神马作文网sat数学（英语）求解析
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sat数学（英语）求解析
tx+12y=-3
the equation above is the equation of a line in the xy-plane,and t is a constant.if the slope of the line is -10,what is the value of
tx+12y=-3
the equation above is the equation of a line in the xy-plane,and t is a constant.if the slope of the line is -10,what is the value of
上面的这个方程是XY平面里的直线方程并且t是个常数,如果直线斜率是-10,t的值是多少?
把方程整理成y=kx+b形式 k即为方程斜率.
12y=-tx-3
y=-(t/12)x-1/4
所以-t/12=-10
t=120
把方程整理成y=kx+b形式 k即为方程斜率.
12y=-tx-3
y=-(t/12)x-1/4
所以-t/12=-10
t=120