13 - 0 = 13 is the answer ?? ;)
        
                    
             
        
        
        
Step 1:
Start by putting 

 in front of each term
![\frac{d}{dx}[y cos x]= \frac{d}{dx}[5x^2]+ \frac{d}{dx}[ 3y^2]](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%7D%7Bdx%7D%5By%20cos%20x%5D%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B5x%5E2%5D%2B%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%203y%5E2%5D)
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Step 2:
Deal with the terms in 'x' and the constant terms
![\frac{d}{dx}[ycosx]= 10x+ \frac{d}{dx} [3y^2]](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bycosx%5D%3D%2010x%2B%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5B3y%5E2%5D%20%20)
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Step 3:
Use the chain rule for the terms in 'y'
![\frac{d}{dx}[ycosx]=10x+6y \frac{dy}{dx}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bycosx%5D%3D10x%2B6y%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%20)
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Step 4:
Use the product rule on the term in 'x' and 'y'


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Step 5:
Rearrange to make 

 the subject


![[cos(x) - 6y]  \frac{dy}{dx}=10x + y sin(y)](https://tex.z-dn.net/?f=%5Bcos%28x%29%20-%206y%5D%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%3D10x%20%2B%20y%20sin%28y%29%20)

 ⇒ Final Answer
 
        
        
        
Conditional probablility P(A/B) = P(A and B) / P(B). Here, A is sum of two dice being greater than or equal to 9 and B is at least one of the dice showing 6. Number of ways two dice faces can sum up to 9 = (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 10 ways. Number of ways that at least one of the dice must show 6 = (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1) = 11 ways. Number of ways of rolling a number greater than or equal to 9 and at least one of the dice showing 6 = (3, 6), (4, 6), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 7 ways. Probability of rolling a number greater than or equal to 9 given that at least one of the dice must show a 6 = 7 / 11
        
             
        
        
        
Answer:
You are correct
Step-by-step explanation:Nice job have a nice day!!!
 
        
             
        
        
        
The function is:
f ( x ) = 2 / ( x² + 3 x - 10 )
x² + 3 x - 10 = x² + 5 x - 2 x - 10 = x ( x + 5 ) - 2 ( x + 5 ) = 
= ( x + 5 ) ( x - 2 )
x + 5 = 0 => x = - 5
x - 2 = 0 => x = 2
Answer:
The vertical asymptotes are: x = - 5 and x = 2.