CPhill

Let me see if I can help you with this...here's what Alan is saying...

Let's "guess" that the "zero" of the function is "4.005"

The "method," in simple terms, is just this:

(Our guess) - (Our Guess put into the function)/ (Our guess put into the derivative) = (The next "guess")

So we have

(4.005)- [4.005cos(4.005) - (1/4.005)]/[ cos(4.005)-(4.005)sin(4.005) + (1/(4.005)²] = (The next "guess")

Now.......if the "next guess" happens to be 4.005, we have found the "zero." If not, we put this "next guess" into the "formula" and evaluate that.  Notice that, when Alan put 4.005 in, he got back 5.16612. That's not equal to 4.055, so we put 5.16612 in the mill and it cranks out 4.76159. Still no "match" from the previous "guess".....

So he puts 4.76159 into the mill and gets back 4.75661...still no match.....then, he puts 4.75661 into the "formula" and gets back 4.7566....AHA!!......we're on to something!! Notice that this "new" answer is really close to the last one!! 

Just to be sure, he puts this "new" answer ("guess") into the formula once more and gets out (wait for it)........4.7566.....note this matches the last "answer," so we've found the "zero.'

See how that works??

Just put your calculator in radian mode and watch your math!!

(Sometimes - particularly if I have a graph of the function - I find it easier just to evaluate the function at various values just to see if I can find something "close" to zero. This is sometimes less cumbersome than Newton's Method - if you can make some good "guesses.")

Hope i've been of some help!!

I saw that I would have to use -1 and -2 to get the +2 (since the middle term was negative).  And then I realized that -2*1 and -1*3 would give me the -5 that I needed.

We don't necessarily have to use the quadratic formula - particulaly if we can factor a quadratic equation. In fact, I always see if I can factor first, before resorting to it.

That's an interesting tecnique, Alan and Bertie......

I'm with Alan...........a calculator seems vital......

Did you mean (64/125)^(-2/3) ??

If so, we have  

(125/64)^(2/3) =

[(125/64)^(1/3)]^(2) =

(5/4)^(2) =

(25/16)

This factors as

(3x-2) (x-1)

If you're tryig to find the "zeroes," just set each factor to 0. You will obtain x = 2/3 and x = 1.

Huh???..........

OK.....2083.33 years per hour.......this answer is as non-sensical as your question....!!!!

V_c =  pi*r^2 * h

V_c = pi * (1^2) * 2 =  (2 * pi) =

about....... 6.28m^3

There's no way to tell unless we know the remaining side.....To see this....note that "B" could act as a "hinge" so that AB and BC might be two sides of an infinite number of possible angles. (Unless these are two legs of a right triangle......if this is so....repost your question  and I'll give you an answer)

I assune you want to find the vertex for this and also want to know if this is a minimum or maximum.

The x coordinate of the vertex is given by -b/(2a), where b = 5 and a =1.

So we have -5/(2(1)) = -5/2 = -2.5.

And "plugging" this back into the function, we find that the y coordinate of the vertex =

(-2.5)^2 + 5(-2.5) + 6 = -.25

Now, if you haven't had calculus yet, just trust me that something called the "second derivative" can be used to determine if the vertex is a minimum or a maximum.

If the second derivative > 0, we have a minimum.....and in this case, the second derivative = 2......so (-2.5, -.25) is a minimum......but....if you don't believe me look at the graph of the function  below......

The vertex is at (-2.5, -.25) and it is indeed a minimum .... just as we expected !!

100000=P(1+(0.08/12))^(12*40)

So...I'm assuming we're looking for some principal invested at 8% and compounded monthly for 40 years which accumulated to $100000??

So .....

P = 100000/(1+(0.08/12))^(12*40) =

$${\frac{{\mathtt{100\,000}}}{{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{0.08}}}{{\mathtt{12}}}}\right)\right)}^{\left({\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{40}}\right)}}} = {\mathtt{4\,119.738\: \!461\: \!084\: \!481\: \!01}}$$

So....the principal was about $4119.74