Refer to the attached image.
Given the rectangle ABCD of length 'l' and height 'h'.
Therefore, CD=AB = 'l' and BC = AD = 'h'
We have to determine the area of triangle AEF.
Area of triangle AEF = Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE
Area of triangle ADF = 
= 
= 

Area of triangle ECF = 
= 
= 

Area of triangle ABE = 
= 
= 

Now, area of triangle AEF =
Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE
= 
= 
=
=

= 27 units
Therefore, the area of triangle AEF is 27 units.
Answer is a because it is what it is jk didnt want to explain
Answer:
Max vol = 2 cubic metres
Step-by-step explanation:
Given that from a square piece of cardboard paper of area size 9 m2 , squares of the same size are cut off from each corner of the paper.
Side of the square = 3m
If squares are to be cut from the corners of the cardboard we have the dimensions of the box as
3-2x, 3-2x and x.
Hence x can never be greater than or equal to 1.5
V(x) = Volume = 
We use derivative test to find the maxima

Equate I derivative to 0

If x= 3/2 box will have 0 volume
So this is ignored
V"(1/2) <0
So maximum when x =1/2
Maximum volume
=
cubic metres
Answer:
c. $0.75 per minute at one rate for the first 5 minutes and $0.25 per minute thereafter
Step-by-step explanation:
The last 5 minutes of the 12-minute call cost ...
$5.50 -4.25 = $1.25
so the per-minute rate at that time is ...
$1.25/(5 min) = $0.25/min . . . . . . . . matches choice C only
__
You know the answer at this point, but if you want to check the rate for the first 5 minutes, you can subtract 2 minutes from the 7-minute call to find that ...
The first 5 minutes cost $4.25 - 2·0.25 = $3.75, so are charged at ...
$3.75/(5 min) = $0.75/min . . . . . . . matches choice C
Answer:
k=2.1
Step-by-step explanation:
7/10k=21
x7 x7 Multiply seven from each side of the equation.
10k=21
/10 /10 Divide ten from each side of the equation. 21/10=2.1
k=2.1