Well, you can just state the first 7 digits of this 48 digit number.

The number will look like this:

$${\mathtt{777\,777\,700\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000}}$$

or

$${\mathtt{777\,777\,777\,777\,777\,777\,777\,777\,777\,777\,777\,777\,777\,777\,777\,777}}$$

Cheers

Yes, there is also something called the power rule, which states that a^m^n=a^(m*n.)

For example, 3^4^5=3^20.

ill give u Them: 1,54,2,27,3,18,6,9

which one is it?

XXX

Sorry wrong answer.  Chris's is the right answer.

I answered properly on your later repost.  Please repost as I have requested.  I would have come back here and fixed it quicker if you had done that.   

i dunno, thats a crazy number

Since 98 and 112, share a GCF of 14, divide the numerator and denominator by 14, and you get 7/8.

Just divide

$${\mathtt{v}} = {\mathtt{20}}{\mathtt{\,-\,}}{\mathtt{t}}{\mathtt{\,-\,}}{{\mathtt{t}}}^{{\mathtt{2}}}$$

Set V equal to 0

$${\mathtt{0}} = {\mathtt{20}}{\mathtt{\,-\,}}{\mathtt{t}}{\mathtt{\,-\,}}{{\mathtt{t}}}^{{\mathtt{2}}}$$

Subtract 20 from zero

$$-{\mathtt{20}} = {\mathtt{\,-\,}}{\mathtt{t}}{\mathtt{\,-\,}}{{\mathtt{t}}}^{{\mathtt{2}}}$$

Flip both sides so they are both positive

$${\mathtt{t}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{t}}}^{{\mathtt{2}}} = {\mathtt{20}}$$

There is proabably a more proper way to do it, but i just cycled through squares of low numbers untill i got the answer >.> It's t=4

$${\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{4}}}^{{\mathtt{2}}} = {\mathtt{20}}$$

4/3×201.1=268.133333333333333333 etc..

20% more than the original number is also 1.2 times the original number. To find the original number, divide 60 by 1.2.

You can add 4b to each side to isolate the b. That gets us 21=11. But the statement is false, so there are no solutions for b.

First subtract both sides by x, which gets -5=6x. Divide both sides by 6 to isolate x, and x=-5/6.

It's about 0.7702. Type cot (55) on the Web 2.0 Calc if you want to check.

9+10=19 if you add you will get the answer

Twice as good as I expected: You've got

$$\textcolor[rgb]{1,0,0}{40/20}$$

CONGRATULATIONS !    

You've earned a cookie:

what formula would i use to solve the problem for area of this triangle: triangle ABC, a=3, angleB=24, angleC=24

$$\\(1) \qquad Area=\dfrac{a\cdot b}{2}\cdot\sin{(C)}\\\\
(2) \qquad b=a \cdot \dfrac{\sin{(C)}} {\sin{(A)}} }\\\\
(3) \qquad \sin{(A)}= \sin{( 180\ensurement{^{\circ}}-(B+C) )} = \sin{( B+C ) }$$

$$Area=\dfrac{a^2}{2} \cdot \dfrac{ \sin{(B)} \cdot\sin{(C)} } {\sin{(B+C)}} = \dfrac{a^2}{2} \cdot
\left( \dfrac{ 1 } { \cot{(B)} + \cot{(C)} } } \right)\\\\\\
Area = \dfrac{3^2}{2} \cdot
\left( \dfrac{ 1 } { \cot{(24\ensurement{^{\circ}})} + \cot{(24\ensurement{^{\circ}})} } } \right)\\\\
Area = \dfrac{3^2}{2} \cdot
\left( \dfrac{ 1 } { 2\cdot \cot{(24\ensurement{^{\circ}})} } } \right)\\\\
\boxed{
Area = \dfrac{3^2}{4} \cdot \tan{(24\ensurement{^{\circ}})} } } \\\\
Area = \dfrac{3^2}{4} \cdot 0.4452286853\\\\
Area = \dfrac{9}{4} \cdot 0.4452286853\\\\
Area = 1.0017645419$$

How many triangles do we have at:

1. Step 6 ?
2. Step 8 ?
3. Step 19 ?
4. Step 50 ?

I. The "empty" triangles in the middle are included:

$$\\
\begin{array}{rcrcrcr}
s_1 &=& 1 &=& 1 &=& 1\\
s_2 &=& 3s_1+1 &=& 4 &=& 1+3\\
s_3 &=& 3s_2+1 &=& 13 &=& 1+3+3^2\\
s_4 &=& 3s_3+1 &=& 40 &=& 1+3+3^2+3^3\\
s_5 &=& 3s_4+1 &=& 121 &=& 1+3+3^2+3^3+3^4\\
s_6 &=& 3s_5+1 &=& 364 &=& 1+3+3^2+3^3+3^4+3^5\\
\cdots\\
s_n &=& 3s_{n-1}+1&=& && \dfrac{3^n-1} {2}
\end{array}$$

$$\\
\rm{a.)}\qquad s_6 = \dfrac{3^6-1} {2} = 364 \\\\
\rm{b.)}\qquad s_8 = \dfrac{3^8-1} {2} = 9841\\\\
\rm{c.)}\qquad s_{19} = \dfrac{3^{19}-1} {2} = 581130733\\\\
\rm{d.)}\qquad s_{50} = \dfrac{3^{50}-1} {2} = 358~ 948~ 993~ 845~ 926~ 294 ~385~124$$

II. The "empty" triangles in the middle aren't included:

$$\\
\begin{array}{rcrcrcr}
s_1 &=& 1 &=& 1 &=& 3^0\\
s_2 &=& 3s_1 &=& 3 &=& 3^1\\
s_3 &=& 3s_2 &=& 9 &=& 3^2\\
s_4 &=& 3s_3 &=& 27 &=& 3^3\\
s_5 &=& 3s_4 &=& 81 &=& 3^4\\
s_6 &=& 3s_5 &=& 243 &=& 3^5\\
\cdots\\
s_n &=& 3s_{n-1}&=& && 3^{n-1}
\end{array}$$

$$\\\rm{a.)}\qquad s_6 = 3^5 = 243
\\\\\rm{b.)}\qquad s_8 = 3^7 = 2187
\\\\\rm{c.)}\qquad s_{19} = 3^{18} = 387420489
\\\\\rm{d.)}\qquad s_{50} = 3^{49} = 239 ~299 ~329 ~230 ~617 ~529 ~590 ~083$$

cos(x)=-5/13

cos(x)= adjacent/hyp

  pitagoras:

-5^2+x^2=13^2

x=13.93----> opposite leg of the triangle

tan(x)=opposite leg/adjacent leg

tan(x)=13.93/-5

6*5=30 cookies