MaxWong

(1)      tan(asin A)=((cosB)^2((tanB)^4-1)+1)/(tanB)^2

1-3+4-6+7-9+10-....-300+301

I will code it like

Look at the numbers with odd-numbered code which I have already bolded.

The numbers of code with the form 2n+1 are with the form of 3n+1.

Look at the numbers with even-numbered code which I set the font Italic.

The numbers of code with the form 2n are with the form 3n.

1-3+4-6+7-9+10-....-300+301

= (1+4+7+10+......+301)-(3+6+9+......+300)

There are \(\color{blue}\frac{301-1}{4-1}+1\) terms in (1+4+7+10+......+301)

which is 101 terms.

There are \(\color{blue}\frac{300-3}{6-3}+1\) terms in (3+6+9+......+300)

which is 100 terms.

1-3+4-6+7-9+10-....-300+301

= (1+4+7+10+......+301)-(3+6+9+......+300)

= \(\color{blue}\frac{(1+301)(101)}{2}-\frac{(3+300)(100)}{2}\)

=303 <------ Final Answer

It will show like

\(\frac{1}{3}+2=\frac{7}{3}=2.33333333333333333333\)

\({{10}^{10}}^{100}\times1000\)

= \({{10}^{10}}^{100}\times10^3\)

=\({10}^{({{10}^{100}+3)}}\)

The problem is: How many people are there in SP?

\(-\frac{2}{3}-(-2\frac{2}{5})\)

=\(-\frac{2}{3}+2\frac{2}{5}\)

=\(2\frac{2}{5}-\frac{2}{3}\)

=\(2\frac{6}{15}-\frac{10}{15}\)

= \(1\frac{21}{15}-\frac{10}{15}\)

=\(1\frac{11}{15}\)

Actually I don't want to know the answer, I have it.

I just want to know if my questions are possible to solve.

7 tenths of 80

= \(\color{red}\frac{7}{10}\times80\)

=\(\color{cyan}\frac{560}{10}\)

=\(\color{blue}\bf56 ----- Final Answer\)

The bottom left corner of the monitor.

\(\color{cyan}k=15+\frac{741}{16}=\color{blue}\frac{981}{16}\)