There are 2 short sides x and 1 long side y.
x+x+y=1,800
2x+y=1,800
y=1,800-2x
The area of the rectangle is A=x(1,800-x).
A is a function with variable x. Moreover it is a quadratic (second degree) function.
The graph of the function is a parabola which opens downwards, because the signs of the leading coefficients of the factors, multiply to minus:
(x)(-x)=-x^2.
This means that the highest point of the parabola, is its vertex V(h,k), so the function takes its largest value for x=h.
The roots of A=x(1,800-x) are x=0 and x=1,800, thus the x coordinate of the vertex, must be the midpoint of 0 and 1,800, that is 900.
Answer:
A=x(1,800-x)
largest area is when x=900
In order to determine whether the equations are parallel, perpendicular, or neither, let's simply each equation into a slope-intercept form or basically, solve for y.
Let's start with the first equation.

Cross multiply both sides of the equation.


Subtract 6x on both sides of the equation.


Divide both sides of the equation by -5.


Therefore, the slope of the first equation is 4/5.
Let's now simplify the second equation.

Add x on both sides of the equation.


Divide both sides of the equation by -4.


Therefore, the slope of the second equation is -5/4.
Since the slope of each equation is the negative reciprocal of each other, then the graph of the two equations is perpendicular to each other.
Answer:
1/4 cup of sugar
Step-by-step explanation:
Answer:
![A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B1%7D%7B9%7D%20%26%20%5Cfrac%7B4%7D%7B27%7D%20%26%20-%20%5Cfrac%7B2%7D%7B27%7D%20%5C%5C%5C%5C%20%5Cfrac%7B8%7D%7B9%7D%20%26%20%5Cfrac%7B5%7D%7B27%7D%20%26%20%5Cfrac%7B11%7D%7B27%7D%20%5C%5C%5C%5C%20-%20%5Cfrac%7B4%7D%7B9%7D%20%26%20%5Cfrac%7B2%7D%7B27%7D%20%26%20-%20%5Cfrac%7B1%7D%7B27%7D%20%5Cend%7Barray%7D%20%5Cright%5D)
Step-by-step explanation:
We want to find the inverse of ![A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7D%201%20%26%200%20%26%20-2%20%5C%5C%5C%5C%204%20%26%201%20%26%203%20%5C%5C%5C%5C%20-4%20%26%202%20%26%203%20%5Cend%7Barray%7D%20%5Cright%5D)
To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.
So, augment the matrix with identity matrix:
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%204%261%263%260%261%260%20%5C%5C%5C%5C%20-4%262%263%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 1 multiplied by 4 from row 2
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%20-4%262%263%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Add row 1 multiplied by 4 to row 3
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%262%26-5%264%260%261%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 2 multiplied by 2 from row 3
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%260%26-27%2612%26-2%261%5Cend%7Barray%7D%5Cright%5D)
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%260%261%26-%20%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B2%7D%7B27%7D%26-%20%5Cfrac%7B1%7D%7B27%7D%5Cend%7Barray%7D%5Cright%5D)
- Add row 3 multiplied by 2 to row 1
![\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%260%26%5Cfrac%7B1%7D%7B9%7D%26%5Cfrac%7B4%7D%7B27%7D%26-%20%5Cfrac%7B2%7D%7B27%7D%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%260%261%26-%20%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B2%7D%7B27%7D%26-%20%5Cfrac%7B1%7D%7B27%7D%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 3 multiplied by 11 from row 2
![\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%260%26%5Cfrac%7B1%7D%7B9%7D%26%5Cfrac%7B4%7D%7B27%7D%26-%20%5Cfrac%7B2%7D%7B27%7D%20%5C%5C%5C%5C%200%261%260%26%5Cfrac%7B8%7D%7B9%7D%26%5Cfrac%7B5%7D%7B27%7D%26%5Cfrac%7B11%7D%7B27%7D%20%5C%5C%5C%5C%200%260%261%26-%20%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B2%7D%7B27%7D%26-%20%5Cfrac%7B1%7D%7B27%7D%5Cend%7Barray%7D%5Cright%5D)
As can be seen, we have obtained the identity matrix to the left. So, we are done.
The leading term is the term with the highest degree (highest exponent number). Here, the leading term is =》B) - 5x²y³
³ is the highest degree in the given expression.
_____
Hope it helps ⚜