In figure 1, the area of square a is 9 square units, the area of square b is 16 square units, and the area of square c is 25 square units. in figure 2, the area of square d is 36 square units and the area of square e is 64 square units, and the area of square f is 100 square units. in figure 3, the area of square g is 25 square units and the area of square h is 144 square units, and the area of square i is 169 square units. what can be concluded about the areas of the squares? a. the sum of the areas of the two smaller squares is less than the area of the larger square. b. the sum of the areas of the two smaller squares is greater than the area of the larger square. c. the sum of the areas of the two smaller squares is equal to the area of the larger square. d. the sum of the areas of the two smaller squares is equal to the area of the triangle.

C. The sum of the areas of the two smaller squares is equal to the area of the larger square.

Step-by-step explanation:

9 + 16 = 25

36 + 64 = 100

25 + 144 = 169

The relations "less than" and "greater than" can be ruled out. These observations are consistent with selection C.

The triangle area is half the product of the square roots of the squares on the legs, so the areas of the triangles are (respectively) 6, 24, 30. These are not related to the sum of the smaller squares, so the last selection can also be ruled out.

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204