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:
$2
Step-by-step explanation:
Let, price of individual child ticket is $x, and price of individual adult ticket is $y.
Now, on Friday we can write an equation is 55x+20y=200....(1).
and on Saturday we can write an equation is 72x+36y=306.....(2).
so, Equation 1: 55x+20y=200
Equation 2: 72x+36y=306
now, we are going to solve both equations.
after solving you will find x=2 and y=4.5
Hence, we can say that the cost of child ticket is $2, and the cost of adult ticket is $4.50.
Answer:
1) solution is y = -7
2.) DNE
3.) Solution is y = -2
Step-by-step explanation:
1.) 2+3 = y + 12
Add the LHS and make y the subject of formula
5 = y + 12
Y = 5 - 12
Y = - 7
The solution of the equation is - 7
2.) 2 + 13 = 1 +8
Since there is no unknown variable and the sum of the numbers in the left hand side ( LHS ) is not equal to the sum of the numbers in the right hand side ( RHS ), it will be concluded that there is no solution in the equation.
3.) y - 7 = 2 - 11
Sum the RHS and make y the subject of formula
Y - 7 = -9
Y = -9 + 7
Y = -2
The solution of the equation is -2
Answer:
FG = 44
EG = 80
Step-by-step explanation:
For FG:
1) 5z - 16 = 3z + 8
2) 5z = 3z + 24
3) 2z = 24
4) z = 12
5) 3(12) + 8 = 44
For EG:
1) 2w + 22 = 4w + 4
2) 2w = 4w - 18
3) -2w = -18
4) w = 9
5) 4(9) + 4 + 2w + 22 = 80
