.010416666666. Use conversion factors to solve.
Answer:
The possible way for the initiative to accomplish its goal without exceeding its budget is use 12.5 hectares for planting trees and 12.5 hectares by purchasing land.
Step-by-step explanation:
Let the variable <em>X</em> represent the amount of land used for planting trees and <em>Y</em> represent the amount of land purchased.
The goal of the environmental initiative is to save at least 25 million hectares of rain forest.
That is:
<em>X</em> + <em>Y</em> = 25....(i)
Now it is provided that:
- The cost of planting trees is $ 400 per hectare.
- The cost of purchasing land is $ 260 per hectare.
- The initiative has a budget of $8,250 million.
Using the above data it can be said that:
400<em>X</em> + 260<em>Y</em> = 8250....(ii)
Solve equations (i) and (ii) simultaneously.
![\ \ \ \ x+y=25]\times 260\\400x+260y=8250\\\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\\\\Rightarrow\\\\260x+260y=6500\\400x+260y=8250\\(-)\_\_\_\_\_\ (-)\_\_\_\_(-)\_\_\_\\\\\Rightarrow\\\\-140x=-1750\\\\x=\frac{1750}{140}\\\\x=12.5](https://tex.z-dn.net/?f=%5C%20%5C%20%5C%20%5C%20x%2By%3D25%5D%5Ctimes%20260%5C%5C400x%2B260y%3D8250%5C%5C%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C%5C%5C%5C%5CRightarrow%5C%5C%5C%5C260x%2B260y%3D6500%5C%5C400x%2B260y%3D8250%5C%5C%28-%29%5C_%5C_%5C_%5C_%5C_%5C%20%28-%29%5C_%5C_%5C_%5C_%28-%29%5C_%5C_%5C_%5C%5C%5C%5C%5CRightarrow%5C%5C%5C%5C-140x%3D-1750%5C%5C%5C%5Cx%3D%5Cfrac%7B1750%7D%7B140%7D%5C%5C%5C%5Cx%3D12.5)
Then the value of <em>y</em> is:

Thus, the possible way for the initiative to accomplish its goal without exceeding its budget is use 12.5 hectares for planting trees and 12.5 hectares for purchasing land.
Answer:
Since them taken same time so the fraction is 1/1
Sum of three numbers is 33 and sum of their cubes is 5049 then three numbers are 7,11,15 or 15,11,7.
As given,
Let a-d, a, a +d be three numbers
Sum =33
⇒a-d + a +a +d =33
⇒ a =11
Sum of cubes= 5049
⇒ (a-d)³ + a +(a +d)³ =5049
⇒ a³ -d³ -3a²d +3ad² + a³ +a³ +d³ +3a²d +3ad² =5049
⇒3a³ + 6ad² = 5049
⇒ d=± 4
Therefore, sum of three numbers is 33 and sum of their cubes is 5049 then three numbers are 7,11,15 or 15,11,7.
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