Answer: The set does not have a solution
Step-by-step explanation:
Adding Equations 1 & 3 we get 5x = 7. This gives x = 7/5
Putting this value of x in eq. 2 we get
-2y + 2z = -1-(7/5) or
2y - 2z = 12/5 or 5y - 5z = 6
Multiplying eq. 1 by 2 we get
4x + 2y - 2z = 6
adding this with eq. 2 we get 5x = 5 or x = 1
As the common solution for x from equations 1&3 does not satisfy eq. 1&2 it comes out that the three equations do not have a common solution.
Same can be verified by using different sets of two equations also.
Dimension of smaller garden :
l = 1/2 ft.
b = 2/3 ft.
Dimension of bigger garden :
L = 4 ft.
Let , breadth be x ft.
We know , area is given by :
Area = L×B.
Area of small garden = 
Area of big garden 
Hence, this is the required solution.
Parameterize
by

with
and
.
Then the surface element is

and the surface integral is

Answer:
Step-by-step explanation:
alternate interior angles
Answer:
The correct option is;
Substitute x = 0 in the function and solve for f(x)
Step-by-step explanation:
The zeros of a function are the values of x which produces the value of 0 when substituted in the function
It is the point where the curve or line of the function crosses the x-axis
A. Substituting x = 0 will only give the point where the curve or line of the function crosses the y-axis,
Therefore, substituting x = 0 in the function can't be used to find the zero's of a function
B. Plotting a graph of the table of values of the function will indicate the zeros of the function or the point where the function crosses the x-axis
C. The zero product property when applied to the factors of the function equated to zero can be used to find the zeros of a function
d, The quadratic formula can be used to find the zeros of a function when the function is written in the form a·x² + b·x + c = 0