She needs  →  23 * 52 = 1196 pages total

She only has 5 * 200  = 1000  sheets

So.....she needs one more package

69? 

Dividing by 1/2 is the same as doubling, so what is 5 and 1/3 doubled?

Like so:

.

You don't list any values for the variables!  However, you should take each set listed and plug them in to the left-hand side of all the equations to see if the result matches the right-hand side.  

(You should find that x = 0, y = -5 and z = -4 gives a solution).

2^2=2×2=4

Darude Sandstorm

use the Gaussian elimination method

\(\begin{pmatrix}1 &0 &0 &| &1 &0 &0 \\ -1 &1 &0 &| &0 &1 &0 \\1 &1 &1 &| &0 &0 &1\end{pmatrix} \Rightarrow\)

\(\begin{pmatrix}1 &0 &0 &| &1 &0 &0 \\ 0 &1 &0 &| &1 &1 &0 \\ 0 &1 &1 &| &-1 &0 &1\end{pmatrix} \Rightarrow\)

\(\begin{pmatrix}1 &0 &0 &| &1 &0 &0 \\ 0 &1 &0 &| &1 &1 &0 \\0 &0 &1 &| &-2 &-1 &1 \end{pmatrix} \Rightarrow\)

and we read off the inverse as

\(\begin{pmatrix}1 &0 &0 \\ 1 &1 &0 \\ -2 &-1 &1\end{pmatrix} \)

as an alternative to the method heureka has used we can use Gaussian elimination and operate on an identity matrix as well.  At the end of the elimation the identity matrix will have been transformed into the inverse matrix.

\(\begin{pmatrix}0 &4 &| &1 &0\\ -6 & 5 &| &0 &1\end{pmatrix} \Rightarrow \\ \begin{pmatrix}-6 &5 &| &0 &1 \\ 0 &4 &| &1 &0\end{pmatrix} \Rightarrow \\ \begin{pmatrix}-6 &5 &| &0 &1 \\ 0 &1 &| &\frac 1 4 &0\end{pmatrix} \Rightarrow \\ \begin{pmatrix}-6 &0 &| &-\frac 5 4 &1 \\ 0 &1 &| &\frac 1 4 &0\end{pmatrix} \Rightarrow \\ \begin{pmatrix}1 &0 &| &\frac 5{24} &-\frac 1 6 \\ 0 &1 &| &\frac 1 4 &0\end{pmatrix} \Rightarrow \\\)

and we read off the inverse as

\(\begin{pmatrix}\frac 5{24} &-\frac 1 6 \\ \frac 1 4 &0\end{pmatrix}\)

  0 4

- 6 5

\(\boxed{~ A=\begin{pmatrix} a&b\\ c&d \end{pmatrix}\\ A^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d&-b\\ -c&a \end{pmatrix} ~}\)

\(\begin{array}{rcl} A&=&\begin{pmatrix} 0&4\\ -6&5 \end{pmatrix}\\ A^{-1}&=&\frac{1}{0\cdot 5- 4\cdot (-6)}\begin{pmatrix} 5&-4\\ -(-6)&0 \end{pmatrix}\\\\ A^{-1}&=&\frac{1}{24}\begin{pmatrix} 5&-4\\ 6&0 \end{pmatrix}\\\\ A^{-1}&=&\begin{pmatrix} \frac{5}{24}&\frac{-4}{24}\\ \frac{6}{24}&\frac{0}{24} \end{pmatrix}\\\\ A^{-1}&=&\begin{pmatrix} \frac{5}{24}&\frac{-1}{6}\\ \frac{1}{4}&0 \end{pmatrix} \end{array}\)

cos(-pi/6) csc(-pi/3) tan(-pi/6) = ?

