I’m trying to correct this problem but i don’t know how can someone me

Remark

It is not a straight line distance from the park to the mall. None of the answers give you that result. And if you know what displacement is, none of the answers are really displacement either. The distance is sort of a "as the crow flies." distance. There's a stop off in the middle of town.

Method

You need to use the Pythagorean Formula twice -- once from the park to the city Center and once from the city center to the mall.

Distance from the Park to the city center.

a = 3   [distance east]

b = 4  [distance south]

c = ??

c^2 = 3^2 + 4^2        Take the square root of both sides.

c = sqrt(3^2 + 4^2)

c = sqrt(9 + 16)          Add

c = sqrt(25)

c = 5

So the distance from the park to the city center is 5 miles

Distance from City center to the mall

a = 2 miles [distance east]

b = 2 miles [distance north]

c = ??

c^2 = a^2 + b^2   Substitute

c^2 = 2^2 + 2^2   Expand this.

c^2 = 4 + 4

c^2 = 8             Take the square root of both sides.

sqrt(c^2) = sqrt(8)

c = sqrt(8)          This is the result

c = 2.8

Answer

Total distance = 5 + 2.8 = 7.8

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations: