I find the choices available too restrictive. 

Very impressive, Melody  !!!

There exist many approximations. But one formula which is exact is: 

pi = 4*atan(1) , with angle measure in radians

Prove that tan(a)/(1-cot(a))+cot(a)/(1-tan(a))= sec(A)*csc(A)+1

$$\\\frac{tan(a)}{(1-cot(a))}+\frac{cot(a)}{(1-tan(a))}= sec(A)*csc(A)+1\\\\
LHS=\frac{\frac{sina}{cosa}}{(1-\frac{cosa}{sina})}+\frac{\frac{cosa}{sina}}{(1-\frac{sina}{cosa})}\\\\
LHS=\frac{\frac{sina}{cosa}}{(\frac{sina-cosa}{sina})}+\frac{\frac{cosa}{sina}}{(\frac{cosa-sina}{cosa})}\\\\
LHS=\frac{sina}{cosa}\div (\frac{sina-cosa}{sina}) +\frac{cosa}{sina}\div(\frac{cosa-sina}{cosa}) \\\\
LHS=\frac{sina}{cosa}\times(\frac{sina}{sina-cosa}) +\frac{cosa}{sina}\times(\frac{cosa}{cosa-sina}) \\\\
LHS=\frac{sin^2a}{cosa(sina-cosa)} +\frac{cos^2a}{sina(cosa-sina)} \\\\
LHS=\frac{sin^2a}{cosa(sina-cosa)} -\frac{cos^2a}{sina(sina-cosa)} \\\\
LHS=\frac{sin^3a}{cosa*sina(sina-cosa)} -\frac{cos^3a}{cosa*sina(sina-cosa)} \\\\$$

$$\\LHS=\frac{sin^3a-cos^3a}{cosa*sina(sina-cosa)} \\\\
LHS=\frac{(sina-cosa)(sin^2a+sinacosa+cos^2a)}{cosa*sina(sina-cosa)} \\\\
LHS=\frac{(sinacosa+cos^2a+sin^2a)}{cosa*sina} \\\\
LHS=\frac{(sinacosa+1)}{cosa*sina} \\\\
LHS=1+\frac{1}{cosa*sina} \\\\
LHS=sec(a)*cosec(a)+1 \\\\
LHS=RHS\qquad QED \\\\$$

Are you going to share a link to the graph????

We can use the sine inverse to find x

So

sine^-1(6/7) = 58.997280866126°  = x

So.... sine (58.997280866126 / 2) = 0.492402907365

Also, x could be 121.002719133874°

And the sine of (121.002719133874 / 2) = about .870

We can use the Law of Cosines to find  DE...so we have

DE^2 = 720^2 + 940^2 - 2(720)(940)cos47

DE^2 = 478847.0198200768

DE = about 692                                                                        

And using the Law of Sines, we can find angle EDF                 

So we have

sin EDF / 940 = sin 47 / 692

sin EDF  = 940sin47/692 = .9934573403495087

And using the sin inverse

sin-1 (.9934573403495087) = EDF= 83.44°

And the remaining angle, DEF = (180 - 47 - 83.44) = 49.56°

shift log 10^(-5.2) 

I just put in the input bar as 

log(10^-5.2)=

$${log}_{10}\left({{\mathtt{10}}}^{-{\mathtt{5.2}}}\right) = {\mathtt{\,-\,}}{\frac{{\mathtt{26}}}{{\mathtt{5}}}} = -{\mathtt{5.2}}$$

Excellent work anon :)

$$\left({\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left(-{\mathtt{5}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,-\,}}\left({\mathtt{3}}\right)\right){\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right){\mathtt{\,\times\,}}{\mathtt{6}}{\mathtt{\,-\,}}\left({\mathtt{2}}\right) = -{\mathtt{44}}$$

Have a look at the graph of the function, here.....

Notice that we can write

16 sin36 / sin(36 + 36) =

16 sin 36 / (sin36cos36 + sin36cos36) =

16 (sin 36) / [(sin36) [cos 36 + cos36] =

16 / (2 cos 36) =

8/cos 36 = 8(1/cos36) =

8sec36 =  about 9.89

(36^2+25^2-23^2)/(2x36x23)

Just copy and paste the above into the onsite calculator.....it gives

$${\frac{\left({{\mathtt{36}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{25}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{23}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{36}}{\mathtt{\,\times\,}}{\mathtt{23}}\right)}} = {\frac{{\mathtt{58}}}{{\mathtt{69}}}} = {\mathtt{0.840\: \!579\: \!710\: \!144\: \!927\: \!5}}$$

as the answer....!!!!

400,000(1.05^x)=1,000,000     divide both sides by 400,000

1.05^x = 1000000/40000

1.05^x = 2.5     take the log of both sides

log 1.05^x = log 2.5      and we can write

x log 1.05 = log 2.5        divide both sides by log 1.05

x = log 2.5 / log 1.05  = about 18.78

That's quite creative, jillybean...!!!

Well presented,  Anonymous.....why don't you become a member???....then, you wouldn't be "Anonymous" anymore !!!

Brenda put her bitter butter in a batery to make her bitter butter better 10 timmes fast ( i made this one up)

Cos B= (36^2+25^2-23^2)/(2x36x23)

This is solving a Law of Cosines problem for an angle

So, using the cosine inverse, we have

cos^-1 (36^2+25^2-23^2)/(2x36x23) = about 32.8°

Lea just emptied her piggy bank and found that she had $2.85 in nickels and dimes. If the number of nickels is 1 more than twice the number of dimes, how many nickels did she have in her piggy bank?

Call the number of dimes , x

Call the number of nickels 2x + 1

So we have, the number of nickels times their value plus the number of dimes times their value = the total value....algebraically, this is....

.05(2x + 1)  + .10x  = 2.85     multiply through by 100 to clear the decimals

5(2x + 1) + 10x = 285             simplify

10x + 5 + 10x = 285   

20x + 5 = 285                        subtract 5 from both sides

20x = 280                              divide by 20 on each side

x = 14    = number of dimes

2(14) + 1  = 29 = number of nickels

the real answer is 19 but their was a vine made a few weeks ago where this dude asks this kid "whats 9+10" and the kid was like its 21 and then it started the 9+10=21 joke

Very nice, saseflower.....!!!!

Call the amount invested at 4% , x

And call the amount invested at 5%, 10000 - x

And using I = Prt, where t= 1, we have

(x).04 + (10000 - x)(.05) = 440  simplify

.04x + 500 - .05x = 440        subtract 500 from both sides  and combine like terms on the left

-.01x  = -60      divide both sides by -.01

x = $6000   and this is the amount invested at 4%

And the amont invested at 5% is just 10000 - x = 10000 - 6000 = $4000

About 420.

9 pluse 10 be 21

1.7320508075688773

8x-2y=-5

-2x+y=4  →  y = 2x + 4

And substituting for y in the first, we have

8x - 2 (2x + 4)  = -5    simplify

8x - 4x - 8  = -5          add 8 to both sides and combine like terms on the left

4x = 3                        divide by 4 on both sides

x = 3/4     

And  y = 2(3/4) + 4  =  6/4 + 4  =  22/4 = 11/2