The question is for any rational function. I give one example in picture below. There we can see that first part of the "rule" is true. we indeed have vertical asymptotes at every x value where denominator is zero in this case x = 3 and x = 1. but second part of the rule is not true. we can see that if we inverse x ( we have factor that looks like a - x we get positive function between asymptotes.
Answer is False
Let the lengths of the sides of the rectangle be x and y. Then A(Area) = xy and 2(x+y)=300. You can use substitution to make one equation that gives A in terms of either x or y instead of both.
2(x+y) = 300
x+y = 150
y = 150-x
A=x(150-x) <--(substitution)
The resulting equation is a quadratic equation that is concave down, so it has an absolute maximum. The x value of this maximum is going to be halfway between the zeroes of the function. The zeroes of the function can be found by setting A equal to 0:
0=x(150-x)
x=0, 150
So halfway between the zeroes is 75. Plug this into the quadratic equation to find the maximum area.
A=75(150-75)
A=75*75
A=5625
So the maximum area that can be enclosed is 5625 square feet.
Temp is explanatory and ice cream sales is the response
Answer:
Yes, it is invertible
Step-by-step explanation:
We need to find in the matrix determinant is different from zero, since iif it is, that the matrix is invertible.
Let's use co-factor expansion to find the determinant of this 4x4 matrix, using the column that has more zeroes in it as the co-factor, so we reduce the number of determinant calculations for the obtained sub-matrices.We pick the first column for that since it has three zeros!
Then the determinant of this matrix becomes:
![4\,*Det\left[\begin{array}{ccc}1&4&6\\0&3&8\\0&0&1\end{array}\right] +0+0+0](https://tex.z-dn.net/?f=4%5C%2C%2ADet%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%264%266%5C%5C0%263%268%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%2B0%2B0%2B0)
And the determinant of these 3x3 matrix is very simple because most of the cross multiplications render zero:
![Det\left[\begin{array}{ccc}1&4&6\\0&3&8\\0&0&1\end{array}\right] =1 \,(3\,*\,1-0)+4\,(0-0)+6\,(0-0)=3](https://tex.z-dn.net/?f=Det%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%264%266%5C%5C0%263%268%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%3D1%20%5C%2C%283%5C%2C%2A%5C%2C1-0%29%2B4%5C%2C%280-0%29%2B6%5C%2C%280-0%29%3D3)
Therefore, the Det of the initial matrix is : 4 * 3 = 12
and then the matrix is invertible
