ChowMein

yep my bad, edited the answer

the intermediate calculations shouldn't be rounded, i just ended it at 1120.59 because i don't want to make it too long, round at the end.

the answer is limited to 3 sig-figs, so the final solution is rounded to 3 sig-figs, not the intermediate calculations

and mb, should be 1.34 mol

For number 2, the answer is limited to 3 sig-figs because of the 2.50g of H₂.

For number 3, the answer is limited to the hundredths place due to the 1.34 mol of CO.

You usually don't want to be limited sig-fig wise when it comes to the molar masses. So, it's best to have the most precise molar mass possible.

$\text{Find theoretical yield}\\ 2*10^3g\;CaCO_3*\frac{1mol\;CaCO_3}{100.09g\;CaCO_3}*\frac{1mol\;CaO}{1mol\;CaCO_3}*\frac{56.08g\;CaO}{1mol\;CaO}=1120.59g\;CaO\\ Percent\;Yield=\frac{Actual\;Yield}{Theoretical\;Yield}*100\%=\frac{1.05*10^3g\;CaO}{1120.59g\;CaO}*100\%=93.7\%$

1)
$\text{Find amounts of each reactant in moles}\\ 2.50g\;H_2*\frac{1mol\;H_2}{2.016g\;H_2}=1.24\;mol\;H_2\\ 30.0L\;CO_2*\frac{1mol\;CO}{22.4L\;CO}=1.34\;mol\;CO\\ \text{Find amount of H}_2\text{ that fully reacts with CO or vice versa}\\ 1.34mol\;CO*\frac{2mol\;H_2}{1mol\;CO}*\frac{2.016g\;H_2}{1mol\;H_2}=5.36\;g\;H_2\\ \text{5.36g of hydrogen is needed to fully react with 30.0L of CO, making hydrogen the limiting reagent}$

2)
$2.50g\;H_2*\frac{1mol\;H_2}{2.016g\;H_2}*\frac{1mol\;CH_3OH}{2mol\;H_2}*\frac{32.042g\;CH_3OH}{1mol\;CH_3OH}=19.9g\;CH_3OH$

3)
$2.50g\;H_2*\frac{1mol\;H_2}{2.016g\;H_2}*\frac{1mol\;CO}{2mol\;H_2}=0.620\;mol\;CO\\ 1.34mol\;CO-0.620mol\;CO=0.72mol\;CO$

1) $d={v_0t}+\frac{1}{2}at^2\\ a_{rocket}=\frac{2d}{t^2}=\frac{2*12.4m}{(2s)^2}=6.2\;m/s^2$

2)$a=\frac{\Delta v}{\Delta t}=\frac{32\;km/h\;-\;2.8\;km/h}{1.5s}=\frac{29.2\;km/h}{1.5s}\\ \text{Use dimensional analysis to convert to m/s}^2\\ \frac{29.2\;km/h}{1.5s}*\frac{1h}{3600s}*\frac{1000m}{1km}\approx5.41\;m/s^2$ 

3) $d=v_0t+\frac{1}{2}at^2\\ a=\frac{2d}{t^2}=\frac{2*18.3m}{(2.74s)^2}\approx4.88\;m/s^2$

4)$v=v_0+at\\ a=\frac{v-v_0}{t}=\frac{46.7m/s\;-\;2.3m/s}{7s}\approx6.34\;m/s^2$

5)$v=v_0+at\\ a=\frac{v-v_0}{t}=\frac{0-6.23m/s}{82s}\approx-0.076\;m/s^2$

if its the whole right side divided by 10 then:

so to sum it up:
1) solve the inequality as if it's a regular equation
2) find the two "boundary" points (what x equals to)
3) [optional] draw a number line and put those points on the line
4) test the values created by the 3 sections (one smaller than the value of the lower boundary, one in between both, one larger than the value of the upper boundary) 
5) if the test passes, then the section containing that value is a part of the solution set
6) count the integers

i guess you can think of it in the form:
ax²+bx+c instead of ax²+bxy+c²

then, factor the resulting quadratic using your methods

yep thats the slope intercept

pretty sure point slope is in the form of
y-y₁=m(x-x₁)
x₁ and y₁ is any point on that line so we can use point (-4,-7) and plug the slope into m to get:

y-(-4)=-6(x-(-7))
y+4=-6(x+7)
is one possible answer