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ne more suggestions????

plzzz post ur comments.....!!!!!

Trevor broke it.......as he was aware of the price increase.......he must have seen the poster on wch Mr.Levine was working on when he broke into the shop the prevoius night......!!!!!

WEBSITES REFERRED....

http://www.thestudentroom.co.uk/wiki/Image:Graphs_Of_Negative_Powers.png

http://www.thestudentroom.co.uk/wiki/Image:Square_Root_Of_X.png

Graphs of negative integer powers

It is now known that the following is true:

$x^{-a} = rac{1}{x^{a}}$

Hence, the graph of a negative power:

$y = x^{-a}$

Is equivalent to the following:

$y = rac{1}{x^{a}}$ ......I THOUGHT U WOULDNT KNOW THIS (just kidding

)

Consider the graph below.

This shows the graphs of y = rac{1}{x}

, and $y = rac{1}{x^{2}}$ .

One can see that both of these graphs have a common point at (1, 1), and also there is a pattern of flattening (it seems) as with the graphs of $y = x^{a}$ .

These graphs do not touch either of the axes, they become infinitesimally close, but never cross them, this is due to the fractional powers of zero.

Consider that the interception with the x-axis occurs only when one has a value of "x" for which "y" is zero. So consider the following:

$y = rac{1}{x^{a}} = 0$

If one is to multiply by $x^{a}$ it produces 1 = 0

which is contradictory, hence there are no values of "x" for which "y" is zero, hence there are no x-axis intercepts.

Consider that for a y-axis intercept, "x" must be zero, hence:

$y = rac{1}{x^{a}} = rac{1}{0^{a}} = rac{1}{0}$

This is undefined, and hence it shows that there is not y-axis intercept.

Not one can consider the flattening pattern. It is rather self evident that as one has a value of above 1, and the power of the function decreases, there will be a flattening; putting this more formally makes it clearer.

$y = x^{-a} = rac{1}{x^{a}}$

Hence, for:

One can predict that as a o infty

, the value of "y" will decrease.

One might also notice that there is a difference in the quadrants which the graphs occupy. It is evident that for:

The value of "y" is always positive (as one is conducting a division of 1 by a positive value, see the previous section for a more detailed explanation).

Conversely, the values of "y" for a graph where:

Can be negative. This results in the graph occupying the same quadrants as the graph of $y = x^{a}$ , for the same constraints on "a".

Differentiation with negative integer indices

There is no special rule for the negative indices (in the context of differentiation). It is merely a case where one can use the general rule for the differentiation of a function of x, of the form:

$f(x) = x^{n}$

One must, however, be careful to ensure that one takes from the power, hence, in the cases that one will discuss in this section of the notes, this will result in no limitation (in contrast to the differentiation of a function whose power is a positive integer; as decrementing a positive integer with a decrementation of 1 will result in a power of 0, and hence a constant value, which differentiates to zero).

Example

1. Calculate the equation to the tangent of the curve y = rac{1}{x}

, at the point where x = 1

.

One will follow the same type of process as one did with the previous examples (previously in these notes) of calculating the equation of the tangent of a curve at a given point.

First one can calculate the differential:

$y = rac{1}{x} = x^{-1}$

$rac{dy}{dx} = -x^{-2} = rac{-1}{x^{2}}$

(It is important that one makes sure one is able to convert between the negative powers, and the reciprocal notation, as this will aid the calculation of
the gradient, numerically speaking).

Now calculate the specific gradient, one can calculate the y-coordinate of the point at this stage also.

Hence one is dealing with the point of contact (1, 1)...........................................!!!!!!!!

Now, find the gradient of the tangent:

$m = rac{-1}{x^{2}} = rac{-1}{1^{2}} = -1$

Hence, the equation is of the form:

Hence, one can now substitute:

Hence, the equation is:

It is a simple process as previously (one could demonstrate the general rule, and how it works through the use of a similar method to that used in the notes upon differentiation, however this is not necessary as it has been shown, using binomial expansion, and other such methods, that the general rule is true for all positive "n"; and one can express these examples in terms of a positive power, in reciprocal form).

Graphs of $y = x^{n}$ for fractional n

Suppose one has the graph:

$y = x^{ rac{1}{n}}$

One can simplify, and rearrange this:

$y = x^{ rac{1}{n}} = sqrt[n]{x}$

Hence:

$y^{n} = x$

This means that if one is to graph the original graph, one can merely "swap" the x-axis with the y-axis (if they are the same scale), and draw a graph of $y = x^{n}$ .

