Let the number of chickens = c.
Let the number of pigs = p.
A chicken has 1 head and 2 legs.
c number of chickens have c heads and 2c legs.
A pig has 1 head and 4 legs.
p number of pigs have p heads and 4p legs.
There are 40 heads.
Equation for heads:
c + p = 40
There are 110 legs.
Equation for legs:
2c + 4p = 110
System of equations:
c + p = 40
2c + 4p = 110
Solve the first equation of the system of equations for c:
c = 40 - p
Substitute 40 - p for c in the second equation:
2c + 4p = 110
2(40 - p) + 4p = 110
80 - 2p + 4p = 110
80 + 2p = 110
2p = 30
p = 15
Now substitute p = 15 in the first equation to find c.
c + p = 40
c + 15 = 40
c = 25
There are 25 chickens and 15 pigs.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The solution is 
Step-by-step explanation:
From the question we are told that
and 
Generally the absolute value of the determinant of the Jacobian for this change of coordinates is mathematically evaluated as
![| \frac{\delta (x,y)}{\delta (u, v)} | = | \ det \left[\begin{array}{ccc}{\frac{\delta x}{\delta u} }&{\frac{\delta x}{\delta v} }\\\frac{\delta y}{\delta u}&\frac{\delta y}{\delta v}\end{array}\right] |](https://tex.z-dn.net/?f=%7C%20%5Cfrac%7B%5Cdelta%20%20%28x%2Cy%29%7D%7B%5Cdelta%20%28u%2C%20v%29%7D%20%7C%20%3D%20%20%7C%20%5C%20det%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B%5Cfrac%7B%5Cdelta%20x%7D%7B%5Cdelta%20u%7D%20%7D%26%7B%5Cfrac%7B%5Cdelta%20x%7D%7B%5Cdelta%20v%7D%20%7D%5C%5C%5Cfrac%7B%5Cdelta%20y%7D%7B%5Cdelta%20u%7D%26%5Cfrac%7B%5Cdelta%20y%7D%7B%5Cdelta%20v%7D%5Cend%7Barray%7D%5Cright%5D%20%7C)
![= |\ det\ \left[\begin{array}{ccc}{-2e^{-2u} cos(5v)}&{-5e^{-2u} sin(5v)}\\{-2e^{-2u} sin(5v)}&{-2e^{-2u} cos(5v)}\end{array}\right] |](https://tex.z-dn.net/?f=%3D%20%7C%5C%20det%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B-2e%5E%7B-2u%7D%20cos%285v%29%7D%26%7B-5e%5E%7B-2u%7D%20sin%285v%29%7D%5C%5C%7B-2e%5E%7B-2u%7D%20sin%285v%29%7D%26%7B-2e%5E%7B-2u%7D%20cos%285v%29%7D%5Cend%7Barray%7D%5Cright%5D%20%20%7C)
So
![\frac{\delta (x,y)}{\delta (u, v)} | = |det \left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] |](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cdelta%20%20%28x%2Cy%29%7D%7B%5Cdelta%20%28u%2C%20v%29%7D%20%7C%20%3D%20%7Cdet%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%7C)
=> 
substituting for a, b, c,d
=> 
=> 
=>
It is 20
39+1
38+2
37+3
36+4
35+5
<span>Until you get to 20+20</span>
Answer:
Perfect positive association
Step-by-step explanation:
Definition: A perfect positive association means that a relationship appears to exist between two variables, and that relationship is positive 100% of the time. Two variables have a positive association when the values of one variable tend to increase as the values of the other variable increase. (+1 indicates a perfect positive linear relationship)
Definition: A perfect negative association means that a relationship appears to exist between two variables, and that relationship is negative 100% of the time. Two variables have negative association when the values of one variable tend to decrease as the values of the other variable increase. (-1 indicates a perfect negative linear relationship)
Values between 0.3 and 0.7 (-0.3 and -0.7) indicate a moderate positive (negative) linear relationship.
From the graph we can see that this relationship shows perfect positive association (both variables increase and we can plot the straight line which will include all points)
Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:
![\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac2n%5Cright%5D%2C%5Cleft%5B%5Cdfrac2n%2C%5Cdfrac4n%5Cright%5D%2C%5Cleft%5B%5Cdfrac4n%2C%5Cdfrac6n%5Cright%5D%2C%5Cldots%2C%5Cleft%5B%5Cdfrac%7B2%28n-1%29%7Dn%2C2%5Cright%5D)
Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,
where 
. Each interval has length 
.
At these sampling points, the function takes on values of
We approximate the integral with the Riemann sum:
Recall that
so that the sum reduces to
Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:
Just to check:
