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Thanks Alan!

90 C 2 ==4005 - products of 2 2-digit numbers

There are 1,089 multiples of 10

Therefore, the probability is:1,089 / 4,005==121 / 445

First divide everything by 3, and we get -5x-y=45. Now add y to both sides, and we have -5x=y+45. Finally, subtract 45 from both sides to get

\(\boxed{y=-5x-45}\)

40 / 24  * 9  = .......... Liters of water needed for 40 pots

You are welcome !

Thank you so much, have a amazing day.

I am not sure of the 1st question, but I think the answer to the second question is 200. I am not 100% sure though.

Expand L side to get

4x     + 2x+12 = 36         subtract 12 from both sides, combine 'x' terms

6x = 24                          divide by 6

x = 4

and you have a wonderful day, as well.  

What is the probability that at least two of the faces match when you roll four fair six-sided dice? Express your answer as a common fraction.  

There's a 1/1 chance that the first die will be any number 

There's a 1/6 chance that the second die will be the same number 

There's a 1/6 chance that the third     die will be the same number  

There's a 1/6 chance that the fourth  die will be the same number  

So the probability is 1/1 • 1/6 • 1/6 • 1/6  =  1/216  

_.

Thank you for helping me with this problem. I hope you have a wonderful day.

2(5x + 8) = 6x + 20

When I did this problem I got 1, but I am not sure that is correct. Please help me     

Original equation                          2(5x + 8) = 6x + 20   

Multiply left side                            10x + 16  =  6x + 20  

Subtract 6x from both sides            4x + 16  =  20  

Subtract 16 from both sides                    4x  =  4  

Divide both sides by 4                               x  =  1   

_.

CPhil finally extracted the Roman Zero from Sisyphus’ bolder.

There are 651 permutations of 5 dice that sum up to 15 as follows:

They begin with these permutations:

11166 , 11256 , 11265 , 11346 , 11355 , 11364 , 11436 , 11445 , 11454 , 11463 , 11526 , 11535 , 11544 ........etc.

And end with these permutations:

63411 , 64113 , 64122 , 64131 , 64212 , 64221 , 64311 , 65112 , 65121 , 65211 , 66111 , Total =  651 ways for a sum of 15 on a roll of 5 dice.

He had earned a break.

You know, this forum is not for homework...

If the two equations have a common root you can equate them (the two equations) to find the two values of x for which it holds (the bx terms cancel out).

Substitute each value of x into either equation and solve for b.  

You should find two values (though not 17675308 or -17675308 according to my calculations).

 Find the set of all possible values of x+y.

Hello Guest!

The set of all possible values of x+y = \(\{ -10,-6,6,10\}\)

  !

The sum of 2 cubics is  ..  (something you need to memorize)

\(a^3+b^3=(a+b)(a^2-ab+b^2)\)

So if you have a denominator of  \((a+b)\) where a or b have cube roots, and you multiply it by  \((a^2-ab+b^2)\) then you will end up with a rational denominator.

\(\frac{1}{\sqrt[3]{2}+\sqrt{16}}=\frac{1}{\sqrt[3]{2}+4}\)

let

\(a= 2^{1/3}\qquad    and\qquad b=4\\so\\ a^2=2^{2/3}\qquad    and\qquad b^2=16\\ ab=4*2^{1/3}\)

\(a^3+b^3=(a+b)(a^2-ab+b^2)\)

\(\frac{1}{\sqrt[3]{2}+4}\times \frac{2^{2/3}-4*2^{(1/3)}+4^2}{2^{2/3}-4*2^{(1/3)}+4^2}\\~\\ =\frac{2^{2/3}-4*2^{(1/3)}+4^2}{2+64}\\~\\ =\frac{\sqrt[3]{4}-4\sqrt[3]{2}+16}{66}\\~\\\)

Here is a video that might help you understand

You are a member because you have a username.  You are not a 'guest'.

Most members have the free version.  I think this just means that since you didn't pay anything, that you see some ads. 

The possible values of n are -31, -13, -17, -1, 1, 13, 17, and 31.

295 / 393, credit to CPhil.