Answer:
1250 m²
Step-by-step explanation:
Let x and y denote the sides of the rectangular research plot.
Thus, area is;
A = xy
Now, we are told that end of the plot already has an erected wall. This means we are left with 3 sides to work with.
Thus, if y is the erected wall, and we are using 100m wire for the remaining sides, it means;
2x + y = 100
Thus, y = 100 - 2x
Since A = xy
We have; A = x(100 - 2x)
A = 100x - 2x²
At maximum area, dA/dx = 0.thus;
dA/dx = 100 - 4x
-4x + 100 = 0
4x = 100
x = 100/4
x = 25
Let's confirm if it is maximum from d²A/dx²
d²A/dx² = -4. This is less than 0 and thus it's maximum.
Let's plug in 25 for x in the area equation;
A_max = 25(100 - 2(25))
A_max = 1250 m²
Answer:
8
Step-by-step explanation:
Because its a 2,2
The answer is....<span>2 × 7 × 11.</span>
Answer:
Step-by-step explanation:
After every Half life , Half of the mass is left.
After 1st Half life = 100 g / 2 = 50 g
After 2nd Half life = 50 / 2 = 25 g
After 3rd Half life = 25 / 2 = 12.5 g
After 4th Half life = 12.5 / 2 = 6.25 g
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>
~AH1807</h3>
Answer:
6307.5
Step-by-step explanation:
Hope that helps!
