Kreyn

3x + 48 = 3x + 3*16

3x + 3*16 = 3(x+16)

20/7*14/15 = 5*2*2/7 * 7*2/(5*3)

2*2*2/3 = 8/3

The law of sines can be summerized pretty well in this picture I found on web (source: www.mathwarehouse.com). If you know an angle and its opposite length, you can solve for every other angle and length for that triangle. 

Ammonia is NH₃ 

The molar mass (reading from the periodic table) is 17.031g/mol

Therefore 3.4 grams of ammonia is equal to 0.1996359579590159 moles of ammonia.

Multiplying this by 6.022 * 10²³ we get 120220773882919374980000 molecules (or 1.2022 * 10²³ molecules).

In 1 molecule of ammonia there are 4 atoms, so in total there are approx. 4.8088 * 10²³ atoms.

The measurement was made using 2 significant figures, so we can only be sure about this result to 2 significant figures, therefore the amount of atoms in 3.4g of ammonia is approx. 4.8 * 10²³ atoms.

The answer is 20.4m

3.4 * 10^9

Thanks,

Don't have a  teacher, just teaching myself haha.

I am 3/5 through the calculus book, nearing integral calculus (it is the next topic). It has started to give me minima/maxima questions (to solve using 2nd derivatives), and trig functions have started popping up (probably reintroducing for integrals).

I'll practice deriving these identities to become more familiar with them, and when solving these problems I'll derive each one instead of memorizing to hopefully become more familiar with them.

Ah. So are the 'double' and 'half' angle identities both derived from the angle addition/difference identities?

So the most important identities to remember would be the sum/difference identities.

sin(a + b) = sin(a)cos(b) + cos(a)sin(b) 
sin(a – b) = sin(a)cos(b) – cos(a)sin(b) 
cos(a + b) = cos(a)cos(b) – sin(a)sin(b) 
cos(a – b) = cos(a)cos(b) + sin(a)sin(b)

As well as:

sin^2(x) + cos^2(x) = 1 (the other identities are easily derived from this).

So most functions with some trig function can be solved using these 2 sets of identities?

This function popped up towards the end of my derivatives chapter, and the book on trig barely covered those identities at all! :( (it mentioned the sin(a+b) identity but never used it).

Thanks!

1° = pi/180 radians

So

50° = 50*pi/180 radians

90° = 90*pi/180 radians = pi/2

180° = 180*pi/180 radians = pi

360° = 360*pi/180 radians = 2pi

etc.

Take the log of base 2 of each side.

2x = log2 15

2x = 3.90...

x = 3.9/2 = 1.9534...

Hmm looks like I was originally correct, just made some calculation errors...

This is my working, I'm sure there is a more efficient way to do it however...

\(\frac{x^2}{16}+\frac{(5-x)^2}{9}=1\)

\(\frac{9x^2+16(5-x)^2}{144}=1\)

\(9x^2+16(5-x)^2=144\)

\(9x^2+16(25-10x+x^2)=144\)

\(9x^2+400-160x+16x^2=144\)

\(25x^2-160x=-256\)

\(25x^2-160x+256=0\)

And then just solve for x using the quadratic formula...

\(x=\frac{160+-\sqrt{25600-25600}}{50}\)

x = 3.2

y = 1.8

Thus point of intersection = (3.2,1.8)

The answer in the book however is (doesn't show any working):

(16/5, 9/5) which of course is the same answer, but there must be a more efficient way to get to this answer if the answer is a fraction.

Is 1/x+1/1-x = 1/x-x^2 ?

Nope, 1/x+1/1-x = (1-x)/(x+1)

Is 1+ 1/x+1 = 2/x+1 ?

Nope, 1 + 1/(x+1) = (x+1)/(x+1) + 1/(x+1) = (x+2)/(x+1)

Is x+ 1/x = x^2+1/x ?

Yes, x + 1/x = x^2/x + 1/x = (x^2+1)/x

Is 1/x-1 - 1/x+1 = 2x/1-x^2 ?

Nope, 1/(x-1) - 1/(x+1) = (x+1)/(x-1) - (x-1)/(x+1) = ((x+1)-(x-1))/(x^2-1) = 2/(x^2-1)

A helpful way to check your answer is to graph the equation (that way you can instantly visually see if they equal each other). 

Thanks, yeah I think i was just overcomplicating it trying to think of some rule that would always work relating the points involved, but in the end it's just equating the 2 equations lol.

\((x+ay)(x+ay)=(x+ay)^2\)

And remember that:

\((a+b)^2 = a^2+2ab+b^2\)

So using this rule, we can just substitute a = x and ay = b

\(x^2+2xay+ay^2\)

You can use the rule to skip the process of using the distributive rule :)

Oh okay, so multiply equation b by 3 and add 9 to each side!

\(y-3=-\frac{4}{3}(x-2)\)

\(3(y-3)=3(-\frac{4}{3}(x-2))\)

\(3y-9=-4x+8\)

\(3y=-4x+17\)

Is there is a rule I can use for different cases. For example If I did not know equation (c), what would I start with to try to equalize the 2 equations? I see that 3 is the y coordinate of the first point. Would you try to multiply each equation by one of the point coordinates and then see which 2 best fits? Does each equation have to be multiplied by the same variable?

Thanks heaps!

Thanks heaps!

Sorry I didn't mean to be rude, just wasn't sure if you would see a comment that I posted on the other thread :D.

I definately appreciate your help!

I'll make an account for next time I have a problem :p.