The perimeter of an airplane ticket is 30 centimeters. The area is 50 square centimeters. What are the dimensions of the ticket?

The length can be 5 and the width 10.

Step-by-step explanation:

perimeter is equal to 2L + 2W

area is equal to L * W

you get 2 equations:

30 = 2L + 2W

50 = L*W

we'll solve by substitution.

you can use either equation to solve for either variable.

we'll use the second equation to solve for L.

second equation is:

50 = L*W

divide both sides of this equation by W to get:

L = 50/W

we now have a value of L that we can substitute in the first equation which we can then use to solve for W.

the first equation is:

30 = 2L + 2W

replace L with 50/W to get:

30 = 2*(50/W) + 2W

simplify to get:

30 = 100/W + 2W

multiply both sides of this equation by W to get:

30W = 100 + 2W^2

subtract 30W from both sides of this equation to get:

2W^2 - 30W + 100 = 0

factor this equation to get:

(2W - 20) * (W - 5) = 0

solve for W to get:

W = 10 or W = 5

if W = 10, then use the equation of 50 = L*W to get L = 5

if W = 5, then use the equation of 50 = L*W to get L = 10

you have 2 possible solutions.

L = 5 and W = 10 or:

L = 10 and W = 5

both are plausible so we'll test them both out.

when L = 5 and W = 10:

the first equation becomes 2*5 + 2*10 = 10 + 20 = 30

the second equation becomes 5*10 = 50

both equations confirm that the solution of L = 5 and W = 10 is good.

when L = 10 andW = 5:

the first equation becomes 2*10 + 2*5 = 20 + 10 = 30

the second equation becomes 10*5 = 5

both equations confirm that the solution of L = 10 and W = 5 is good.

the ticket can have either:

a length of 5 and a width of 10 or:

a length of 10 and a width of 5.

since the length is usually longer than the width, you are free to choose that the ticket most likely has a length of 10 and a width of 5.

alternatively, you can simply say that the dimensions are 5 by 10 and let the recipient of that information determine which is the length and which is the width.

mathematically, the length can be 10 and the width 5 or the length can be 5 and the width 10.

there is only one way to arrange all the eight relative to each other because they are not distinct. we have

now there are nine places to place an (and not more than one in one place).   because we have four symbols, we need to choose four of these nine places to insert an . order does not matter because the are not distinct.

thus there are 1 · c(9,4) ways, or, if we calculate it,

thus there are 126 ways to form such "words."