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
Step-by-step explanation:
A)y=8
B)k=10
5y=72-32
5y=40
y=8
1/2k+24=29
k+48=58
k=58-48
k=10
Answer:
Plot points at (0,1) and (-3,3) and draw a line going through both points.
Step-by-step explanation:
Let's start by graphing the y intercept.
y=mx+b
m is the slope. b is the y intercept. Since the equation is y=-2/3x+1, we can conclude the y intercept is 1. We graph a point at (0,1).
If you didn't know the y intercept is where the line intercepts the y-axis.
Now, from the point (0,1) we go up 2 and to the left 3 as it is a negative slope. We reach (-3,3). Plot a point there. Then draw a line going through both points. There's your line!
−2.72 or -68/25
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Answer:
No it is not -3 is an integers
