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.
The computed value is x=14%. I just completed the quiz.
What was the amount of ziti consumed?
The tension does not approach infinity.
<span>Let's analyze free body diagrams (FBDs) for each mass, considering the direction of motion of m₁ as positive.
For m₁: m₁*g - T = m₁*a
For m₂: T - m₂*g = m₂*a
Assuming a massless cord and pulley without friction, the accelerations are the same.
From the second equation: a = (T - m₂*g) / m₂
Substitute into the first:
m₁*g - T = m₁ * [(T - m₂*g) / m₂]
Rearranging:
m₁*g - T = (m₁*T)/m₂ - m₁*g
2*m₁*g = T * (1 + m₁/m₂)
2*m₁*m₂*g = T * (m₂ + m₁)
T = (2*m₁*m₂*g) / (m₂ + m₁)
Taking the limit as m₁ approaches infinity:
T = 2*m₂*g
This aligns with intuition since the greatest acceleration m₁ can have is -g. The cord then accelerates m₂ upward at g while gravity acts downward, leading to a maximum upward acceleration of 2*g for m₁.</span>
