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發問:
1)A cyclist travels 40 km at a speed of xkm/h . Find the time taken in terms of x . Find the time taken when his speed is reduced by 2km/h . If the difference between the time is 1 hour , find the original speed x. 2)ABCD(rectangle corner) is a rectangle with AB =12cm(the length) and BC =7cm(the width)... 顯示更多 1)A cyclist travels 40 km at a speed of xkm/h . Find the time taken in terms of x . Find the time taken when his speed is reduced by 2km/h . If the difference between the time is 1 hour , find the original speed x. 2)ABCD(rectangle corner) is a rectangle with AB =12cm(the length) and BC =7cm(the width) .AK=BL=CM=DN=x cm . If the area of KLMN is 54cm^2 find x. 有一個長方形用ABCD黎話4個角,咁長方形入面又有一個斜ge長方形用KLMN黎表示斜長方形ge 4個角.長方形ge長係12cm,闊係7cm.斜長方形面積係54cm^2 .AK,BL嗰D係兩個長方形相隔最少ge
最佳解答:
1) time taken to travel 40km at a speed of x km/h = 40/x hrs If speed becomes (x-2)km/h time taken = 40/(x-2) Time difference = 1 Therefore, 40/(x-2) - 40/x = 1 40x - 40(x-2) = x(x-2) 80 = x^2-2x x^2-2x-80 = 0 (x-10)(x+8)=0 x=10 or x= -8 Since x is larger than 0, x=-8 is rejected Therefore, the original speed = 10km/h 2) Area of ABCD = 12x7 = 84cm^2 Area of 4 Triangles AKN, BLK, CML, DNM = Area of ABCD - Area of KLMN Therefore, x(7-x)/2 + x(12-x)/2 + x(7-2)/2 + x(12-x)/2 = 84-54 x(7-x) + x(12-x) = 30 19x-2x^2 = 30 2x^2-19x+30 = 0 (2x-15)(x-2)=0 x= 15/2 = 7.5 or x=2 Since x is less than 7, x= 7.5 is rejected Therefore, x = 2
其他解答:
1. time taken in terms of X 40/x hrs 40/x+1=40/(x-2) (x-10)(x+8)=0 x=10 or -8(rejected) x=10 original speed = 10km/h
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(maths)Problem slove by quadratic equations發問:
1)A cyclist travels 40 km at a speed of xkm/h . Find the time taken in terms of x . Find the time taken when his speed is reduced by 2km/h . If the difference between the time is 1 hour , find the original speed x. 2)ABCD(rectangle corner) is a rectangle with AB =12cm(the length) and BC =7cm(the width)... 顯示更多 1)A cyclist travels 40 km at a speed of xkm/h . Find the time taken in terms of x . Find the time taken when his speed is reduced by 2km/h . If the difference between the time is 1 hour , find the original speed x. 2)ABCD(rectangle corner) is a rectangle with AB =12cm(the length) and BC =7cm(the width) .AK=BL=CM=DN=x cm . If the area of KLMN is 54cm^2 find x. 有一個長方形用ABCD黎話4個角,咁長方形入面又有一個斜ge長方形用KLMN黎表示斜長方形ge 4個角.長方形ge長係12cm,闊係7cm.斜長方形面積係54cm^2 .AK,BL嗰D係兩個長方形相隔最少ge
最佳解答:
1) time taken to travel 40km at a speed of x km/h = 40/x hrs If speed becomes (x-2)km/h time taken = 40/(x-2) Time difference = 1 Therefore, 40/(x-2) - 40/x = 1 40x - 40(x-2) = x(x-2) 80 = x^2-2x x^2-2x-80 = 0 (x-10)(x+8)=0 x=10 or x= -8 Since x is larger than 0, x=-8 is rejected Therefore, the original speed = 10km/h 2) Area of ABCD = 12x7 = 84cm^2 Area of 4 Triangles AKN, BLK, CML, DNM = Area of ABCD - Area of KLMN Therefore, x(7-x)/2 + x(12-x)/2 + x(7-2)/2 + x(12-x)/2 = 84-54 x(7-x) + x(12-x) = 30 19x-2x^2 = 30 2x^2-19x+30 = 0 (2x-15)(x-2)=0 x= 15/2 = 7.5 or x=2 Since x is less than 7, x= 7.5 is rejected Therefore, x = 2
其他解答:
1. time taken in terms of X 40/x hrs 40/x+1=40/(x-2) (x-10)(x+8)=0 x=10 or -8(rejected) x=10 original speed = 10km/h
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