Let E be the midpoint of OA and F be the midpoint of OB.
If point O is placed between A and B, then points E and F are placed <span>on different sides from O. The segment OE has length 12÷2=6 cm and segment OF has length 9÷2=4.5 cm. Than EF=6+4.5=10.5 cm.
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If
point O is placed outside the segment AB, then point B is closer to
point O than point A, because OA>OB. Since OE=6 cm and OF=4.5 cm,
EF=6-4.5=1.5 cm.
Answer:
Whats the question?
Step-by-step explanation:
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:
A
Step-by-step explanation:
Sorry wrote down the wrong thing the first time. Try doing 2 plus 3 times 5.