An architectural firm must decide on the size of a scale model to present to a client. The area of the base A (in ) is related to the weight of the model W (in lbs) according to the equation . If the area of the base is currently and is increased by (while keeping the model to scale), determine the corresponding change in the weight of the model.

The weight increases from 32 lbs to 64 lbs

Step-by-step explanation:

The question is not complete, the complete question is:

An architectural firm must decide on the size of a scale model to present to a client. The area of the base A (in ft^2) is related to the weight of the model W (in lbs) according to the equation W^2 = 4A^3. If the area of the base is currently 16 ft^2? and is increased by 16ft^2 (while keeping the model to scale), determine the corresponding change in the weight of the model

Solution:  

Given that the square of the area of the base (A) is directly proportional to the cube of the weight of the model (W). The relationship is given as:

W² = 4A³

This means a the weight increase, the area also increases. To the determine the corresponding change in the weight of the model, we have to determine the weight when the area is 16 ft² and increased by 16 ft².

At A = 16 ft²

W² = 4A³

W² = 4(16³)

W² = 1024

W = √1024 = 32

W = 32 lbs

When the area is increased by 16 ft² A = 16 ft² + 16 ft² = 32 ft²

W² = 4A³

W² = 4(32³)

W² = 4096

W = √4096 = 64

W = 64 lbs

The weight increases from 32 lbs to 64 lbs

step-by-step explanation:

20% of 120 is 24. if you subtract 24 from 120 you get 96

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