For example, see the graph below....

(Compare the similarity between a graph of $y = x^{2}$ ).

The above graph depicts $y = x^{ rac{1}{2}} = sqrt{x}$ .

These graphs will all pass through the point (1, 1). Also, as one might expect (from the previous transformatory method of depiction, or prediction thereof) those fractions whose denominator and numerator are even will produce positive "y" values, and the odd values will produce both positive, and negative values of "y".

Other than this information, one must be aware that these powers will use the same rules for differentiation as usual.

Example

1. Calculate the equation of the tangent to the curve y = sqrt{x}

at the point (1, 1).

This is a simple task as the point has been given in the question....................!!!!!!!!!

First one must calculate the derivative:

$rac{dy}{dx} = rac{1}{2}x^{ rac{-1}{2}} = rac{1}{2 sqrt{x}}$

Hence, the gradient of the tangent at the point (1, 1), is calculated thus:

m = rac{1}{2 sqrt{x}} = rac{1}{2 imes sqrt{1}} = rac{1}{2}

Hence, the equation is of the form:

Hence:

Now, one can substitute:

Hence the equation is:

its simple....
suppose there r 'x' identical resistors in a row r 'y' such rows....!!!!

hence, equivalent resistance is xR/y.....
therefore,

xR/y = 5
2x = y.....................................(equation 1)

now, each row of x resistors produces an EMF of xR*1 = xR volts....

there r y such cells....

hence,
equivalent EMF = 5*4 = (xyR/xR)/(y/xR)
therefore,

x = 2....!!!!!

y = 4....!!!!! (from equation 1)

hence, there r 8 resistors....2 in each row....!!!!

Actually, area is not considered as a vector......!!!!
A unit vector perpendicular 2 the surface area is the quantity wch is considered....!!!!!

u wrote all that..............!!!!!!!!

MIND-BLOWING

1)H₃PO_{3..........................................}phosphorous acid

2) HClO_{3...........................................}chloric acid

3) H₂SO_{3..........................................}sulfurous acid

4) HBrO_{3..........................................}bromic acid

HClO₄ has the ClO₄¯ polyatomic ion and its name is perchlorate.

Since the "ate" suffix is used, it gets changed to "ic."

The name of HClO₄ is perchloric acid.

HClO₃ has the ClO₃¯ polyatomic ion and its name is chlorate. Any time you know the "ate" ending is used on the polyatomic, you use "ic" when you write the corresponding acid formula.

The name of HClO₃ is chloric acid.

HClO₂ has the ClO₂¯ polyatomic ion in it. The name of this ion is chlorite.

Since the "ite" suffix is used, it gets changed to "ous."

The name of HClO₂ is chlorous acid.

LETS C HOW 2 NAME SOME INORGANIC ACIDS....!!!!

HClO is an acid involving a polyatomic ion. You MUST recognize the polyatomic ion in the formula. There is no other way to figure out the name. If you don't recognize the polyatomic, then you're sunk without a trace.

The polyatomic ion is ClO¯ and its name is hypochlorite. Any time you see the "ite" suffix, you change it to "ous" and add the word acid.

The name of HClO is hypochlorous acid.

Rules for Naming Acids/Writing their Formulas

Rule 1: all acids must contain hydrogen

Rule 2: acids with `ic' suffix represent natural `ate' polyatomic ions

HBrO₃ bromic acid

Rule 3: when all oxygen atoms are removed, add `hydro' prefix to name

HBr hydrobromic acid HCl hydrochloric acid

Rule 4: when an extra oxygen is added, add a `per' prefix to name

HBrO₄ perbromic acid HClO₄ perchloric acid

Rule 5: when 1 oxygen is taken away (from `ate' ion number), change the `ic' suffix to `ous'

HbrO₂ bromous acid HClO₂ chlorous acid

Rule 6: when 2 oxygens are taken away (from `ate' ion number), change the `ic' suffix to `ous' and add a `hypo' prefix

HBrO hypobromous acid HClO hypochlorous acid

When working with inorganic acids, always start with the `ate' polyatomic ion.

example: CO₃ has a 2- charge so 2 hydrogens will be needed:

+1 2-
H₂CO₃ this would be named: carbonic acid

The six most common acids we use are given next. They all end in `ic'

HbrO₃ bromic acid H₃PO₄ phosphoric acid

HClO₃ chloric acid H₂SO₄ sulfuric acid

HNO₃ nitric acid H₂CO₃ carbonic acid

Use these as starting points for the alternative acid (and their names)

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