Rom

\(\text{Here's a bit of a derivation of the monthly payment}\\ \text{ignore months and years for a moment and just let $r$ be the periodic rate}\\ \text{$n$ will be the number of periods. Let $\alpha = (1+r)$}\\ \text{every period we take the principal, apply interest, and subtract a payment}\\ p(k+1) = \alpha p(k) -pmt,~p(0) = loan\\ \text{If you work through this you find that}\\ p(n) = \alpha^n loan - pmt \sum \limits_{k=0}^{n-1} \alpha^k = \alpha^n loan - pmt\dfrac{\alpha^n-1}{\alpha -1}\)

\(\text{We want $p(n)=0$ and we need to solve for $pmt$}\\ \alpha^n loan = pmt \dfrac{\alpha^n -1}{\alpha-1}\\ loan = pmt \dfrac{1-\alpha^{-n}}{\alpha -1}\\ pmt =loan \dfrac{(\alpha-1)}{1-\alpha^{-n}} = loan \dfrac{r}{1-(1+r)^{-n}}\)

\(\text{Here we plug in $loan=1600000,~n=360,~r=\dfrac{0.03}{12}=0.0025$}\\ pmt = \$1600000\dfrac{0.0025}{1-(1.0025)^{-360}} = \$6745.66\)

\(1)~\text{you have to inspect the terms in the two polynomials and find which ones, $\\$ when their exponents are added, sum to two}\\ (-5x^2, 1)\\ (-7x,-6x)\\ (1,-x^2)\\ \text{and that's all of them. Performing the multiplication and summing them all we get}\\ (-5 + 42-1)x^2 = 36x^2\\ \text{You are correct, well done!}\)

\(2)~3x^4 + 2x-1 +h(x) = 5x^2 -6x-1\\ h(x) = (5x^2-6x-1) - (3x^4+2x-1)\\ h(x) = -3x^4 + 5x^2 -8x\)

The question is poorly worded.

Is the end result a single piece of chain in which all but the first and last link are joined to only two other links,

and the first and last link joined to only one other link, or not?

If not.  Break one link, attach all the pieces of chain at that link and reweld it.  Fifteen cents.

If so, Melody's solution above is correct.

\(\dfrac 2 5+\dfrac 2 5 + \dfrac 8 9 = \\ \dfrac{18}{45}+\dfrac{18}{45}+\dfrac{40}{45} = \\ \dfrac{76}{45} = 1 \dfrac{31}{45}~pizza\)

\(\text{it helps to start by dividing the range into portions and mapping that back to the domain}\\ \text{the range of $f(x)$ is the entire real line which can be expressed as $\\ (-\infty,\infty) = (-\infty, 1)\cup [1,\infty)$}\\~\\ \text{on $(-\infty, 1),~f^{-1}(x) = \sqrt{1-x}+1$}\\ \text{on $[1,\infty),~f^{-1}(x) = 2-x$}\\~\\ f^{-1}(-3) = 3\\ f^{-1}(0) = 2\\ f^{-1}(3) = -1\\ \text{These sum to 4}\)

\(\text{for Cedric}\\ fv = \$12000(1+0.05)^{15} = \$24947.10\\~\\ \text{For Daniel}\\ fv = \$12000(1 + (0.07)15) = \$24600.00\\~\\ \text{The difference to the nearest dollar is $\$347$}\)

\(\text{most of this can be ignored. Two important points}\\ 1)~p = p[\text{make the free throw}] = 0.546\\ 2)~\text{$n = 2$, i.e. he takes two free throw shots at a time}\\ \text{The number of free throws, and thus the points he makes $\\$ has a binomial distribution, with the above parameters, $n$ and $p$}\\ \text{it's well known that the expectation of the binomial distribution is $\mu = np\\$ which in this case is $\mu = 2(0.546)=1.092$}\)

use of the phrase homecoming is fine... it's the first part of the sentence that's open to misinterpretation....

\(\text{Equilibrium solutions are found where $\dfrac{dx}{dt}=0$}\\ \text{Here these occur at $x=6,~x=2$}\\~\\ \text{To determine their stability we look at the second derivative}\\ \dfrac{d^2x}{dt^2} = (-1)(x-2) + 1(6-x) = 8-2x\\ \text{at $x=6,~\dfrac{d^2x}{dt^2}=8-12=-4$}\\ \text{A negative second derivative indicates an unstable equilibrium point}\\ \text{at $x=2,~\dfrac{d^2x}{dt^2}=8-4=4$}\\ \text{A positive second derivative indicates a stable equilibrium point}\\ \)