A gift box has dimensions of: 8 1 2 inches, 5 1 2 inches, and 2 1 2 inches, respectively. how many cubes with side lengths of 1

Answer:

I think the first one

Step-by-step explanation:

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.

0.6%. Dividing 100% by 2,500 individuals indicates that each individual accounts for 0.04%. Multiplying 0.04 by 15 equals 0.6

Answer:

The 10th term in the geometric progression is 29.

Step-by-step explanation:

Given: In a geometric series, [T3 = 18] and [T6 = 486].

To find: The term [T10]?

Solution:

A geometric sequence takes the form [a, ar, ar^2,...]

Where, a represents the first term, and r denotes the common ratio.

The nth term is expressed as [Tn = a * r^(n-1)]

From the information provided: [T3 = a * r^2 = 18]

And [T6 = a * r^5 = 486]

By dividing the second equation by the first:

[(a * r^5) / (a * r^2)] = 486 / 18

[r^3 = 27]

Taking the cube root provides: r = 3.

Inserting r into one of the equations allows us to solve for a.

Substituting r gives: [T3 = a * r^2 = 18]

Thus, the first term is a = 2, and the common ratio is r = 3.

The 10th term in the geometric progression is computed as:

[T10 = a * r^(10-1)]

[Thus, T10 = 29.]

To round the figure 47,125 to the closest tenth, hundredth, and thousandth, it's essential to recognize the positions of these places. The tens position is the second digit from the right. The hundreds position is the third, and the thousands position is the digit immediately following the comma. For rounding to the tens place, you must inspect the digit directly behind it (the ones place) and evaluate. If this digit is between 0-4, you won’t round the tens place, but if it falls between 5-9, rounding is permissible. The same regulation applies when rounding up for the hundreds and thousands places. Consequently, the number arrived at is - 47,135.