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$${{\mathtt{0.12}}}^{{\mathtt{9}}} = {\mathtt{0.000\: \!000\: \!005\: \!159\: \!780\: \!4}}$$

One person can be 'anchored" in one of two seats on one side.......the other seven people can be arranged in 7! ways.....so....

2 x 7!  =

2 x 5040 = 10,080 ways

nevermind, there was a problem.

Consider that the three are anchored as a "block' .....  they can be seated in 3! = 6 ways

And the other 5 people can be arranged  in 5! =120 ways

So

6 x 120  = 720 different arrangements

i posted something there. Please help. Sorry

i think this is wrong. But I have no idea how.

not sure if this is the same question. Sorry if it is!! So sorry!!

(9+3)-3x+34^2-8(2-9)=38x    simplify

12 - 3x + 34^2 - 8(-7) = 38x

12 - 3x +  1156 + 56 = 38x

1223 - 3x = 38x      add 3 x to both sides

1224 = 41x        divide both sides by 41

x =  about 29.8536585365853659

5

B***s do not matter but boxes matter

we have following pssible scenarios: 4-0, 0-4, 3-1, 1-3, 2-2 

You can consider it like Box A gets 0-4 b***s(5 possibilities) and Box B gets the rest.

Mathemathical explanation may be C(4,2) - 1 (1 for (2,2) scenario which both boxes have same amount and do not have to count the scenario that b***s switch and get (2,2) again) 

Assuming you can't even tell which box is on the left vs. right as you move the b***s around, and that all b***s must be put into a box...

3

i) You can put all 4 in one, with none in the other
ii) You can move one of those b***s (doesn't matter which one because they're indistinguishable) to the box that's empty in answer i) thus having 3 in one and 1 in the other.
iii) You can move one more to even it out.

... and that's it because as soon as you move another, it's indistinguishable from ii), etc.

Yea i do what's the question?

2+7=9 

9-7=2

My favourite colour is silver :)

This won't work

The equation would be something like

5x + 4 = 953   and if we subtracted 4 from both sides, we have

5x = 949        and x won't be an integer...!!!

 76-70-73-68-70-66

The pattern is....

-6 + 3 - 5 + 2 - 4    so the next number should be 66 + 1 = 67

I assume we have

log₂ 6

This just equals.....

log 6 / log 2     ....and you can evaluate that on the onsite calculator....

I'm assuming that we have :

3a/[(a-b)(a+b)] - 3/(b-a)     note... -3/(b-a)  =  -3 / [- (a - b) ] =  + 3/(a - b)     the negative signs "cancel"

This gives us

3a/[(a-b)(a+b)] + 3/(a-b)       multiply the second fracton by (a + b) on top and bottom

3a/[(a-b)(a+b)]  + 3(a + b)/ [(a - b) (a + b)]=   {combine everything over a common denominator}

[3a + 3(a + b)] / [(a - b)(a + b) ] =

[3a + 3a + 3b ] /   [(a - b)(a + b) ] =

[6a + 3b] / [a^2 - b^2] =

3[2a + b] / (a^2 - b^2)

There cannot be a "largest number in the world" because you can add 1 to any number that exists.

So say the "largest number" was 1,000,000,000. You can add 1 to make it 1,000,000,001.