Jamal wants to put a rectangular frame on the above picture of the snowboarder. He has 18 cm of string to make the frame. Does he have enough for the frame? Explain why your answer is correct.

Nope. Too big.

To determine if Jamal has enough string to make the frame for the picture, we first need to calculate the perimeter of the rectangular frame.

The perimeter of a rectangle is calculated by adding the lengths of all four sides.

Given that Jamal has only 18 cm of string, we should consider that he needs to use this string for all four sides of the frame.

Since it is not mentioned in the question, we must make assumptions about the dimensions of the snowboarder picture. Let's assume the picture is a standard 4:3 ratio. This means the width of the picture is 4/3 times the height.

To calculate the perimeter, we need two dimensions: height and width.

Let's say the height of the picture is H cm. Therefore, the width of the picture is (4/3)H cm.

The perimeter of the frame is then:
2H + 2(4/3)H
= 2H + (8/3)H
= (6/3)H + (8/3)H
= (14/3)H

Now, we can set up an equation using the perimeter of the frame to find the height:
(14/3)H = 18

To isolate H, we divide both sides of the equation by (14/3):
H = 18 / (14/3)
H = 9 / (7/3)
H = 9 * (3/7)

Simplifying further:
H = 27/7

Now, we have the height of the picture. To find the width, we multiply the height by (4/3):
W = (4/3) * (27/7)

Simplifying further:
W = 108/21
W = 36/7

We have now found the height and width of the picture. To find the total perimeter, we add the heights and widths together:
Perimeter = 2H + 2W
= 2(27/7) + 2(36/7)
= 54/7 + 72/7
= 126/7

Since we want to compare the perimeter to Jamal's available string length of 18 cm, we need to convert the perimeter to cm:
Perimeter (in cm) = (126/7) cm
≈ 18 cm

Now, we can compare the perimeter of the frame to Jamal's available string length. Since the perimeter is approximately equal to 18 cm, Jamal does have enough string to make the frame.

Therefore, the answer is yes, Jamal has enough string for the frame.