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標題:
計數勁in!!!!
發問:
if a man walks for x hours at a speed of x km/h and (x-1) hours at speed of (x-1) km/h he will travel a total distance of 41 km ,find the value of x????
最佳解答:
Here it is: X x X + (X-1)x(X-1)=41 X^2+X^2-2X+1=41 2X^2-2X=40 2(X^2-X)=40 X^2-X=20 X^2-X-20=0 (X+4)(X-5)=0 X=4 or X= -5 There are two values for X. If you have finding the speed, then x should be at 4 km/h. 2006-11-18 20:18:51 補充： Sorry,,,,if you ARE finding the value for the speed, should use X =4
x*x+(x-1)*(x-1)=41 x^2+(x-1)^2=41 x^2+[x^2-2x(1)+1^2]=41 x^2+(x^2-2x+1)=41 x^2+x^ -2x+1=41 2x^2-2x+1-41=0 2x^2-2x-40=0 (2x+8)(x-5)=0 x=5 or x= -4(rejected) So x=5 P.S: hour 冇可能係負數,所以唔係 -4|||||x*x+(x-1)^2=41 2x^2-2x+1=41 2x^2-2x-40=0 x^2-x-20=0 (x-5)(x+4)=0 x=5 or -4
計數勁in!!!!
發問:
if a man walks for x hours at a speed of x km/h and (x-1) hours at speed of (x-1) km/h he will travel a total distance of 41 km ,find the value of x????
最佳解答:
Here it is: X x X + (X-1)x(X-1)=41 X^2+X^2-2X+1=41 2X^2-2X=40 2(X^2-X)=40 X^2-X=20 X^2-X-20=0 (X+4)(X-5)=0 X=4 or X= -5 There are two values for X. If you have finding the speed, then x should be at 4 km/h. 2006-11-18 20:18:51 補充： Sorry,,,,if you ARE finding the value for the speed, should use X =4
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其他解答:x*x+(x-1)*(x-1)=41 x^2+(x-1)^2=41 x^2+[x^2-2x(1)+1^2]=41 x^2+(x^2-2x+1)=41 x^2+x^ -2x+1=41 2x^2-2x+1-41=0 2x^2-2x-40=0 (2x+8)(x-5)=0 x=5 or x= -4(rejected) So x=5 P.S: hour 冇可能係負數,所以唔係 -4|||||x*x+(x-1)^2=41 2x^2-2x+1=41 2x^2-2x-40=0 x^2-x-20=0 (x-5)(x+4)=0 x=5 or -4
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