The diagonals of a convex quadrilateral are mutually perpendicular. the sum of the lengths of the diagonals is 12. we want to find the maximum possible area of such a quadrilateral. let us denoted by x and y the lengths of the two diagonals. then the area of the quadrilateral is the following function of x and y:

33

Step-by-step explanation:

Step 1: The Diagonals of the Convex Quadilateral are Mutually Perpendicular this makes the shape a Rhombus Shape

Step 2: The formula for a area of a rhombus is the multiply the two lengths and divide by two.

In this Question we have x×y÷2 or xy/2

Step 3: The sum of the lengths of the diagonals is 12

x+y=12

y=12-x

Therefore, the area, as a function of x alone = x(12-x)/2

Step 4: To find the maximum area, let's find the derivative of the area function in respect to x and then equate to 0.

d/dx(x(12-x)/2)=0

d/dx(12x - x^2) = 0

12-2x=0

12-0=2x

12=2x, divide both sides by 2

x= 6

Step 5: Calculate the Maximum Area

The maximum area will then be 6(12 - 6)/2

= (72-6)/2 = 66/2= 33

So, the maximum area is 33.

i think it's ф or x=6,-6

step-by-step explanation:

-x^2=-36

x^2=-36

ф

or

if you mean by : -(x^2)=-36

it'll be: x^2=36

x=6,x=-6

hope it'll ! have a good day!

you for asking!

darn right!

step-by-step explanation:

steve kerr's sons name is let that skin in.