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  1. Rectangle 相關
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  1. 1 ) Rectangle 周界 is 68 cm , 求 biggest area . The biggest area of... 432 cm ^2 , 求 biggest 周界 . The flatter size of the rectangle , the longer perimeter of it is. Taking cm as the smallest...

    分類:科學及數學 > 數學 2010年06月30日

  2. 因為rhombus and rectangle are both a type of parallelgram so you should first prove they are...equal sides so we need to prove the four sides are equal for rectangle , the definiton of rectangle is parallelgram which have four right angles so...

    分類:科學及數學 > 數學 2008年05月03日

  3. Let width is x cm and length is x+5 cm. 2(x+5+x) = 62 x+5+x = 31 2x+5 = 31 2x = 26 x = 13 The length of the rectangle is 13+5 cm = 18cm .

    分類:科學及數學 > 數學 2013年08月14日

  4. To draw a rectangle , you can use the CDC:: Rectangle () method. Its syntax is...void CExoView::OnDraw(CDC* pDC) { pDC-> Rectangle (20, 20, 226, 144); } 圖片參考:http://functionx.com/MFCFundamentals/...

  5. a) let P be (x,y) x = (1(-1) + r(5))/(1 + r) = (5r - 1)/(1 + r) y = (1(3) + r(-8))/(1 + r) = (3 - 8r)/(1 + r) ans: P is ((5r - 1)/(1 + r) , (3 - 8r)/(1 + r)) b) assume line AB and the line x - 2y - 5 = 0 intercept at P Put P into the eqn., we have (5r - 1)/(1 + r) - 2(3 - 8r)/(1 + r) - 5...

    分類:科學及數學 > 數學 2008年05月15日

  6. a) Please refer to additional mathematics textbook on co-ordinate geometry and area of triangle. b) Since R is on the y-axis, its co-ordinates will be (0,y). 1) Area of trapezium QR with the x-axis = 3(2+y)/2. 2) Area of trapezium RP with the x-axis = 4...

    分類:科學及數學 > 數學 2008年05月15日

  7. a) Eqn. of line AB: (5 - -5)/(-3 - 11) = (y-5)/(x - -3) -10/14 = (y-5)/(x + 3) 14(y-5)= -10(x + 3) 14y - 70 = -10x - 30 10x + 14y - 40 = 0 5x + 7y - 20 = 0 -------(1) Put x=y into (1), 5y + 7y - 20 = 0 12y = 20 y = 5/3 x = y = 5/3 P is (5/3 , 5/3) Put y=0 into (1), 5x + 7 x 0 - 20 = 0 5x = 20...

    分類:科學及數學 > 數學 2008年05月15日

  8. As follows~~~ As follows~~~ 圖片參考:http://i182.photobucket.com/albums/x4/A_Hepburn_1990/A_Hepburn01May052007.jpg?t=1209989295

    分類:科學及數學 > 數學 2008年05月15日

  9. Using similar triangle, y/8 = (4-x)/4 y = 8 - 2x So area of rectangle = xy = x(8 - 2x) = -2(x^2 - 4x) = -2(x^2 - 4x + 4 - 4) = -2(x - 2)^2 + 8 Thus the maximum area is 8cm^2

    分類:科學及數學 > 數學 2011年01月12日

  10. (a) width of original rectangle = 224/x width of new rectangle = 224/x + 1 = (224+x)/x (b) area of new rectangle : (x-1...

    分類:科學及數學 > 數學 2019年09月19日

  1. Rectangle 相關
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