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Uh, OK.....I'll take your word for it.....Is there a question there??

Scientific Notation is a style of equation where a number that is too big to be practical (i.e. a number that has many zeros after it) is shifted around until it is between 1 and 10.

For example, a number like 986000000 would be 9.86*10^8 in scientific notation. Scientific notation is recognizable by the (term)*10^n, where 'n' is however many places the decimal moves.

Hi Bioschip,

It is a way of writing very big or very small numbers in a way that is easy to read and interprete (when you get used to it)

for instance

5678=5.678x1000 = 5.678 x 10³

this might not seem very helpful but what about

12,345,732,000,000,000,000,000,000

this equals 1.2345732 x 10²⁵   which is approximately 1.2 X 10²⁵      

FOR ME THIS IS MUCH EASIER TO READ.  I do not have to count all the zeros.

( I have not looked at little numbers but I can if you want me to)

Basically, it's just a shorthand notation. It comes in handy for expressing extremely "large" or "small" numbers.

For instance, suppose we had 9,100,000,000,000

Instead of having to write this "long" number out, we can write it as......  9.1 x 10¹²

The rule is to move the decimal point to the right of the first non-zero digit and count how many places we moved it - in this case, 12 - and that's the exponent that goes on the "10." Moving the decimal point to the left results in a positive exponent. Moving it to the right results in a negative exponent, as in the following example:

.0000000000002345 = 2.345 x 10^-13

It also comes in handy for multiplying/dividing large or small numbers, in certain situations.

This article explains it more thoroughly than I can:

2 + 2 = 2 * 2 = 2^2 = sqrt(16)

yes - of course - thanks Chris.

(I took away a negative number - and hence added)

this would not have happened if i had just found 

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{18.434\: \!948\: \!822\: \!922^{\circ}}}$$ in th first place and worked from there.

The quadrants are correct but the angles should be 161.565051177078° and 341.565051177078° since the angles lie in those quadrants.

What do you think it should be Chris?

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = -{\mathtt{18.434\: \!948\: \!822\: \!922^{\circ}}}$$

NOTE:  inverse tan is the same as arc tan - they are different notations meaning exactly the same thing.

$$\mathrm{\ }$$

This is the answer in the 4th quadrant.

Now Tan is negative in the 2nd and 4th quadrant

2nd quad would be 180-18.43....=

$${\mathtt{180}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{198.434\: \!948\: \!822\: \!922}}$$   

WRONG

this is wrong - see my later post should be

$${\mathtt{180}}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{161.565\: \!051\: \!177\: \!078}}$$

---------------------------------------------------------

4th quadrant (also)

$${\mathtt{360}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{378.434\: \!948\: \!822\: \!922}}$$     

WRONG - should be

$${\mathtt{360}}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{341.565\: \!051\: \!177\: \!078}}$$

So to the nearest degree    $$\theta = 162^0\:\: or\:\:342^0 \mbox{ where }0^0\le\theta\le 360^0$$

NOTE: It may have been easier for you to just found  $$tan^{-1}(1/3)$$   and then worked it out for the required quadrants afterwards

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annoymous,

I know it is very confusing (I really dislike this notation-in my oppinion the atan notation is much nicer) but

 $$tan^{-1}(ratio)$$   DOES NOT MEAN  $$\frac{1}{tan(ratio)}$$

To do this place the 30 over 1.

-2/5 * 30/1  Then mulitiply straight across top and bottom.

-60/5  Divide.

-12 is your answer.

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {cot}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{theta}}$$

Make sure you set your calculator to radians or degrees depending on what your problem states and solve

On the on-site calculator, hit the "2nd" key and the "atan" button. Make sure you specify degrees or radians (found at the bottom left). "Atan" stands for "arctangent," which is the same thing as the "tangent inverse.'

