How do you round 282 to the nearest whole number
1 answer:
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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>
<h3>Answer: -2w^2 + 25w = 25 or -2w^2 + 25w - 25 = 0</h3>
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Explanation:
Refer to the diagram below. The width is w. We have two opposite and parallel sides equal to this. The other two parallel congruent sides are L = 25-2w meters long. We start with the total amount of fencing, and then subtract off the two width values, so 25-w-w = 25-2w.
The area of the rectangle is
Area = length*width
Area = L*W
Area = (25-2w)*w
Area = 25w - 2w^2
Area = -2w^2 + 25w
Set this equal to the desired area (25 square meters) to get
-2w^2 + 25w = 25
and we can subtract 25 from both sides to get everything on one side
-2w^2 + 25w - 25 = 0
side note: The two approximate solutions of this equation are w = 1.0961 and w = 11.4039 (use the quadratic formula or a graphing calculator to find this)