Straight

Hello,

64^x = 16^(x−1),

We can write 64^x to and 16^(x−1) to:

(${4}^{3x}={4}^{2x-2}$),

($3x=2x-2$),

($3x-2x=-2$),

($x=-2$)

Straight

If both two-digit numbers A8 and 3B have two digits and the product of these factors should be a 4-digit number, then there could not be a 1 at A, because the largest possible result from the factors A8 and 3B would be 702. The following possibilities would not be possible through a small product that is smaller than 1000:

18 and 30 ... 18 and 39; 28 and 30 ... 28 and 35, so together there would be 15 possibilities that are not possible because of their small products that are less than 1000. Now note down an important point:

To get C430 as a product, the factor 3B must (except for a few exceptions) be on either one of these digits 0 or 5. This important point now gives rise to more impossible factors, namely: 28 and 30 ... 28 and 39, since we said that 3B should end in 0 or 5, it was already in the first lists that they would not be possible to form a product that is 1000 and the factor 3B should end in 0 or 5. Now, according to the new rules, there can be no 1 and no 2 in the factor A. At this moment, they are trying to deal with the factor A8, so this factor is now called 38, and as I said, the second factor 3B can only have a 0 or 5, because these factors form a product that is at least divisible by 5. 38 and 30 add up to 1140 - doesn't fit and 38 and 35 - doesn't fit either. And this is how we always proceed. Of course, I'm not writing the new rules and the steps to find possible solutions, but I'm saying that such numbers are a possibility.
These are the following possibilities:

98 and 35

I think that's the onliest possibility, where it applies:

A=9

B=5

C=3

Straight

I will answer your question...

In box A there were (at first) 250 balls and 

in box B were (at first) 40 balls ($250+40$).

I suggested you to get trys and then get an answer,

but if you're trying please note that the balls on

each side must be divided by 5.

Use for example:

($162-128=34$)

($156-134=22$)

You see the differences beetween the first and the second chart they're only ($34-22=$) ($12$).

And the differences are ALWAYS 12, so ($22-12=10$) and 10 is the difference we should get.

I subtract now 260 balls to 250 balls (like I did) and add these 10 to the 30 balls ($260-10=250, 30+10=40$)

I know that we are getting the difference of 10, but still I make a chart for a proof:

The difference is now ($150-140=$) 10 and that is what we should get.

Straight

Absolutely no one will help you and that's

because you're asking many questions,

you're just doing this for your homework,

"help me, i don't know, don't know what to do,..."

If you're not stopping, then I unfortunately report you.

Stop with it.

Straight

$2:3= \frac{2}{3}= 0.66...$, so $2:3$ is 0.66... (period)

Straight

That's mega easy,

why are you asking the question,

that should be a joke.

Straight

You are right, so it stands on the book too!

but you mus divide ($30:2$) and thats 15, so there are 15 ways.

there are 6 different 2 cloured squares

yes you are right, at first there are 6 possibilities, but at second only 15 ways because if you have the blue square and the red square,

you cannot have then the red square and blue square (this is not a second possibility!), this does not make sense. if you have both ones, you have the both ones and not 

again.

so i think there are 15 possibilities.

could you please answer again?

:) 

happy b-day hectictar