Suppose that a wave forms in shallow water. Then the depth d of the water (in meters) and the velocity v of the wave (in meters per second) are related by the equation =v9.8d. If a wave formed in shallow water has a velocity of 5.2 meters per second, what is the water's depth? Carry your intermediate computations to at least four decimal places, and round your answer to the nearest tenth.

The clear question is;

Suppose that a wave forms in shallow water. Then the depth d of the water (in meters) and the velocity v of the wave (in meters per second) are related by the equation v=√9.8d . If the wave formed in shallow water has a velocity of 5.2 meters per second, what is the water’s depth? Carry your intermediate computation to at least four decimal places, and round your answer to the nearest tenth

Water's depth;d ≈ 2.8 m

Explanation:

We are told that the relationship between the velocity and depth of the water is given by;

v = √9.8d

Where;

v is the velocity of the wave and d is the depth of the water

Now, we are given velocity; v = 5.2 m/s.

Thus, plugging it into the relation above, we have;

5.2 = √9.8d

Let's take the square of both sudes to get;

5.2² = 9.8d

27.04 = 9.8d

Let's divide both sides by 9.8 to get;

9.8d/9.8 = 27.04/9.8

d = 2.7592m

We are told to approximate to nearest tenth.

Thus, d ≈ 2.8 m

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form