An airplane takes off in a straight path. When it has traveled 5 km, it is 3 km high. To the nearest kilometer, what is the horizontal distance, d that the airplane has flown over?

Use the same process I showed you in your last post.

http://www.jiskha.com/display.cgi?id=1304041347

To find the horizontal distance the airplane has flown over, we can use the Pythagorean theorem.

The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

In this case, the horizontal distance is the base of the right triangle, and the height is the altitude of the airplane. Let's represent the horizontal distance as 'd' and the altitude as 'h'.

According to the problem, when the airplane has traveled 5 km, it is 3 km high. Therefore, we have the following information:

Horizontal distance (d) = 5 km
Altitude (h) = 3 km

Applying the Pythagorean theorem, we have:

d² + h² = (5 km)²

Simplifying this equation, we obtain:

d² + 3² = 5²

d² + 9 = 25

Subtracting 9 from both sides of the equation, we have:

d² = 25 - 9

d² = 16

Taking the square root of both sides, we find:

d = √16

Since we are interested in finding the distance to the nearest kilometer, we round the result to the nearest whole number:

d ≈ 4 km

Therefore, the horizontal distance that the airplane has flown over, to the nearest kilometer, is 4 km.