The formula for the box's volume is V = (length)(width)(height). If we denote the side length of the cutouts as x, we establish V = (28 - 2x)(22 - 2x)(x). Explanation: We intend to excise x by x squares from each corner of a 28 by 22-inch poster board, leading to bottom dimensions of 28 - 2x and 22 - 2x, with a height of x. This expression can be left as is or multiplied and simplified if desired. If we select x = 2 (a random choice as you did not specify), the box's volume calculates to V = (28 - 2*2)(22 - 2*2)(2), rendering V = (24)(18)(2) cubic inches. Since x measures length, it must be greater than zero. Furthermore, the base width of the box can't fall below zero, establishing the inequality for x: 22 - 2x > 0, meaning 11 - x > 0, or x < 11. If we check with x = 10, then V = (28 - 20)(22 - 20)(10). Is this greater than zero? YES. Thus, x < 11 is indeed a reasonable domain in this context.
Response:
6 hours worked as a tutor
Detailed clarification:
40 x 9.25 = 370$
425 - 370 = 55
55/10 = 5.5
so 5.5 but since you work a full hour, round up to 6 hours worked as a tutor
The distance from point Y to the flag post measures 38.13 m. Step-by-step explanation: Assuming point Y is located at the intersection of both lines shown. Point X is positioned 34 meters east of point Y. The flagpole at point X is observed at a bearing of N18°W, meaning it creates an angle of 18° to the west from the north at point X. Conversely, at point Y, the flagpole has a bearing of N40°E, which makes a 40° angle towards the east from the north.
Considering ∆ AXY as a right triangle, the angle FXY is established. Then, concerning ∆ BYX as another right triangle, the angle FYX is also determined. To find the third angle ∠YFX in triangle FYX, the angle sum property of triangles can be applied:
∠YFX + ∠FYX + ∠FXY = 180°
Thus, we have: ∠YFX + 50° + 72° = 180° leading to ∠YFX = 58°.
Now we can calculate the distance FY using the sine rule.
