(04.02 MC) A contractor is building a new subdivision on the outside of a city. He has started work on the first street and is planning for the other streets to run in a direction parallel to the first. The second street will pass through (1, 5). Find the equation of the location of the second street in standard form. coordinate grid with a segment labeled Street 1 that extends between the points negative 5 comma 6 and three comma negative 2; a point labeled Street 2 is at 1 comma 5 x + y = 6 2x + y = 7 x − y = 6 2x − y = 7

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240

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."