sat数学问题,06.1s2
问题描述：
sat数学问题,06.1s2
tx+12y=-3
the equation above is the equation of a linein the xy-plane, and t is a constant. if the slope of the line is -10,what is the value of t?
步骤+翻译~~~谢谢
答
上述的方程式是一个在xy轴上的直线方程式,并且t是一个常数.如果这条线的斜度（y=ax+b的a）是-10,t的值是多少?
tx+12y=-3
12y=-3-tx
y=-1/4-(t/12)x
y=-(t/12)x-(1/4)
-(t/12)=-10
t/12=10
t=120