ChowMein

draw a triangle to find sin(x) and sin(y)

(little drawing thing)

sin(x) = 4/5, sin(y) = 3/5 from the picture

cos(x-y) = cos(x)cos(y) + sin(x)sin(y)
= (3/5)(4/5) + (4/5)(3/5) = 12/25 + 12/25 = 24/25

S₁₀ = S₉ + a₁₀

a₁₀ = S₁₀ - S₉

S₁₀ = 10(11)(12) = 1320

S₉ = 9(10)(11) = 990

a₁₀ = 1320 - 990 = 330

\(1)\frac{7}{4n}-\frac{2n+4}{15n-3}=\frac{7}{4n}-\frac{2n+4}{3(5n-1)}\\ 2)\frac{21(5n-1)}{12n(5n-1)}-\frac{8n^2+16n}{12n(5n-1)}\\ 3)\frac{105n-21-16n-8n^2}{12n(5n-1)}\\ 4)\frac{89n-21-8n^2}{12n(5n-1)} \)
further explanations:
step 1 --> i just factored out the 3 from the right expression's denominator to make it easier to work with

step 2 --> trying to get both expressions to have common denominators, i multiply the left expression's numerator and denominator by 3(5n-1); i also multiply the right expression's numberator and denominator by 4n

step 3 --> i subtract the numerator of the left and right expressions since the denominators are now the same

step 4 --> simplify and get choice #3

a)
initial velocity in the y direction = v₀sinθ = 70sin(30°)=35m/s
bc of symmetry and parabola stuff, final velocity in the y direction = -35m/s

use v=v0+at, solve for t ---> t = (v-v0)/a = (-35m/s-35m/s)/(-9.8m/s²) = 7.14286s = 7.14s

b)

initial velocity in the x direction = v₀cosθ = 70cos(30°)=60.62178m/s

you found the time it takes to hit the ground in part a, so use that in this equation:
d = vt, where d = distance/displacement, 
d = 60.62178m/s * 7.14286s = 433.01ft

1)
\(\omega = 7rad/s\\ r=15cm\\ v_{tan}=\omega r\\ v_{tan}= 7rad/s*(15cm)=105cm/s*(60s/1min)=6300cm/min\)
3)
\(\Delta \theta=\frac{3\pi}{4}rads\\ t=2s\\ \omega = \frac{\Delta \theta}{t}=\frac{\frac{3\pi}{4}rads}{2s}=\frac{3\pi}{8}rad/s\)

basically all of them work as they can be converted to each other. it just depends on what the question asks for.

0.72mol, 16.1L, 20.1g all work and can be converted to and from each other

If its asking for it in grams, then we can extend the dimensional anaysis-ing to find how much CO is left over in grams. This is also equal to the amount of moles/liters of CO left over. They should all be correct as they are all equivalent amounts.
\(2.50g\;H_2*\frac{1mol\;H_2}{2.016g\;H_2}*\frac{1mol\;CO}{2mol\;H_2}*\frac{28.01g\;CO}{1mol\;CO}=17.37g\;CO\;consumed\\ 30.0L\;CO*\frac{1mol\;CO}{22.4L\;CO}*\frac{28.01g\;CO}{1mol\;CO}=37.5g\;CO\;originally\\ 37.5g-17.37g=20.1g\;CO\;left\;over\)

Try multiplying by 2e3 instead of 2

You can extend the dimensional anaysis-ing to find how much CO is left over in liters as well. This will be equal to the amount of moles of CO left over. 

\(2.50g\;H_2*\frac{1mol\;H_2}{2.016g\;H_2}*\frac{1mol\;CO}{2mol\;H_2}*\frac{22.4L\;CO}{1mol\;CO}=13.9L\;CO\;consumed\\ 30L\;CO-13.9L\;CO=16.1L\;CO\; remaining\)

yep, mb, sorry for these careless mistakes.
1.34*2*2.016=5.40g H2.

Hydrogen remains the limiting reagent.