3D Geometry and Coordinate Distance

The correct answer is A (3 feet). The minimum distance between two vehicles at a parking place must be at least 3 feet. This space allows for doors to open and for maneuvering in and out of parking spaces safely.

Using the distance formula d = sqrt((2-(-1))^2 + (7-3)^2) = sqrt(3^2 + 4^2) = sqrt(9+16) = sqrt(25) = 5. This forms a 3-4-5 right triangle.

Scale = real distance : map distance. Convert both to the same unit: 400 feet = 400 × 12 = 4800 inches. Scale = 4800 inches : 3 inches = 4800:3 = 1600:1, which is written as 1:1600. This means 1 inch on the map represents 1600 inches in reality.

Using formula, distance = |(0+0-9)/rt (1+1)| 

It is given that n miles are represented by s inches. So, 1 mile is represented by s/n inches. So, (n + 1) miles will be represented by s(n + 1)/n inches.

On a number line, - 2 is 8 steps behind 6 and 14 is 8 steps ahead of 6. That means both integers are at the same distance from 6 on the number line.

One kilometer is approximately 0.621371 miles. Therefore, 0.62 is the correct conversion factor.

Let the actual distance be x km . More distance on the map, more is the actual distance 0.6 : 80.5 : : 6.6 : x غ 0.6 x = 80.5 × 6.6 , x = 885.5

PQ = 10 units ^{$\sqrt{(10-2)^2 + (y+3)^2} = 10 \Rightarrow \sqrt{64+ y^2 + 6y + 9} = 10$}_.  Squaring both sides: 64 + y² + 6y + 9 = 100 _{^{$\Rightarrow$}} y² + 6y - 27 = 0 y² + 9y - 3y - 27 = 0     _{^{$\Rightarrow$}} y(y + 9) - 3(y + 9) = 0 (y + 9) (y - 3) = 0  _{^{$\Rightarrow$}} y +9 = 0 _{^{$\Rightarrow$}} y= -9 and y -3 = 0 _{^{$\Rightarrow$}} y = 3 

The perpendicular distance of the point (6, – 5) from the x-axis will be y-coordinate but with a positive sign as distance is never negative. So, the perpendicular distance of (6, – 5) from the x-axis is 5 units.

The perpendicular distance from y–axis of any point is x co-ordinate but distance is always positive. So, the perpendicular distance of (– 7, 9) from y–axis is 7 units.

The perpendicular distance from y–axis of any point is x co-ordinate but distance can never be negative. So, the perpendicular distance of (13, 12) from y–axis is 13.

The perpendicular distance of a point (- 2, - 4) from x-axis will be y co-ordinate but with a positive sign, as the distance is never negative. So, the perpendicular distance of (- 2, - 4) from x-axis is 4 units.