Inverse Proportion (2.10.3)

Question 1

• If two quantities are inversely proportional (or indirectly proportional), then when one quantity increases by a certain percentage, the other quantity decreases by the same percentage.

Answer 1

• If two quantities are inversely proportional (or indirectly proportional), then when one quantity increases by a certain percentage, the other quantity decreases by the same percentage.

Question 2

• If some value x is inversely proportional to some value y, then the equation x = k/y must be true.

Answer 2

• If some value x is inversely proportional to some value y, then the equation x = k/y must be true.

Question 3

• If two inversely proportional values are given, then those values can be used to find the constant of variation, k, by using the equation x = k/y or k = xy.

Answer 3

• If two inversely proportional values are given, then those values can be used to find the constant of variation, k, by using the equation x = k/y or k = xy.

Question 4

Inverse proportion works with light in that the farther away the viewer is from the light source, the weaker the
illumination from the light appears to be. So, as the distance from the light increases, the light’s illumination decreases. It is given that the intensity of the illumination of light at an object is inversely proportional to the square of the distance from the light source to the object.
First, define the variables. Let L be the illumination and let d be the distance. So, since the intensity of the illumination from the flashlight L at an object is inversely proportional to the square of the istance d from the flashlight to the object, it follows that L is inversely proportional to d 2. Therefore, the equation L = k/d 2 must be true. The constant of variation k must be found in order to determine the flashlight’s illumination at 12 meters. The problem states that when the distance is 5 meters, the illumination is 70 candela. So, substitute these values into the equation L = k/d 2 and then solve for k. Now write the inverse proportion equation where k = 1750. Substitute 1750 for k in L = k/d 2. L = 1750/d 2 Use this equation to find the flashlight’s illumination at 12 meters. Substitute 12 for d and solve for L.

Answer 4

Inverse proportion works with light in that the farther away the viewer is from the light source, the weaker the illumination from the light appears to be. So, as the distance from the light increases, the light’s illumination decreases. It is given that the intensity of the illumination of light at an object is inversely proportional to the square of the distance from the light source to the object.
First, define the variables. Let L be the illumination and let d be the distance. So, since the intensity of the illumination from the flashlight L at an object is inversely proportional to the square of the istance d from the flashlight to the object, it follows that L is inversely proportional to d 2. Therefore, the equation L = k/d 2 must be true. The constant of variation k must be found in order to determine the flashlight’s illumination at 12 meters. The problem states that when the distance is 5 meters, the illumination is 70 candela. So, substitute these values into the equation L = k/d 2 and then solve for k. Now write the inverse proportion equation where k = 1750. Substitute 1750 for k in L = k/d 2. L = 1750/d 2 Use this equation to find the flashlight’s illumination at 12 meters. Substitute 12 for d and solve for L.