kingjewel1

Lovely!! How do you come about the formula to use? I'm still not sure how to find c.

True. Lets say that a=1.

Ahhh my fault! sorry wrote the wrong function. y=x+1 excuse me

The functions only intersect twice.

But i can't factorise because i don't know where it cuts the x axis. Maybe this helps: the curve passes through where y=6/x and y=x+1meet. I set them equal that's where i found the points A and B. Edit function is y=x+1

ax^2+b^x+c is the standard function of a quadratic. If i know that a quadratic with this format passes though points A(-3,-2) and B(2,3). How would i go about working out a, b and c? This's got me stumped. Thanks

I take your point "The Rebel". I actually have the functions drawn out infront of me to make sure they do intersect where i thought. Like this: functions <click>. I calculated the area in two stages in the end. Thanks guys for all your help

yeah sure you could do that. it's just that i wanted to do it all in one go by joining the functions. just like eg area between x^3 and x^2 in the positive quadrant X= 0 and 1 intx^3-x^2dx betweent 0 and 1 =x^4/4-x^3/3 sub in 1 = 1/12 Just the same as doing intx^3dx - intx^2dx between 0 and 1 why doesn't this method work for the other functions? this is buggin me

i have examples where i've equalled two intersecting functions and then integrated the result... i'll see if i can dig some out. I was wanting to find the total area bounded by the quartic and the quadratic. I integrated between 0 and 0.9 because x=0.9 is where the two functions intersect in the positive quadrant. somebody mentioned integrating with respect to y but then i'm not sure how to do that. making y the subject from this guy or would i not just need to int(1/(1/x^4-x^4)) . y=1/x^4-x^4-x^2 how do i make y subject here? excuse my ignorance

thanks for your help

you see, i've been setting the functions equal to each other thus 1/x^4-x^4=x^2 and integrating this with respect to x between 0 and 0.9 which is it's positive intersecting point. why can't i do this?

i'm trying to work out the area between 1/x^4-x^4 and x^2 and the x axis and the y axis 1/x^4-x^4 and x^2 can you give me a hand please? i've integrated between 0 and 0.9 but it gives me a negative area. :S what am i doing wrong? eventually i wanted to work out the area between -x^2 and the quartic. -x^2 aren't these curves beautiful

thanks. i could see the -1 as being factor but i thought there might be a way to figure out the other one(s): relatively easily i've looked on the net but i don't seem to be coming up with any answer. :S

how do i solve this guy? thanks

how do i work out that if a machine is worth $15000 and if it's price decreases by 20% per annum that it will cost $9600 in two years using Sn=a(r^n-1)/r-1? using this formula it doesn't work.... thanks guys

1.a potential problem with human cloning is when the technology falls into say... bin laden's hands and he makes a super human race. 2. religious ethics say that God created every living cell: cloning goes against that grain for some people. 3.what's the point of human cloning? "there are enough of us as it is": say others 4.clones could become dispensible: where they are just used for a purpose and then disgarded. pick a reason

When does 1 and 1 make 3? Happy New Year

thanks Matt

LOL yup my examiners appear to have certain retention so..... the problem is that i've seen all three answers from 3 different sources. i mean you might aswell use the right answer if that's what they're looking for... would c) be good enough though?

how do i know when to incorporate the Kw constant into the calculation of the pH of an acid? eg. i have 1x10^-8 mol dm^-3 HNO3. a ridiculously small amount of acid in water... this means that the H30+ from the dissociation of the water actually is greater than this acid: 1x10^-7. my question is how do i know not to incorporate it into an strong acid equation.

hi there, i was hopeful someone could clear this up for me. if i have a solution of H2SO4(aq) and i want to calculate its pH would i count the deprotonations of both H+ ions as equal? eg. if the concentration of my solution is 0.3mol dm-3 what is my pH? is it? a) -lg2[0.3]=0.22 or b) -lg[0.3]=0.52 the thing is the 2 proton is obscured by the first so to say. in a way 1:10 concentration. so... would it be... c) -lg[0.33]=0.48? i like the last one but i've seen both the first two before tell us what you think thanks

at 1000K the equilibrium constant for H20(g)+C(s)<=>H2(g)+CO(g) is 3.72atm. If total pressure is 25atm what would the partial pressures be? Could someone tell me if i've got the right answer in: 69% reacted. therefore Partial pressures (Pg): Pg(H20)=3.875 Pg(H2)=8.625 Pg(CO)=8.625 i used x=reacted H20 1-x ; C 1 ; H2 x; CO x; total=3 Xg= 1-x/2; unity; x/2; x/2; for the compounds respectively. i'm not sure how to treat carbon as it is solid. do i count it as 1 when calculating the equilibrium total moles? thanks in advance. my Alevel is in a week n i'm not sure of this one

hi there. would you mind helping me a lil with this one. it's sinx=1/2+cosx i think its really easy to solve but i can never get a positive value for it. i tried substituting Cos2x=1-2sin^2x also tried it it with sec^2x=1+tan^2x i know there's 1 positive and 1 negative value at least. between 0 and 2pi thanks in advance

if you think of a circle as a polygon with infinite sides, it will have infinite triangles-2. so if you use the formula to calculate the are of a triangle n n=number of sides A=1/2abSIN© you can substitute C for (n-2)pi/n if you then multiply the area of 1 triangle by infinity-2 you will have pi or close enough. try it out with 99999 sides

Hi there, i see pi as being the area of a circle with radius 1unit. considering a circle is a equilateral polygon with infinite sides: ergo area of polygon= (∞-2).1/2.sin(∞-2).pi/∞ As has been stated only the boundaries can be worked out see if it helps **Happy Holidays Guys!!