Solve T = L(5 + RS) for S 0
2 answers:
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<span>the limit as x approaches -3 of [g(x)-g(-3)]over(x+3) is the same as the derivative, or slope, of g(x) at the point x=-3, or g'(-3).
Since you are given the equation of the tangent line, the answer is just the slope of that line.
</span><span>2y+3=-(2/3)(x-3)
</span><span>6y+9=-2(x-3)
6y+9=-2x+6
6y=-2x-3
y= (-2x-3)/6
slope is -2/6 = </span>
Since you are given the equation of the tangent line, the answer is just the slope of that line.
</span><span>2y+3=-(2/3)(x-3)
</span><span>6y+9=-2(x-3)
6y+9=-2x+6
6y=-2x-3
y= (-2x-3)/6
slope is -2/6 = </span>