一道sat数学题 tx+12y=-3 t is a constant ,if the slope of the line is -10,what the value of

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一道sat数学题 tx+12y=-3 t is a constant ,if the slope of the line is -10,what the value of
一道sat数学题 tx+12y=-3 t is a constant ,if the slope of the line is -10,what the value of

一道sat数学题 tx+12y=-3 t is a constant ,if the slope of the line is -10,what the value of
问题:tx+12y=-3,t是常数,如果这个线性函数的斜率是-10,t是几?
原式可化为(-t/12)x-1/4=y 斜率为-t/12,因为斜率为-10
所以可得-t/12=-10 解得t=120