Straight

Helli abmusical2021,

i accept your apology, at least you apologised, the others would probably keep writing or unfortunately insulting.

Thank you.

Straight

Hello abmusical2021,

you don't have to answer again a question that has already been answered, only if you got another answer out of it - not in this case .

That answer would be better if it has a calculation step - like in this case \(\rightarrow\) Someone has answered a question with (!) calculation step and you just write it down. It looks like you stole it from that person with (!) the calculation path. Also, it's not called disliking directly if you know you made a mistake, they would apologise there, and you are also aware of what you are disliking, even though you know it's only true.

Straight

Hello abmusical2021,

Stop stealing the solutions from the other questions that have already been answered!

And that doesn't mean that you have to disliked everything if you made the mistake and you are aware that it is not allowed.

Please stop it.

Straight

Hello Guest,

please ask your question in the forum,

thank you.

But please check your typing, so write again what you mean, please.

Straight

Hello Guest,

This question has already been answered in the following link:

https://web2.0calc.com/questions/who-is-smarter-than-a-6th-grader

Straight

Hello abmusical2021,

stop it, that's not cool...

Straight

Hello Guest,

almost right, but there are 36 positive integers, less than 1000 and not equal to or larger.

Anyway, everyone makes mistakes ... 

Straight

Hello Guest,

Evaluate the infinite geometric series: 1 + 2/7 + 4/49 + 8/343 + ...

\({\color{red}Solution:}\)

\(1 + \frac{2}{7} + \frac {4}{49} + \frac {8}{343} + \mbox { } ...\)

\( \mathcal{ \mbox {Like Guest said it right } } \mbox { } (^2/_7)^0 + (^2/_7)^1 + (^2/_7)^2 + (^2/_7)^3 + \mbox { } ...\)

Use the double summation sign \(\sum\) .

\(\displaystyle\sum_{n=1}^{\infty} \Bigl( \frac {7}{2} \Bigr)^{1 - n} = \frac {7}{5} = 1.4 \) .

\(\mathit{ Straight}\)

Hello Guest,

The prime factorization of 2007 is 3^2*223. How many ordered pairs of positive integers (x,y) satisfy the equation xy^2=2007?

\(xy^2=2007\)

\(x = 223\)

\(y = \pm \mbox { } 3\)

\(x = 2007\)

\(y= \pm \mbox { } 1\)

These are the onliest possibilities, \((x, \mbox { } y = 223, \mbox { } \pm \mbox { } 3)\) and \((x, \mbox { } y = 2007, \mbox { } \pm \mbox { } 1)\) .

Straight

Hello Guest,

The sum of two numbers is 235, and the difference of the same two numbers is 41. Find these numbers.

\(a+b=235\)

\(a-b=41\)

\(\mbox {You can change the variables by the first equation.}\)

\(b+a=235\)

\(a-b=41\)

\(b+a=235\)

\(-b+a=41\)

\(\mbox {Subtract a from the first equation.}\)

\(b=-a+235\)

\(-b+a=41\)

\(\mbox {Substitute } b=-a+235 \mbox { in the second equation.}\)

\(b=-a+235\)

\(2a-235=41\)

\(\mbox {Add } 235 \mbox { by } 41\)

\(b=-a+235\)

\(2a=276\)

\(\mbox {Solve for a}\)

\(b=-a+235\)

\(a=138\)

\(\mbox {Substitute } a \mbox { into the first equation. }\)

\(b=97\)

\(a=138\)

\(\mbox {This equation was solved with the variables a and b having the values } (a,b)=(138,97) \mbox { . }\)

Straight