Answer:
C.
Step-by-step explanation:

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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.
2 divides both numbers, so divide both numbers by 2 to get 3 and 7. These are relatively prime, so 2 is the gcd. The lcm is 6*14/2=42.
Answer:
Train A = 128
Train B = 68
Step-by-step explanation:
We can set up a system of equations for this problem
Let A = # of tons of Train A
Let B = # of tons of Train B
A + B = 196
A = B + 60
Now, we plug in A for the first equation, using substitution
(B+60) + B = 196
2B + 60 = 196
Subtract 60 from both sides
2B = 136
Divide both sides by 2
B = 68
Plug in 68 for B in the 2nd equation
A = 68 + 60
A = 128
Checking work: 128 + 68 = 196 :D hope this helped
The answer would be A. Commutative property of addition