onyango wishes to fence a rectangular research plot using 100 m of wire .one end of the plot has a wall already erected .calculate the maximum area that he can fence

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

d (5 x (-20)) * (100 x (0.7))

step-by-step explanation:

5÷1/100 * -.7 *-20

copy dot flip

5*100 * -.7 *-20

rearrange

5*-20   * 100 *-.7

-100   *-70

7000

total amount to be paid = $19.5

step-by-step explanation:

given ,

original price = $25.99

discount = 25%

discount amount = 25% of original price

discount amount = 25/100   * 25.99

discount amount = $6.49

hence , the total amount to be paid = original price - discount amount

                                                            = $25.99 - $6.49

                                                            =$19.5

total amount to be paid = $19.5