Apicture that measures 8 cm by 6 cm is to be surrounded by a mat. the mat is to
be surrounded by a mat. the mat is to be the same widt
all sides of the picture. the combined area of the mat and the picture is double the arca
what is the width of the mat to the nearest tenth? include a diagram. (5)

1.4 inches

Step-by-step explanation:

The picture is a rectangle 8 cm by 6 cm. The area of a rectangle is length * width. The area of the picture is 8 cm * 6 cm = 48 cm^2

After the mat is applied, the area doubles, so the new area will be 2 * 48 cm^2 = 96 cm^2.

Let the width of the mat be x. The mat has the same width all around the rectangular picture, so it adds x on each side of the length and x on each side of the width.

old length: 8

new length: 2x + 8

old width: 6

new width: 2x + 6

Area of the new rectangle with mat = new length * new width

area = (2x + 8)(2x + 6)

The new area is 96, so that give us an equation.

(2x + 8)(2x + 6) = 96

Use FOIL on the left side:

4x^2 + 12x + 16x + 48 = 96

Combine like terms, and subtract 96 from both sides:

4x^2 + 28x - 48 = 0

Divide both sides by 4:

x^2 + 7x - 12 = 0

To factor the trinomial, we need two numbers that add to 7 and multiply to -12. There are no such numbers, so we need to use the quadratic formula.

x = [-b +/- sqrt(b^2 - 4ac)]/(2a)

x = [-7 +/- sqrt(7^2 - 4(1)(-12)]/[2(1)]

x = [-7 +/- sqrt(49 + 48)]/2

x = [-7 +/- sqrt(97)]/2

x = 1.4 or x = -8.4

Since the mat cannot have a negative width, the negative solution is discarded.

The width of the mat is 1.4 inches.

answer: ouch. thats difficult

step-by-step explanation: