Day 4: as a landscaper, one of your clients wants a rectangular garden in which the length is five feet
more than twice the width. the perimeter of the garden is 82 feet. in square feet, what is the area of
the garden?

The Area of Rectangular Garden is 1044  feet²    

Step-by-step explanation:

According to question

The perimeter of the garden = 82 ft

Let the length be L ft

The width be W ft

Now as per question

L = 5 + ( 2× W )

∵ Perimeter of Rectangle = 2 × ( Length + Width )

Or , Perimeter of Rectangle = 2 × ( L+ W )

Or, 82 = 2 × ( L+ W )

Or,  82 = 2 × [ 5 + ( 2 ×W ) + W ) ]

Or,  82 = 2 × ( 5 +3W )

Or,  41 = 5 + 3W

Or, 41 - 5 = 3W

So, 3W= 36

∴    W = = 12 feet

I.e Width = 12 feet

And L = 5 + ( 2× W )

Or, Length = 5 + 24 = 29 feet

Now The Area of Rectangle = Length × width

So, The Area of Rectangle = 29 ft × 36 ft

The Area of Rectangle is 1044 feet²

Hence The Area of Rectangular Garden is 1044  feet²        Answer

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

child does not get an education so she can use the molla on junk