\n\nA = 1/2 * 4 * 6 = 12\n\nIf you don't know the diagonals, but you know the length of the sides, you could construct the diagonals and calculate their dimensions using the Pythagorean theorem:  c^2 = a^2 + b^2, and then solve from there.  Keep in mind that you'll have to do some extra geometric analysis to find the length of each leg of the right triangles that are created by the intersection of the diagonals before you can use the theorem.
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
291732
The areas bounded by any s.d. can be looked up ona table, but be aware that the tables can be set to give out the answer in several different ways. One such table is on the link below. To answer your question you would have to multiply each number by 2, eg, for 3 s.d. the table gives out 0.4987 (the area between 0 and 3 s.d.) so what you are looking for is 0.9974. You don´t do "calculations" to find these numbers. That would require solving an integral that so far no one has been able to do so analitically. What you do is work a numeric solution like simpsons integration or a polinomial approximation like Abramowitz´s