Ali paints with watercolors on a sheet of paper 20 in. wide by 15 in. high. he then places this sheet on a mat so that a uniformly wide strip of the mat shows all around the picture. the perimeter of the mat is 86 in. how wide is the strip of the mat showing around the picture?

The wide of the mat strip showing around the picture is 2 in.

Step-by-step explanation:

Please refer to the diagram attached at the bottom of these answer.

Since the mat's perimeter is 86 in,

     (eq.1)

Now, the paper is laid son there is a uniformly wide strip of the map showing all around the picture, so all four distances from the perimeter of the mat to the paper are the same; we will call this distance "a".

As you can see in the diagram:

W - 15 in = 2a

L - 20 in = 2a

So:

W - 15 in = L - 20 in      (eq.2)

So with eq.1 and eq.2 we have a system of two equations and two unknowns.

We will solve for L in eq.1:

Solving for W in eq.2:

[tex]W=L-20in+15in=L-5in[\tex]

Replacing L in this solution:

[tex]W=(43in-W)-5in[\tex]

[tex]W=38in-W[\tex]

[tex]W+W=38in[\tex]

[tex]2W=38in[\tex]

[tex]W=\frac{38in}{2}=19in[\tex]

As we said before, W - 15 in = 2a, then:

[tex]19in-15in=2a[\tex]

[tex]4in=2a[\tex]

[tex]a=\frac{4in}{2}=2in[\tex]

Which is the answer to the question.

We can also solve eq.1 to find L:

L = 43 in - W = 43 in - 19 in = 24 in

the answer would be 150 books

step-by-step explanation:

if you divide 300 by 2 you will get 150

answer: on graphs, the open and closed circles, or vertical asymptotes drawn as dashed lines us identify discontinuities.

as before, graphs and tables allow us to estimate at best.

when working with formulas, getting zero in the denominator indicates a point of discontinuity.

when working with piecewise-defined functions, check for discontinuities at the transition points where one piece ends and the next begins.

step-by-step explanation: