Answer:
50%
Step-by-step explanation:
Let the total number of trees be T and suppose that this is made up of eucalyptus trees E and other kinds of trees, O.
Then E+O=T, and since 99% of the trees are eucalyptus trees, EE+O=0.99.
Therefore
0.99E+0.99O=E
0.01E=0.99O
E=99O.
Suppose we reduce the number of eucalyptus trees by some amount x and that the remaining number of eucalyptus trees are now 98% of the total trees.
Then we would have E−x(E−x)+O=0.98.
Therefore,
0.02(E−x)=0.98O
0.02E−0.02x=0.98O
0.02x=0.02E−0.98O
x=E−50(0.98)O
x=E−49O
x=E−49(E99)
x=5099E≈0.505050...×E.
So we end up removing a little over half of the eucalyptus trees. EDIT: Forgot to mention - removing 50.505050...% of the eucalyptus trees means that we are removing 0.50505050....×0.99T which is 0.5T or half the trees in the neighborhood.
Straight line depreciation, the value of the car is given by the following equation, where V is the value of the car, and O is the original value of the car, and x is the number of years since buying the car. V=O-2x This equation follows the form of a straight line, that is: y=mx+b
Answer:
Step-by-step explanation:
No its not possible I cant explain right now because I'm in class but I will later
