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:
WHAT THE FRACTION IS THIS!!!!
Step-by-step explanation:
i dont know what kind of math this is but this takes geomatry and mathematical observations to get it correct.
Ok what happened what do you need help on
Answer:
235
Step-by-step explanation:
42+1
43
3 times 43
129
129 times 2
258
258-23
235
Picture related to the question is attached below
Answer:
Kendra should have divided by 2 instead of multiplying by 2
Step-by-step explanation:
40 percent = percentage who plan to attend
100% = total percentage of student
From the question ;
When reducing percentages, division is employed and this the equation should be :
40/2 ÷ 100/2
40/2 * 2/100
20 * 1/50
20/50
= 0.4
0.4 * 50
= 20 students