He will pay a $70 administrative fee for both items and a $35 delivery charge for both items = $105.  Since you didn't specify a tax rate, we can't calculate the taxes due. All we can say definitively is that he will pay at least $105 at the time of purchase.

$${\mathtt{1}}{\mathtt{\,-\,}}{\left({\frac{{{\mathtt{7}}}^{{\mathtt{1}}}}{{\mathtt{10}}}}\right)}^{{\mathtt{50}}} = {\mathtt{0.999\: \!999\: \!982\: \!015\: \!349\: \!6}}$$

Or did you mean this???

$${\mathtt{1}}{\mathtt{\,-\,}}{\left({{\mathtt{7}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{10}}}}\right)}\right)}^{{\mathtt{50}}} = -{\mathtt{16\,806.000\: \!000\: \!000\: \!000\: \!000\: \!9}}$$

6t² = 3    Divide both sides by 6

t² = (1/2)      Take the square root of both sides

t = ±√(1/2)    (If you want to "rationalize" the denominator, we get  ±[√(2)] / 2 .

Remember we have to take both roots when applying the square root property in solving equations !!

OK...let's see if it's "factorable.'

(3x +1) (x + 5)    Nope !!

(3x +5) (x +1)     Nope !!

These are the only possibilities we have, because 3 and 5 are prime.

Thus, the expression isn't "factorable."

A famous question in Zen practice........it's the dichotomy of "two and one". We understand "duality." What then, is "non-duality??"

I'll answer your question if you will answer mine......

In a certain village, every man who doesn't shave himself is shaved by the barber.

The question is.....Who shaves the barber??

For, if he doesn't shave himself, then he is shaved by himself. And, if he does shave himself, then he is not shaved by himself.......

I've thought about this problem for several days. Here's my take on the answer.

As Rom pointed out, the interval [-(7/2), (1/4)] does indeed make the "log" in the numerator undefined.

But I believe the answer to the problem lies in the denominator. In effect, we have the domain of an "inside" function -  (4x-1) - to consider, and the domain of an "outside" function - log(4x-1) - to consider. Note that any values between (1/4) and (1/2) are OK - they just make the denominator negative. But also notice that any values less than (1/4) aren't defined for log(4x-1). And x=(1/2) isn't defined, either.

So, I believe the correct answer to this problem are the two intervals (1/4, 1/2) and (1/2, ∞)

Anybody have another take??

Your (linear) equation has the form y = mx + b, where "m" is the slope and "b" is the y intercept.

So m = -2

And that's the slope.

@@ End of Day Wrap - Fri 9/5/14   Sydney, Australia Time 03:30 (It is really Saturday morning)

I didn't do a wrap last night because access to the forum was denied to me.  I was denied access for 24 hours there abouts.  

Andre Massow, site owner and developer, has now found the cause and he sent me this message an hour or so ago.

"The problem is fixed. The reason for the error message was a programming mistake in my watchlist implementation. In your case the error appeared because you had a question in your watchlist which was delete later (by the question creator). The watchlist could not find the deleted question and crashed. "
Andre has temporarily deactivated this feature.  I couldn't get the sticky topics to open either. Maybe he has deactivated these as well.

Thanks you Andre for finding and overcoming this problem for me.  

There have been a lot of questions in the last 2 days and the following members provided great answers; Alan, reinout-g, CPhill, GoldenLeaf, DavidQD, FoxTears, Kitty<3, tfrancis15, all-them-marvel-feels and many anonymous people.  Thank you all.  Your efforts are greatly appreciated!

I do have more to say but the thread/s I wanted to share are 'locked' in a sticky topic and it is very late.  3:30am.  I have a little more I want to do on the forum before I 'hit the sack', so that is enough for this wrap.

Thank you all for making this forum such a success.

Melody.

melodymathforum@gmail.com

Thanks Chris - as Chris has said

It is the line and everything above the line

Without knowing what values  x, y and z  take on, there's nothing to "evaluate."

This question makes no sense