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:
4600
Step-by-step explanation:
We can write a proportion to find the total amount who attend university using the information given. A proportion is two equivalent ratios set equal to each other. Since 70% live on campus, then 30% live off campus and we are told that number is 1,380.

We will cross multiply the numerator of one ratio with denominator of the other. And then solve for y.
30y=100(1380)
30y=138000
y=4600.
There are 4600 students who attend the university.
Theres about a 7,200,000 difference between household dogs vs household cats
Answer:
x=-8
(-8,-7)
Step-by-step explanation:
–9x + 9y = 9
–9x + 9(-7) = 9 (y=-7)
-9x-63=9
Both side +63
-9x=63+9
-9x=72
Both side divide -9
x=-8
y=-7
(-8,-7)