\(\begin{array}{rcll} \cos( \frac{-\pi}{6} )\cdot \csc( \frac{-\pi}{3} ) \cdot\tan( \frac{-\pi}{6} ) &&\\\\ =\cos( \frac{-\pi}{6} )\cdot \frac{1}{\sin( \frac{-\pi}{3} )}\cdot \tan( \frac{-\pi}{6} )\\\\ \boxed{~ \cos( \frac{-\pi}{6} ) = +\cos( \frac{\pi}{6} ) = \frac{1}{2}\sqrt{3} \\ \sin( \frac{-\pi}{3} ) = -\sin( \frac{\pi}{3} ) = -\frac{1}{2}\sqrt{3} \\ \tan( \frac{-\pi}{6} ) = -\tan( \frac{\pi}{6} ) = -\frac{1}{3}\sqrt{3} ~}\\\\ =( \frac{1}{2}\sqrt{3} ) \cdot \left( \frac{1}{-\frac{ 1 }{2}\sqrt{3}} \right)\cdot (-\frac{1}{3}\sqrt{3})\\\\ =\left( \frac{ \frac{1}{2}\sqrt{3} }{-\frac{ 1 }{2}\sqrt{3}} \right)\cdot (-\frac{1}{3}\sqrt{3})\\\\ =( -1 )\cdot (-\frac{1}{3}\sqrt{3})\\\\ =\frac{1}{3}\sqrt{3}\\\\ \mathbf{\cos( \frac{-\pi}{6} )\cdot \csc( \frac{-\pi}{3} ) \cdot\tan( \frac{-\pi}{6} ) =\frac{1}{3}\sqrt{3}} \end{array}\)

dB = 20*Log(p1/p2) I need to solve for p1 what is the formula?

\(\begin{array}{rcll} dB &=& 20\cdot \log{ \left( \frac{p_1}{p_2} \right) }\\\\ dB &=& \log{ [ \left( \frac{p_1}{p_2} \right)^{20} ] } \qquad &| \qquad 10^{()}\\\\ 10^{dB} &=&10^{ \log{ [ \left( \frac{p_1}{p_2} \right)^{20} ] } } \\\\ 10^{dB} &=& \left( \frac{p_1}{p_2} \right)^{20} \qquad &| \qquad 1^{\frac{1}{20}}\\\\ 10^{\frac{dB}{20}} &=& \frac{p_1}{p_2} \qquad &| \qquad \cdot p_2 \\\\ p_2\cdot 10^{\frac{dB}{20}} &=& p_1\\\\ \boxed{~ \mathbf{ p_1 } \mathbf{=} \mathbf{p_2\cdot 10^{\frac{dB}{20}} } ~}\\\\ \end{array}\)

Oh, you cannot use Heron's Rule.

that is for finding the area when you have the sides.

OR

somtimes you may be able to use it fo find a side when given an area 

BUT this question has nothing to do with area :))

Manas has asked me to explain my answer more fully. (by private message)

I drew up the diagram. Let X be the centre of the cirlce.  (I fogot to put that on the diagram) 

Draw in each radii and draw in t he intervals between the centre ifthe circle and the vertices

The angle between a radius and a tangent is 90 degrees 

So you have now split your triangle up into 6 right angled triangles

cr=cp=8   tangents subtended from a point intersect the circle at equal distances.

br=bQ=6

ap=aq = ?

\(tan\beta = 4/6\\ so\\ \beta = tan^{-1}(4/6) = 33.6900675\; degrees \)

\(tan\delta=4/8\\ \delta=tan^{-1}(4/8)=26.5650511\; degrees\)

\(2\alpha+2\beta+2\delta=180\;\;angle\;sum\;of\;a\;triangle\\ \alpha+\beta+\delta=90\\ \alpha=90-\beta-\delta\\ \alpha=90-33.6900675-26.5650511\\ \alpha=29.74488\\ tan\;\alpha=\frac{4}{distance\;Aq}\\ distance\;Aq=\frac{4}{tan\;29.74488}\\ distance\;Aq=\frac{4}{tan\;29.74488}\\ distance\; Aq\approx6.99888\\ Aq=Ap\approx 7\;units \)

If you still do not understand manas you can say so but you need to ask specific questions. 

If you do not understand any of it then perhaps it is just way over your head. :)

Thank you, CPhill

[a+b]+[c] when a=-3, b=7, c=-15    .....so we have.....

[ -3 + 7 ] + [ -15]   =

[ 4 ] + [ -15]  =

-11

(1 + 1 + 1 ) !  =

3!  =

6

(0!   +  0!   +  0!) !  =

(1 + 1 + 1) !  =

3!  =

6

(2/3) = two-thirds

(7/8) =seven-eights or 7 over 8.

etc. etc.

Here is an example.  You can just type it into the input box on the calc if you want.   :))

7+sqrt(3/4)=

its 1000000000... its 10 multiplied by itself 8 times. its simple. im assuming your in 5th or 6th so its 1000000000

Ok thanks :)

Got It Thanks its 5.2143 your a big help 

Ummm IDK ._.

How many weeks are there in 1 year? Ask your Mom, if you don't know. Once you get that number, then multiply it by $10. OK?.