If there is a girl in front, behind, to the left, or to the right of a boy, then the two of them flirt. one student may be in multiple flirting couples; for example, a student in a corner of the classroom can flirt with up to two others, while a student in the center can flirt with as many as four others. suppose that desks are occupied by boys and girls with equal probability and mutually independently. what is the expected number of flirting couples? hint: linearity.

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

or

step-by-step explanation:

we need to solve for 'x' the following equation:  

we have that:

⇒

⇒ ⇒

factorizing, we get that:

  ⇒

then, the solutions are: or

nothing extraneous has been found!

what is the question? i am guessing you are looking for the area.

to find an area, you have to do length x width.

in this case, it would be 12 ft x 4 ft = 48 sq. ft