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:
x=3
Step-by-step explanation:
please mark me as brainliest
The answer is 343
Because volume is length x width x height which would be 7 x 7 x 7
Answer:
x=4, y=2
Step-by-step explanation:
2x - y = 6
x + y = 6
Add these together to eliminate y
2x - y = 6
x + y = 6
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3x = 12
Divide by 3
3x/3 = 12/3
x=4
But we still need y
x+y =6
4+y =6
Subtract 4 from each side
4+y-4 = 6-4
y=2
Answer:
8
Step-by-step explanation:
b=8
c=2
(8)(2)-(2)^3 = 16-8 =8