3D Geometry and Coordinate Distance

The penalty spot is exactly 12 yards (approximately 11 meters) from the goal line. This distance was established in the original penalty kick law of 1891 and remains unchanged. It's measured from the center of the spot to the goal line.

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.

Geostationary orbit is at approximately 35,786 km above Earth's equator. This specific altitude allows a satellite to orbit at the same rate as Earth's rotation, making it appear stationary relative to a point on the ground. The orbital period is 24 hours.

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.