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We are equal in 2U maths (these types of questions seem hard to me, but doing it doesnt seem too bad)

These questions are super hard! I'll give it a go: So the value of the company's shares are falling each day After monday, shares have fallen 2.7% (0.027 as decimal) After tuesday, shares have fallen 3.09% (0.0309 as decimal) Let A be the value of shares Opening time, the value of the...

thanks guys

Common denominate or multiply alll by 15, then make x the subject

\frac{d}{dx}\left ( xlnx \right ) = lnx + 1 Now integrate both sides with respect to x, \therefore \int \frac{d}{dx}\left ( xlnx \right ) = \int\left ( lnx + 1 \right )dx \therefore \int \frac{d}{dx}\left ( xlnx \right ) = \int lnxdx + \int dx Make integral of lnx the subject...

Re: Chess thread? ok

Re: Chess thread? Still busy Menomaths?

http://community.boredofstudies.org/showthread.php?t=311248 Read post #13, if you will do not understand

Identify which is which The very top one goes down 3 units from 0 to T1 The middle one down 6 units from 0 to T1 The very bottom one goes up 3 units from 0 to T1 Therefore since we can conclude that CO is the TOP one, H2 is the middle one and CH3OH is the bottom one At T1, u get it so i...

y = \frac{1}{4}lnx + \frac{1}{2} x-intercept occurs when y = 0, So we let y = 0, \therefore 0 = \frac{1}{4}lnx + \frac{1}{2} Make 'lnx' the subject and put it on the left so its easier to see, lnx = -2 \therefore x = e^{-2} (THIS PART, changing logs to exponential, DO U UNDERSTAND?)

No problem!

the reason why u have 'C' is that the original function could have a constant added to it eg. y = x+1, y' = 1 If u integrate back, u get: y = x + C eg. y = x, y' = 1 If u integrate back, u get: y = x + C C is determined by a relation between x and y Another example, y' = x^2 /2 Integrate...

cool!

Going from 25 - y^2 to (5-y)(5+y) is using difference of two squares (just to let the OP know)

y' = \frac{dy}{dx } \\ \frac{dy}{dx} = \frac{1}{4x} = \frac{1}{4}\cdot \frac{1}{x} dy = \frac{1}{4}\cdot \frac{1}{x}dx \int dy = \int \frac{1}{4}\cdot \frac{1}{x}dx \int dy = \frac{1}{4}\int \frac{1}{x}dx \therefore y = \frac{1}{4}lnx + C When y = 1, x = e2 (is this e squared? im gonna...

thanks

When comparing the reactivity of oxygen and ozone, do i compare BOND ENERGIES?

I hate the library

Yeh!

re: HSC Chemistry Marathon Archive nvm