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Coordinates of C are (13, 16)

A mole of perfect gas with an initial temperature of 40ºC expands at constant pressure until it reaches a volume twice the initial one.

to calculate:

1) the final gas temperature

2) the work done by the gas

pv = nrt          n is constant    r is constant    p is constant

v/t = constant

v/(273.15 + 40)  =  2v / ( 273.15 + t2)

313.14  (2) = 273.15 + t2

t2 = 80 C

                                                                                                           30 – 2  

You can get the side of the squares by using the Y-axis distances.    ———   =  7  

                                                                                                                4  

So if the two squares in the middle were corner to corner  

then the horizontal distance from A to B would be 4 • 7 = 28  

But, by subtracting the X-axis distances,  

the horizontal distance from A to B is only 27 – 3 = 24  

So the third square up is displaced to the left by 28 – 24 = 4 units  

So the X-axis component to C is 3 + 7 + 3 = 13  

    3 = the starting distance over from the Y-axis  

    7 = the length of the side of the bottom square  

    3 = how far C is from the top left corner of the second square up  

And the Y-axis component to C is 2 + 7 + 7 = 16  

    2 = the starting distance up from the X-axis  

    7 = the length of the side of the bottom square  

    7 = the length of the side of the second square up  

                                              C is located at (13,16)  

Put given line into   y = mx + b form to find   m = -5/8

    perpindicular   is  - 1/m   = 8/5

then using 10,10

y- 10 = 8/5 (x-10)    re-arrange this into y = mx + b form to compute your final answer

(7*n + 53 ) mod 8=0.........................(1) 

7*n + 53 > 500, solve for n...............(2)

n =8m + 5    and n >=8, where m =0, 1, 2, 3........etc.

Therefore, m = 8 and:

8 x 8 + 5 =69 - original number of gold coins in each of the 7 bags.

69 x 7 =483 - total of all original coins.

[69 - 2] =67 - coins in each of the 7 original bags after removing 2 coins from each bag.

[53 + (7 x 2) ] =53 + 14 =67 gold coins in the 53-coin bag after removing 2 coins from each of the original 7 bags.

8 x 67 =536 coins after balancing the 8 bags with 67 coins in each bag.

Lol I just happened to be on this page when you posted this.

Oh wow that's correct! Thank you!

BTW How did you see this question so fast though? You answered it like 3 minutes after I posted it!

Here's my best attempt at this question... hope I get it!

We know the probability of Sans throwing 79 bones is $\left(\frac{1}{6}\right)^{79}$
We also know that the probability of you dodging 60 or more of them is $\left(\frac{1}{12}\right)^{60}$
But he can choose which 60 to dodge in $\binom{79}{60}$
Therefore the answer is: (Lol you can just plug that into a calculator, but I'm not gonna do it since this page isn't saved) \[\left(\frac{1}{6}\right)^{79}*\binom{79}{60}*\left(\frac{1}{12}\right)^{60}\]

Hope this helped! 

if  f(3) = 6     then   f^-1 (6) = 3     

then     f^-1 (3)  = 2  

This is my attempt at this question:

Any such distribution of the strawberry candy can be encoded in a string such as \[XX|XXX|X|XX\]
of 8 copies of $X$ along with three dividers. (The above example corresponds to giving 2, 3, 1, and 2 candies to the four children, in order.) Since we are allowed to give a child no candy if we wish, the three dividers and the 8 X's can be in any order (in particular we're allowed to have two dividers next to each other). There are then $\binom{11}{3} =165$ ways of arranging 8 X's and 3 |'s, and thus 165 ways of distributing the strawberry candy.

Similarly, distributing the rhubarb candy corresponds to arranging 7 Y's and 3 |'s, so we have $\binom{10}{3} = 120$ ways of distributing the rhubarb candy.

This yields a total of $165\cdot 120 = \boxed{19800}$ ways of distributing the candy.

I hope this helped! 

The three weeks part is irrelevant

176.25 m  / 11.75 m / hr  = 15 hours

Just use your calculator     arcsin (6/7) = theta

6/12 = 1/2

x^10 / x^2 = x^10-2                       x⁸ /2

But how did you GET that answer? 

536  coins all together

Start with 7 bags of 69 each =483 coins to start

quadratic model representing revenue for a hockey arena is given by R= (800-100 x)(12+4x)where x present the number of tickets sold and r is the revenue

You can generate the number of terms beginning with 3 and ending with 9 using the following "closed form" of the sequence: a_n = 1/6 n (n + 1) (n + 2):

(1, 4, 10, 20, 35, 56, 84) ==210 integers.

If the ratio of BD:CD = 2 : 5    then                  10² - (2x)² = 17² - (5x)²

                                                                        x = 3          side  BC = 21

                                              ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

In case of  3 : 5 ratio, we have:                    10² - (3x)² = 17² - (5x)²

                                                                                 x = 3.43693

                                                                       BC = 27.49544    Impossible!

3210  4210  4310  4320  4321  5210  5310  5320  5321  5410  5420  5421  5430  5431  5432  6210  6310  6320  6321  6410  6420  6421  6430  6431  6432  6510  6520  6521  6530  6531  6532  6540  6541  6542  6543  7210  7310  7320  7321  7410  7420  7421  7430  7431  7432  7510  7520  7521  7530  7531  7532  7540  7541  7542  7543  7610  7620  7621  7630  7631  7632  7640  7641  7642  7643  7650  7651  7652  7653  7654  8210  8310  8320  8321  8410  8420  8421  8430  8431  8432  8510  8520  8521  8530  8531  8532  8540  8541  8542  8543  8610  8620  8621  8630  8631  8632  8640  8641  8642  8643  8650  8651  8652  8653  8654  8710  8720  8721  8730  8731  8732  8740  8741  8742  8743  8750  8751  8752  8753  8754  8760  8761  8762  8763  8764  8765  9210  9310  9320  9321  9410  9420  9421  9430  9431  9432  9510  9520  9521  9530  9531  9532  9540  9541  9542  9543  9610  9620  9621  9630  9631  9632  9640  9641  9642  9643  9650  9651  9652  9653  9654  9710  9720  9721  9730  9731  9732  9740  9741  9742  9743  9750  9751  9752  9753  9754  9760  9761  9762  9763  9764  9765  9810  9820  9821  9830  9831  9832  9840  9841  9842  9843  9850  9851  9852  9853  9854  9860  9861  9862  9863  9864  9865  9870  9871  9872  9873  9874  9875  9876  Total =  210 such numbers

Like so

I'll leave you to compute the decimal version.

(1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 108, 144, 216, 432) >>Total = 20 Divisors

Add up all the divisors EXCEPT 432

Just insert "L" at the beginning and "Q" at the end of the following permutations:

{A, E, S} | {A, E, U} | {A, S, E} | {A, S, U} | {A, U, E} | {A, U, S} | {E, A, S} | {E, A, U} | {E, S, A} | {E, S, U} | {E, U, A} | {E, U, S} | {S, A, E} | {S, A, U} | {S, E, A} | {S, E, U} | {S, U, A} | {S, U, E} | {U, A, E} | {U, A, S} | {U, E, A} | {U, E, S} | {U, S, A} | {U, S, E} (total: 24)

176.25 / 11.75 = 15 - hours

(1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99, 198) >>Total = 12 Divisors

[1+2+3+6+9+11+18+22+33+66+99] = 270 - Sum of the proper divisors of 198.