A box is 10 inhigh, 18 inand 12 inwide. What is the longest poster you could fit in the box?

a. 24.2 in

b. Paths of the internal diameter, where the longest poster is placed cross each other

Step-by-step explanation:

a. The longest poster that can be fitted into the box is the size of the internal diagonal of the box, that is

√((8² + 6²) +15²) = 3·√65 = 24.2 in

b. Since the above give the diagonal of the box, there can only be one poster of the dimension 24.2 in as due to path interference of the longest poster that is placed at the internal diagonal of the volume of the box

However, for the surface diagonal we have the dimension given by;

√(8² + 15²) = 23.4 in. Therefore since the side diagonal is parallel to the side of the box, one can fit several 23 in. posters which are lesser than the length of the diagonal of the side of the box side by side in the box.

Step-by-step explanation:

CarryOnLerning

MarkAsBrainliest Please

step-by-step explanation:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

step-by-step explanation:

given that megan owns a clothing boutique.

the profits of 2010, 11, are recorded and shown as follows:

years since     2010 0     1     2     3       4       5

profits made in 1000's   39       84     111     120   111     84

we find that profit gradually increases form 2010 to 2013 upto 120,   then starts decreasing.

the profits for 2012 and 2014 are the same while that of 2011 and 2015 are also the same.

this shows that there is symmetry about the line profit = 120000

hence answer is:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

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