Melody

<OAP=<OBP=90 degrees              A radius and the tangent it touches are at right angles to one another.

OA=OB                                           Equal radii

OP                                                  common side

therefore

\(\triangle AOP \equiv \triangle BOP \qquad \mbox{Right angle Hypotenuse and one other side test.}\)

b)

If <AOP=64,      <BOP=64            corresponding angles in congruent triangles

<BOP+<BOD = 180                     adjacent supplementary angles

    64  +<BOD = 180

            <BOD =  116 degrees

arc tan is the same as inverse tan

hit

[2nd] [atan] ....

Underneath the messages there is a "block messages from ..." button.

Hit that and all their messages will disappear.

Maybe that will fix your problem :)

That really good MWizzard 

I'd like to show you a little bit more :)

This is your code.

\({\left ( {2n}^{3} \right )}^{4} = {\left ( 2 \times {n}^{3} \right )}^{4} = {\left ( 2 \times {n}^{3} \right )}^{4} = {2}^{4} \times {\left ( {n}^{3} \right )}^{4} = 16 \times {\left ( {n}^{3} \right )}^{4} = 16 {\left ( {n}^{3 \times 4} \right )} = 16 {\left ( {n}^{12} \right )} = 16 {n}^{12}\)  

I expect you know that the left and right functions only make the brackets bigger :)

Most of those parentheses do nothing.

I have copied your code but left out all the unnecessary parentheses.

\left ( 2n^3 \right )^4= \left ( 2 \times n^3 \right )^4 = \left ( 2 \times n^3 \right )^4= 2^4 \times \left ( n^3 \right )^4 = 16 \times \left ( n^3 \right )^4 = 16 \left ( n^3 \times 4 \right ) = 16 \left ( n^{12} \right ) = 16 n^{12}

\(\left ( 2n^3 \right )^4= \left ( 2 \times n^3 \right )^4 = \left ( 2 \times n^3 \right )^4= 2^4 \times \left ( n^3 \right )^4 = 16 \times \left ( n^3 \right )^4 = 16 \left ( n^3 \times 4 \right ) = 16 \left ( n^{12} \right ) = 16 n^{12} \)

Now I am going to add some \\ commands  these indicate to go to a new line

\left ( 2n^3 \right )^4\\= \left ( 2 \times n^3 \right )^4 \\= \left ( 2 \times n^3 \right )^4\\= 2^4 \times \left ( n^3 \right )^4 \\= 16 \times \left ( n^3 \right )^4\\ = 16 \left ( n^3 \times 4 \right )\\ = 16 \left ( n^{12} \right )\\ = 16 n^{12}

\(\left ( 2n^3 \right )^4\\= \left ( 2 \times n^3 \right )^4 \\= \left ( 2 \times n^3 \right )^4\\= 2^4 \times \left ( n^3 \right )^4 \\= 16 \times \left ( n^3 \right )^4\\ = 16 \left ( n^3 \times 4 \right )\\ = 16 \left ( n^{12} \right )\\ = 16 n^{12}\)

This was the question

3^(x+1)=4^(x-1) 

This is the beginning of Admin's answer

3^(x+1)=4^(x-1)      | *4 

What Admin meant is that you have to multiply both sides by 4.

\(4*   3^{(x+1)}=4^{(x-1)} *4\\ 4*   3^{(x+1)}=4^{x-1} *4^1\\ 4*   3^{(x+1)}=4^{x-1+1} \\ 4*   3^{(x+1)}=4^{x} \\ 4*   3^{(x+1)}=4^{x} \\\)

Then he took the log of of both sides

\(ln(4*  3^{(x+1)})=ln(4^{x}) \\ ln(4)+ln( 3^{(x+1)})=ln(4^{x}) \\ ln(4)+(x+1)ln( 3)=xln(4) \\ ln4+xln3+ln3=xln4 \\ ln4+ln3=xln4-xln3 \\ ln4+ln3=x(ln4-ln3) \\ ln12=xln(4/3)\\ \frac{ln12}{ln(4/3)}=x\\ x=\frac{ln12}{ln(4/3)}\)

This is the same as admins answer of       -ln12/(ln3-ln4) = 8.6376

I am assuming that A,c and E are collinear.

Use pythagorean theorum :)

\(AE^2=8.5^2+17^2\\\)

AE^2=8.5^2+17^2\\

Here is a nefty trick for finding the square of numbers ending in 5

85^2      

 = 8^2+8    with 25 tacked on the end.

= 64+8    with 25 tacked on the end.

= 72    with 25 tacked on the end.

=7225

so  8.5^2=72.25

17^2=289   (I just know that)      

But I could do it as (10+7)^2 = 100+140+49=289

so

\(AE^2=8.5^2+17^2\\ AE^2=72.25+289\\ AE^2=361.25\\ AE^2=\frac{36125}{100}\\ AE^2=\frac{25*1445}{100}\\ AE^2=\frac{25*5*289}{100}\\ AE=\frac{5*\sqrt5*17}{10}\\ AE=\frac{85\sqrt5}{10}\\ AE=8.5\sqrt5\\\)

That is in  8.5sqrt5    cm of course.

Will has given you a good decimal approximation.  Thanks Will

(3u^2v^3)^3

\((3u^2v^3)^3 =3^3u^6v^9 =27u^6v^9\)

592/74 = 8

Shouldn't that be  ?        

(-1)^(1/2)(592/74)(3.14159265359...) and it was delicious!

Tara had 1 5/8 pounds of dried nuts. She ate 1/4 pound of dried nuts each day for 6 days. How many pounds of dried nuts did Tara have left after 6 days?

(1+5/8)    -   (1/4)*6 =     that is how many  :)

Thanks guest.

You did a good job os explaining why 1 book costs $19

The next step is 95/19 = 5 books