Determine the zeros of the expression -16t^2-8t+24.

1 answer:

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$-16t^2-8t+24=-8(2t^2+t-3)=-8(2t^2+3t-2t-3)\\\\=-8[t(2t+3)-1(2t+3)]=-8(2t+3)(t-1)\\\\\text{the zeros}\\\\2t+3=0\to t=-1.5\\t-1=0\to t=1\\\\Answer:\ \boxed{-1.5\ and\ 1}$

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Pls help! I'm confused. Can you explain this if you can.

alexdok [17]

Answer:

From a table, for an ordered pair (0, y), y will not be zero. From a graph, the y-intercept will not be zero. From an equation, it will have the form, y = mx + b where b is ≠ 0.

Step-by-step explanation:

From a table, for an ordered pair (0, y), y will not be zero. If there is not a constant rate of change in the data displayed in a table, then the table represents a nonlinear nonproportional relationship.
From a graph, the y-intercept will not be zero. This means that it doesn't contain or go through the origin.
From an equation, it will have the form, y = mx + b where b is ≠ 0. (not equal to zero). If an equation is not a linear equation, it represents a nonproportional relationship. A linear equation of the form y = mx + b may represent either a proportional (b = 0) or nonproportional (b ≠ 0) relationship. Therefore, when b ≠ 0, the relationship between x and y is nonproportional.