Suppose that a rectangular picture is surrounded by a frame of uniform width of x

Mathematics

1 answer:

user100 [1]3 years ago

3 0

Answer:

Ok so the area of the frame is 18*12 = 216cm^2

Let’s suppose the width of the frame is x

The total area (painting and frame) is (2x+18)(2x+12) and this is equal to 432

4x^2 + 60x + 216 = 432

4x^2 + 60x - 216 = 0

x^2 + 15x - 54 = 0

(x+18)(x-3) = 0

Therefore x = {-18,3}

Since width of the frame has to be positive, the width has to be 3cm Step-

i hope i helped!

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7. Which dimensions result in the minimum perimeter for a rectangle with area 42.0 cm2?

natali 33 [55]

9514 1404 393

Answer:

7. d. l = 6.48 cm, w = 6.48 cm

8. d. square

Step-by-step explanation:

For a given area, the perimeter can always be shortened by reducing the length of the long side and increasing the length of the short side. When you get to the point where you can't do that, then you have the minimum perimeter. You will reach that point when the sides are the same length: the rectangle is a square.

7. In light of the above, the best dimensions are √42 ≈ 6.48 cm for length and width.

8. In light of the above, the shape is a square.

_____

The attached graph shows the length of one side (x) and the associated perimeter. The other side is 42/x, which will also be 6.48.