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:
1
Step-by-step explanation:
Step 1: Simplify both sides of the equation.
x+24+13=2x−16
x+ 1 2 + 1 3 =2x+ −1 6
(x)+( 1 2 + 1 3 )=2x+ −1 6 (Combine Like Terms)
x+ 5 6 =2x+ −1 6
x+ 5 6 =2x+ −1 6
Step 2: Subtract 2x from both sides.
x+ 5 6 −2x=2x+ −1 6 −2x
−x+ 5 6 = −1 6
Step 3: Subtract 5/6 from both sides.
−x+ 5 6 − 5 6 = −1 6 − 5 6
−x=−1
Step 4: Divide both sides by -1.
−x −1 = −1 −1
x=1
 
        
                    
             
        
        
        
Answer:
THE ANSWER IS THE PIC lolol
 
        
             
        
        
        
Answer:
all i know is that number #2 is A 
adding it all up is hard to figure out but #2 will show you that lay out so it mite be easier but i need help on this as well
Step-by-step explanation:
 
        
             
        
        
        
Your answer would be c. 18,6