an 8in by 12in picture has a frame of uniform width around it the area of the frame is equal to the area of the picture. what is the width of the frame

  2 inches

Step-by-step explanation:

The area of the frame will be the difference between the area of the picture with frame, and the area of the picture without. We want that difference to be equal to the area of the picture without the frame.

Let x represent the width of the frame. The overall length and width of the picture will be increased by 2x when the frame is added. Then we have ...

  a = LW = (8+2x)(12+2x) . . . . area of picture and frame

  b = LW = (8)(12) . . . . . . . . . . area of picture

  a -b = b . . . . . . desired relationship between areas

Substituting and expanding the expressions for 'a' and 'b', we find ...

  a -2b = 0 . . . . subtract b

  (8 +2x)(12 +2x) -2(8)(12) = 0

  4x² +40x -96 = 0 . . . simplify

  x² +10x -24 = 0 . . . . . divide by 4

  (x -2)(x +12) = 0 . . . . factor

The positive value of x that makes this true is x = 2.

The width of the frame is 2 inches.

what does muffins have to do with anything?

step-by-step explanation:

-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