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It doesnt make XOHY a rectangle, but i had showed that XOHY was rectangular here: The only reason why i wrote $We have shown that XOHY is rectangular,$ was because i wanted to reiterate what i said as a reason for the next line \therefore OX = HY\,\,$(opp. sides of rectangle are equal)$...

Thanks. I dont have any questions... but yes it would be awesome if you could post another ;)

P(x) = ax^5+bx^4+cx^3+dx^2+ex+f P(-x) = -ax^5+bx^4+-cx^3+dx^2 -ex+f $For even functions,$\,\,P(x)=P(-x) $ie:$ ax^5+bx^4+cx^3+dx^2+ex+f \equiv -ax^5+bx^4+-cx^3+dx^2 -ex+f \therefore ax^5 \equiv -ax^5 \therefore cx^3 \equiv -cx^3 \therefore ex \equiv -ex ax^5 \equiv -ax^5\,\,$if and...

Here is the diagram for the above question:

(a) $Join the three centres of the circles to form an equilateral triangle with side length '2a'.$ $Using the formula for the area of a triangle:$ A = \frac{1}{2}\cdot b \cdot h = \frac{1}{2} \cdot 2a \cdot 2a = 2a^2 $There are two methods to proceed from here, one uses year 12...

Um, first you have a fraction, that is [3x+4]/[2rt(x+2)], then you use the product rule, where you are supposed to use the quotient rule.

$let$\,\, u = x\,\, $and$\,\, v = (x+2)^{\frac{1}{2}}\,\, $Then:$ u' = 1 v' = 1 \cdot \frac{1}{2} \cdot (x+2)^{- \frac{1}{2} } = \frac{1}{2} \cdot(x+2)^{- \frac{1}{2} } \frac{d}{dx}\left( x\sqrt{x + 2} \right) = u \cdot v' + v \cdot u' = x \cdot \frac{1}{2} \cdot(x+2)^{- \frac{1}{2} }...

You have to be more specific, for example, a programmer, a hardware engineer, a software engineer, a graphics designer...

$This question works all on the fact that the angle at the centre of a circle is twice the angle at the circumference and also that the angle sum of a quad is 360 degrees. Can you figure it out now?$

$First we have to understand what the diagram looks like, the key word to understand this is 'produced at X'. This means that the two chords do not intersect inside the circle, but outside! So now you just draw the diagram with this information.$ $Construct the lines AO, BO, DO, CO and BD$...

\angle OXP = \angle HYP = 90^{\circ}\,\,$(OX$\,\,\perp\,\,$AB and\,\,$HY$\perp\,\,$AB)$ \therefore XO \parallel YH\,\,$(co-interior angles are supplementary)$ $Furthermore,$ \angle XOH = \angle YHO = 90^{\circ}\,\,(XY \parallel OH,\,\,$co-interior angles are supp.)$ $As all angles are right...

Doing even general maths can be of uttermost importance, it opens many many more doors. Doing maths will increase her thinking, her logical processing and will aid her tremendously with arithmetic. If she ever has a change of heart, then at least she will have a multitude of many more options.

$Let this point be (x,y)$ $Using the distance formula$ d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2} \sqrt{(x - 4)^2 + (y - y)^2} = \sqrt{(x - x)^2 + (y - 1)^2} (x - 4)^2 + (y - y)^2 = (x - x)^2 + (y - 1)^2 (x-4)^2 = (y-1)^2 (x-4)^2 - (y-1)^2 = 0 (x-4 + y -1)(x-4-y+1) = 0 x+y-4+1 =...

maths in maths is REAL maths :haha:

I also got the same answer as you, the textbook must be wrong. The answer they gave is wrong as it doesnt satisfy f(-1) = 0

I was more aiming this thread for year 10 students picking their year 11 subjects.

This thread is all about senior subject selections! What subjects will you chose for next year?

More please :P

(This is the diagram) $Name their intersection Z \angle AZO = \angle BZO = 90^{\circ}\,\,$(ACDB \perp OH)$ AO = BO\,\,$(equal radii in circle O)$ OZ\,\,$is common$ \therefore\,\,$By RHS$, \triangle AZO \cong \triangle BZO \therefore AZ = BZ\,\,$(corresponding sides in congruent...

This is an example of a transcendental equation. The best method, and the method which you would be required to use in the HSC, is to graph the two functions, then estimate their values from the intersection of the graph.