Rom

I know nothing of the class you're taking and just reading through these it seems pretty clear that a, b, and d are disadvantages.  So I guess (c) is the answer.

I suppose though I don't see any reason to use "n".

The word row is more descriptive of what's going on.

$\dfrac{4x^2-4}{x^2+4x+3} = \dfrac{4(x^2-1)}{(x+3)(x+1)}=\dfrac{4(x+1)(x-1)}{(x+3)(x+1)}$

True

True

This is a bit tricky.  Clearly there are two terms in both the numerator and denominator as written but we should cancel out the (x+1) term in both the numerator and denominator and obtain

$\dfrac{4(x-1)}{x+3}$

This has only a single term in both the numerator and denominator.

The answer your teacher wants to see depends on them.

$nseats(row) = 12 + 4(row-1)$

1) Do you mean gravitational potential energy near the Earth?  If so the two factors are the objects mass and it's height above the Earth.

2) Increase an object's kinetic energy by increasing it's velocity or it's mass or both.

$2(x-4)^2 + 8 = 2(x^2 - 8x+16)+8 = 2x^2-16x+32+8 = 2x^2-16x+40 \\ a=2,~b=-16,~c=40\\ 3ax^2+3bx+3c = 6x^2-48x+120\\ 6(x^2-8x+20) = 6((x-4)^2+4) = 6(x-4)^2 + 24 \\ h=4\\ \text{Having gone through all that it's pretty clear to see that this should be so }\\ \text{since multiplying all the constants by 3 just has the effect of scaling the parabola}\\ \text{it's location remains unchanged. The x coordinate of the vertex of the original parabola}\\ \text{was 4, and thus we expect that the scaled parabola's vertex also has an x coordinate of 4}$

$y^8-17y^4+16=0\\ (y^4-16)(y^4 -1) \\ (y^2-4)(y^2+4)(y^2-1)(y^2+1) = \\ (y-2)(y+2)(y+2i)(y-2i)(y+1)(y-1)(y+i)(y-i) \\ \text{and thus roots at }\\ y=\pm 2,~\pm 2i,~\pm 1,~\pm i$

$\dfrac{vote_m}{votes_w}=\dfrac{125}{235}\\ \dfrac{vote_m}{3960+vote_m}=\dfrac{25}{47}\\ 47vote_m = 25vote_m + 99000\\ 22vote_m=99000 \\ vote_m = 4500\\ vote_w=8640\\ vote_t = 12960$

$\text{Let's count the combination that have neither Penelope or Quentin first}\\ \text{there will be }\dbinom{20}{3}3! =6840 \text{ ways to assign the officers w/o P or Q}\\ \text{now if we have P we have Q so that just leaves }\dbinom{20}{1}=20 \text{ combinations}\\ \text{or }20\cdot 3!=120 \text{ different ways off assigning the officers using P and Q}\\ \text{this gets us a total of }6960 \text{ ways of assigning officers}$

oh I missed the "4 letters or less"