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3 years ago

What is 600+60+6? Please tell me i dont even kno what it is.

Mathematics

2 answers:

faltersainse [42]3 years ago

8 0

Bust out a calculator my dear mammal!
600 +
060 +
006 =
666

;)
Have a good one!
Benni OUT!

Elodia [21]3 years ago

6 0

The answer is 666 cause 600+60 is 660 + 6 would be 666

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Find the sum of the geometric series: 32 - 8 + 2 - 1/2 +...

ZanzabumX [31]

Time to work it out
32 - 8 + 2 - 1/2
32 - 8 + 2 = 26
now you have
-1/2 + 26
so we convert 26 to fraction form to subtract
26/1 - 1/2
we must find the common denominator
52/2 - 1/2
51/2
fifty-one over 2 (fraction) is your answer! :)

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3 years ago

Write in factored form: 13x - 26

iogann1982 [59]

Factored form is x-2

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3 years ago

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Some people think it is unlucky if the 13th day of a month falls on a Friday. Show that in every calendar year (non leap or leap

MakcuM [25]

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 d, February will start 31 days later. Following our pattern above, this will put it at d + 3 (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 d+3 also. April will start 31 days after that, so that puts us at d+3+3=d+6. 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 d+1. June starts 31 days after that, so d+1+3=d+4. July starts 30 days after that, so d+4+2=d+6. 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 d+2. September starts 31 days after that, so d+2+3=d+5. October starts 30 days after that, so d+5+2=d+7, which is just d. November starts 31 days after that, so d+3. December starts 30 days after that, so d+5. 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 d, d, d+1, d+2, d+3, d+3, d+3, d+4, d+5, d+5, d+6 and d+6. 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) d, d+3, d+4, d, d+2, d+5, d, d+3, d+6, d+1, d+4, and d+6. 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.

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3 years ago

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Ivenika [448]

15 is the answer i divided

6 0

3 years ago

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Suppose a computer manufacturer has the total cost function

S_A_V [24]

Answer:

(a) P(x) = 300 x - 3600

(b) P(340) = $ 98400

Step-by-step explanation:

Part (a):

The Profit function is the difference between the revenue function (R(x)) and the Cost (C(x)) function:

P(x) = R(x) - C(x)

P(x) = 384 x - [84 x + 3600]

P(x) = 384 x - 84 x - 3600

P(x) = 300 x - 3600

Part (b):

The profit on 340 items is:

P(340) = 300 (340) - 3600

P(340) = 102000 - 3600

P(340) = $ 98400

Part (c):

To avoid losing money, the profit P(x) has to be larger or equal than zero. That is:

P(x) $\geq$ 0

300 x -3600 $\geq$ 0

300 x $\geq$ 3600

x $\geq$ 3600/300

x $\geq$ 12

So at least 12 items must be sold to avoid losing money.

7 0

3 years ago