We will set a variable, d, to represent the day of the week that January starts on. For instance, if it started on Monday, d + 1 would be Tuesday, d + 2 would be Wednesday, etc. up to d + 6 to represent the last day of the week (in our example, Sunday). The next week would start over at d, and the month would continue. For non-leap years:
If January starts on <u>d</u>, February will start 31 days later. Following our pattern above, this will put it at <u>d</u><u> + 3</u> (28 days would be back at d; 29 would be d+1, 30 would be d+2, and 31 is at d+3). In a non-leap year, February has 28 days, so March will start at <u>d</u><u>+3</u> also. April will start 31 days after that, so that puts us at d+3+3=<u>d</u><u>+6</u>. May starts 30 days after that, so d+6+2=d+8. However, since we only have 7 days in the week, this is actually back to <u>d</u><u>+1</u>. June starts 31 days after that, so d+1+3=<u>d</u><u>+4</u>. July starts 30 days after that, so d+4+2=<u>d</u><u>+6</u>. August starts 31 days after that, so d+6+3=d+9, but again, we only have 7 days in our week, so this is <u>d</u><u>+2</u>. September starts 31 days after that, so d+2+3=<u>d</u><u>+5</u>. October starts 30 days after that, so d+5+2=d+7, which is just <u>d</u><u />. November starts 31 days after that, so <u>d</u><u>+3</u>. December starts 30 days after that, so <u>d</u><u>+5</u>. Remember that each one of these expressions represents a day of the week. Going back through the list (in numerical order, and listing duplicates), we have <u>d</u><u>,</u> <u>d,</u><u /> <u>d</u><u>+1</u>, <u>d</u><u>+2</u>, <u>d+3</u><u>,</u> <u>d</u><u>+3</u>, <u>d</u><u>+3</u>, <u>d</u><u>+4</u>, <u>d</u><u>+5</u>, <u>d</u><u>+5</u>, <u /><u /><u>d</u><u>+6</u><u /><u /> and <u>d</u><u>+6</u>. This means we have every day of the week covered, therefore there is a Friday the 13th at least once a year (if every day of the week can begin a month, then every day of the week can happy for any number in the month).
For leap years, every month after February would change, so we have (in the order of the months) <u></u><u>d</u>, <u>d</u><u>+3</u>, <u>d</u><u>+4</u>, <u>d</u><u />, <u>d</u><u>+2</u>, <u>d</u><u /><u>+5</u>, <u>d</u><u />, <u>d</u><u>+3</u>, <u>d</u><u /><u>+6</u>, <u>d</u><u>+1</u>, <u>d</u><u>+4</u>, a<u />nd <u>d</u><u>+</u><u /><u /><u>6</u>. We still have every day of the week represented, so there is a Friday the 13th at least once. Additionally, none of the days of the week appear more than 3 times, so there is never a year with more than 3 Friday the 13ths.<u />
Let us make a list of all the details we have
We are given
The cost of each solid chocolate truffle = s
The cost of each cream centre chocolate truffle = c
The cos to each chocolate truffle with nuts = n
The first type of sweet box that contains 5 each of the three types of chocolate truffle costs $41.25
That is 5s+5c+5n = 41.25 (cost of each type of truffle multiplied by their respective costs and all added together)
The second type of sweet box that contains 10 solid chocolate trufles, 5 cream centre truffles and 10 chocolate truffles with nuts cost $68.75
That is 10s+5c+10n = $68.75
The third type of sweet box that contains 24 truffles evenly divided that is 12 each of solid chocolate truffle and chocolate truffle with nuts cost $66.00
That is 12s+12n=$66.00
Hence option C is the right set of equations that will help us solve the values of each chocolate truffle.
Answer: f(x) = x-7
explanation: the function is already being translated down 5, so subtracting 2 more units leaves you with this new equation