We are given with the polynomial above and is asked to evaluate the end behavior of the function. The end behavior of a polynomial function is the behavior of the graph<span> of f ( x ) as x approaches positive infinity or negative infinity. As x approaches negative infinity, the function goes to negative infinity while when x</span> approaches positive infinity, the function goes to <span>positive</span> infinity
Isosollese triangle has 2 equal sides that are longer than legnth of base
perimiter=15.7
equation is 2a+b=15.7
so if the longer side is 6.3, wat is legnth of base
we know that identical sides are bigger so 6.3 is one of the identical sides
2a means the 2 idneical sides
2(6.3)+b=15.7
12.6+b=15.7
subtract 12.6 from btohs ides
b=3.1
answer is base=3.1 cm
Answer
<span>A. Rectangle
</span><span>D. Parallelogram
Explanation
The four choices are all quadrilaterals with with at least a pair of parallel sides.
A rectangle, a square, and a rhombus have opposite sides been parallel.
For a square and a rhombus, the diagonals are perpendicular.
The diagonals of a rectangle and a parallelograms are not perpendicular.
</span><span>
</span>
Answer:
Step-by-step explanation:
Given data
Total units = 250
Current occupants = 223
Rent per unit = 892 slips of Gold-Pressed latinum
Current rent = 892 x 223 =198,916 slips of Gold-Pressed latinum
After increase in the rent, then the rent function becomes
Let us conside 'y' is increased in amount of rent
Then occupants left will be [223 - y]
Rent = [892 + 2y][223 - y] = R[y]
To maximize rent =

Since 'y' comes in negative, the owner must decrease his rent to maximixe profit.
Since there are only 250 units available;
![y=-250+223=-27\\\\maximum \,profit =[892+2(-27)][223+27]\\=838 * 250\\=838\,for\,250\,units](https://tex.z-dn.net/?f=y%3D-250%2B223%3D-27%5C%5C%5C%5Cmaximum%20%5C%2Cprofit%20%3D%5B892%2B2%28-27%29%5D%5B223%2B27%5D%5C%5C%3D838%20%2A%20250%5C%5C%3D838%5C%2Cfor%5C%2C250%5C%2Cunits)
Optimal rent - 838 slips of Gold-Pressed latinum
Answer:
See diagram attached.
Step-by-step explanation:
The line of reflection of preimage ABC reflected onto image A'B'C' can be found by the perpendicular bisector of lines joining the corresponding points of the image and preimage, for example, AA', or BB'.
See diagram attached.