Marvin is painting three squares shaped like the one below if it takes him exactly 4 cans of paint to paint square 3 five cans of paint to paint square to how many cans of paint should he need to paint square one

Complete question:

Marvin is painting three squares shaped like the one below if it takes him exactly 4 cans of paint to paint square 3, five cans of paint to paint square 2 to how many cans of paint should he need to paint square one

Picture of squares is attached.

9 cans of paint.

To paint square 1, he needs 9 cans of paint.

Step-by-step explanation:

Given that to paint square 3, four cans of paint are needed.

Therefore, let's take area of the square 3 as 4.

i.e a² = 4

Solving further, we have :

a = 2

We can say, each side of square 3 = 2 units

To paint square 2, five cans of paint are needed. Let's take the area of 2 to be 5.

i.e b² = 5

Solving further, we have :

We can say, each side of square 2 = units

Using Pythagoras theorem, we have that the area of the square of a hypotenuse in a 90° angle is equal to the sum of the areas of the other two squares.

Therefore,

h² = 4 + 5

h = 3 units.

We now know one side of the square 1 is 3 units.

Area of square 1 = 3²

Area = 9.

Therefore to paint square 1, he needs 9 cans of paint.

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50