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:
g(h(x)) = [x + 3]^2 or x^2 + 6x + 9
Step-by-step explanation:
g[h(x)] signifies that h(x) is the input to g(x).
Writing out g(x) = x^2 and replacing "x" with [x + 3], we get:
g(h(x)) = [x + 3]^2 or x^2 + 6x + 9
Answer:

Step-by-step explanation:
The answer would be C because the first figure is half of every number on the second figure making it similar it’s not congruent because there not the same size