Look at picture above. I already did some work, so....

Answer:25 miles

Step-by-step explanation: Its 25 because if you divide 125 with 5 you get 25 miles very simple math you probably in 5th or 4th grade. Np btw

The longest flagpole that could be shipped in a box that measures 2 ft by 2 ft by 12 ft is 2 feet long.

How to obtain the minimum value of a function?

To find the minimum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.

Putting those values of x in the second rate of function, if results in negative output, then at that point, there is maxima. If the output is positive then its minima and if its 0, then we will have to find the third derivative (if it exists) and so on.

We are given that;

The measurement of box= 2*2*12ft

Now,

To determine the longest flagpole that could be shipped in a box that measures 2 ft by 2 ft by 12 ft, we need to find the maximum length of a flagpole that can fit inside the box without exceeding any of the box's dimensions.

The longest flagpole that could fit inside the box would be equal to the shortest dimension of the box, which is 2 feet. This is because the flagpole would need to fit inside the box in a diagonal orientation in order to maximize its length.

Therefore, by the maxima and minima answer will be 2 feet long.

Learn more about maxima and minima of a function here:

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Answer:

y = 2 (x + 3)² − 4

Step-by-step explanation:

y = 2x² + 12x + 14

First, factor the leading coefficient from the first two terms:

y = 2 (x² + 6x) + 14

Complete the square.  Take half of 6, square it, then add and subtract.

(6/2)² = 3² = 9

y = 2 (x² + 6x + 9 − 9) + 14

Simplify:

y = 2 (x² + 6x + 9) − 2(9) + 14

y = 2 (x² + 6x + 9) − 18 + 14

y = 2 (x² + 6x + 9) − 4

Factor the perfect square.

y = 2 (x + 3)² − 4

Answer:

x = .98

Step-by-step explanation:

-.5(17x - 24) = 3.4x - 6 + 6.3x

-8.5x + 12 = 9.7x - 6

-18.2x = -18

18.2x = 18

x = .98