log₉(log₉(x))=0  means log₉(x) = 9⁰, so log₉(x) = 1 

log₉(x) = 1 means x = 9¹, so x = 9

.

9+10 = 10+9

   Do  NOT  ask  again, please!

1 kl  x [1000m / 1 kl ] x [  39.37 in / 1 m ]  x [ 1 ft / 12 in ]  x [ 1mi / 5280 ft ] =  about .621 miles

If a triangle has the side lengths of 8, 22, and 16.113921. One of the angle measures is 35 degrees. What are the measures of the other angles?

a = 8,    b = 16.113921,     c = 22        angles:     A = 35,     B = ?    and    C = ?     height (h) = ?

        h =sin35 * 8

        h = 4.59

        sin(B) = h/16.113921

             B = 16.55^o

             C = 180^o - ( A +B )

             C = 180^o - 51.55^o

             C =  128.45^o

log₆(2x-5)+1=log₆(7x+10)    note that "1 " can be written as log₆(6)  .....so we have

log₆(2x-5)+log₆(6) =log₆(7x+10)  and we can write

log₆(2x-5)(6) =log₆(7x+10)     and now, we can just solve for this

(2x - 5)(6) = (7x + 10)  simplify

12x - 30  = 7x + 10      add 30 to both sides, subtract 7x from both sides

5x = 40     divide by 5 on each side

x = 8 

The sum of the sides or the angles?

The answer they are looking for is 21. It's from a Vine.

False. Swag is an irrational number. So sweg is not >, <, or = to swag.

You need to get the same denominator by multiplying the numerators and deniminators by the other denominator. $$\left({\frac{\left({\mathtt{11}}{\mathtt{\,\times\,}}{\mathtt{10}}\right)}{\left({\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{10}}\right)}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{\left({\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{12}}\right)}{\left({\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{12}}\right)}}\right)$$ You then have $$\left({\frac{{\mathtt{110}}}{{\mathtt{120}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{84}}}{{\mathtt{120}}}}\right)$$ which makes it easier to solve.

If you realize it or not, you use math everyday. Such as to determine if a car is slowing down at a red light or the amount amount of time it took you to get from one place to another.

incorrect there are to many sweg to handle so swag < sweg yis 

The correct answer would be 4.

1200=200+a * 4   subtract 200 from each side

1000 = 4a  divide by 4 on each side

250 = a

4.9t^2+16.2t-59.8=0

$${\mathtt{4.9}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{16.2}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,-\,}}{\mathtt{59.8}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{35\,863}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{81}}\right)}{{\mathtt{49}}}}\\
{\mathtt{t}} = {\frac{\left({\sqrt{{\mathtt{35\,863}}}}{\mathtt{\,-\,}}{\mathtt{81}}\right)}{{\mathtt{49}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = -{\mathtt{5.517\: \!863\: \!036\: \!299\: \!136\: \!9}}\\
{\mathtt{t}} = {\mathtt{2.211\: \!740\: \!587\: \!319\: \!545\: \!1}}\\
\end{array} \right\}$$

34.8 = 10.9 *t + 4.9t^2   let's mulftiply throgh by 10 to clear the decimals

348 = 109t + 49t^2   rearrange

49t^2 + 109t - 348 = 0

Using the onsite solver, we have

$${\mathtt{49}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{109}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,-\,}}{\mathtt{348}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\frac{{\mathtt{87}}}{{\mathtt{49}}}}\\
{\mathtt{t}} = -{\mathtt{4}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{1.775\: \!510\: \!204\: \!081\: \!632\: \!7}}\\
{\mathtt{t}} = -{\mathtt{4}}\\
\end{array} \right\}$$

1200= 200+a*4

1200/4= 200+a*4/4

300=200+a

300-200= 200-200+a

100=a

59.8=16.2*t+4.9t^2

rearrange

4.9t^2+16.2t-59.8=0

now solve it useing the quadratic formula.

y+3-(7+y)=-4

I look at the left hand side

y+3-(7+y)

= y+3-1(7+y)

=y+3 -1*7+-1*y

=y+3-7-y

=3-7

=-4

=right hand side  aways!

So y can be absolutely any number and this statement will always be true!

Thanks Alan, 

I will take a little time to absorb this.

this      dA= 2pi L dR     was a typo,  I did mean dr       

the rest i will need to think about.    

Thanks MG.

Do you actually look at the posts that I include?  

I always wonder how many people really look at what I write :/

I know many members like to see their names included.  

It is always nice to see some recognition when we put effort into something.  :)

Google might. tell me what u find

Thanks 4 sking this question, because people need a place to share there feelings with out fear of judgement/ everyone in school looking at them weird. 

There are not any rational numbers that will do this.

Self worth and coping with bullying/family issues