The answer is <span>(x, y)→(x - 9, y - 3)
proof
according to the figure H (3, -1) and H' (-6, -4)
</span><span>-6= 3 -9, and - 4= -1 -3, </span>
Answer:
a. dy/dx = -2/3
b. dy/dx = -28
Step-by-step explanation:
One way to do this is to assume that x and y are functions of something else, say "t", then differentiate with respect to that. If we write dx/dt = x' and dy/dt = y', then the required derivative is y'/x' = dy/dx.
a. x'·y^3 +x·(3y^2·y') = 0
y'/x' = -y^3/(3xy^2) = -y/(3x)
For the given point, this is ...
dy/dx = -2/3
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b. 2x·x' +x^2·y' -2x'·y^3 -2x·(3y^2·y') + 0 = 2x' + 2y'
y'(x^2 -6xy^2 -2) = x'(2 -2x +2y^3)
y'/x' = 2(1 -x +y^3)/(x^2 +6xy^2 -2)
For the given point, this is ...
dy/dx = 2(1 -0 +27)/(0 +0 -2)
dy/dx = -28
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The attached graphs show these to be plausible values for the derivatives at the given points.
Answer:
If we look at the table we notice that 2 + 4 = 6, 3 + 4 = 7, 4 + 4 = 8 and so on so the equation is y = x + 4.
3y= 5x-2
so the slope is 5/3