I need help with #12 b
1 answer:
Answer:
y = 2x -33
Step-by-step explanation:
The line segment from the center (6, 4) to the given point (16, -1) is a radius of the circle. It is perpendicular to the tangent. Hence you need to write the equation of a line through point (16, -1) that is perpendicular to the line through that point and (6, 4).
The slope of the radius is ...
... ∆y/∆x = (-1-4)/(16-6) = -5/10 = -1/2
The slope of the tangent is the negative reciprocal of that, so is ...
... m = -1/(-1/2) = 2
In point-slope form, the tangent line through (16, -1) is
... y = m(x -h) +k
... y = 2(x -16) +(-1)
Simplifying, this is ...
... y = 2x -33
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Answer:
∫₂⁵ ln(x) dx
Step-by-step explanation:
lim(n→∞) ∑ᵢ₌₁ⁿ (3/n) ln((2n + 3i) / n)
lim(n→∞) ∑ᵢ₌₁ⁿ (3/n) ln(2 + (3/n) i)
The width of the interval is b−a = 3, and there are n rectangles. So the width of each rectangle is 3/n, and the height of each rectangle is ln(2 + (3/n) i).
The ith term is 2 + (3/n) i, so a = x₀ = 2. Therefore, b = 2+3 = 5.
So the region is the area under f(x) = ln(x) between x=2 and x=5.
∫₂⁵ ln(x) dx
