Expert Answers to Algebra 2 Problems

If y varies directly as x and inversely as the square of z and if y=20 when x=50 and z=5 how do you find y when x=3 and z=6?

How do you determine the constant of proportionality if P varies inversely as Q and Q=2/3 when P=1/2?

the length of a rectangle is 3 in. less than twice its width. if the perimeter of the rectangle is 8 in., what are the dimensions of the rectangle?

If y varies inversely as x, how do you find the constant of variation and the inverse variation equation given y=90 when x =6?

55 people showed up on the party. there were 3 less women than men. How may men were there?

Is ${\displaystyle \frac{y}{xz}=-4}$ a direct variation, inverse variation, joint or neither?

How do you find the constant of proportionality and write an equation if q is inversely proportional to p, and q = 3 / 2 when p = 60?

A truck's odometer reads 72,551.6 miles at the beginning of a trip and 72,831.0 miles at the end. If the trip takes 11.3 gallons of gas, what is the gas mileage in miles per gallon?

The length of a rectangle with constant area varies inversely as the width. When the length is 18 ft, the width is 8 ft. How do you find the length when the width is 9 ft?

If z varies directly as x and inversely as y, and z = 8 when x = 12 and y = 3, the value of the constant of proportionality is what number?

The deepest point in the pacific ocean is 676 m more than twice the deepest point in the arctic ocean. If the deepest point in the pacific is 10,920 m, how many meter is the deepest point in the arctic ocean?

A number is decreased by 40​% to 240. What is the original​amount?

How do you find "k" if y varies inversely as x and if y=24 when x=3?

Integration by parts: $\int <!--\; \int \; -->x\ln <!--\; \; -->x^{2}dx$
Problem: $\int <!--\; \int \; -->x\ln <!--\; \; -->x^{2}dx$
So what I did first was make $u=\ln <!--\; \; -->x^2$ and $dv=x$
Then I solved by getting the derivative of $u$ and the anti derivative of $dv$ and I got $du=1/x^2$ and $v=x^2/2$ then I did the formula
$\int <!--\; \int \; -->udv=uv-<!--\; -\; -->\int <!--\; \int \; -->vdu$
which then after plugging in the numbers and simplifying got me
$\frac{x^2}{2}\ln <!--\; \; -->x^2-<!--\; -\; -->\frac{1}{2x}+C$
Is this the right way to do the problem and answer?

Is ${\displaystyle \frac{1}{2}xy=2}$ an inverse or direct variation?

Suppose that a y varies inversely with x, and y=6 when x=8. What is an equation for the inverse variation?

Corita ran the 400-meter dash in 56.8 s. This waas 1.3 s less than her previous time. What was her previous time?

How to generate $\log$ function that intersects at $(0,1)$ and $(1,0)$?
I apologize for any incorrect or missing formatting, first time posting in the math stack exchange. It's been a few years since I've done any kind of calculus, so I remember nothing at all, which is probably the reason why I find my self stumped so early in my calculations.
I'm looking to generate a logarithmic algorithm that will follow a plot that will pass the following points: $(0,1)$ and $(1,0)$. My end goal is to generate a programming function that will return a very simple repulsion force ceofficient value based on a distance between two points. I would like, however, that the force drop (logarithmic?) the further these two points are, and grow (exponentially?) the closer they get. This is in no way an accurate calculation I'm looking for, but merely a non-linear function to return me a scalar coefficient between 0 and 1 that I can manipulate later on.
I want the function to cross at $(1,0)$ and $(0,1)$ so I get a coefficient between $0$ and $1$. In my function, any Y value lower than $0$ will be floored at $0$ and likewise for $Y$ values over $1$ (where $x<0$, which is possible since my $X$ value in my algorithm will likely be an offsetted value of the distance between two points).
That being said, with the following few assumptions:
1.$y=0$ when $x=1$
2.$y=1$ when $x=0$
3.$y<0$ when $x>1$
4.$y>1$ when $x<0$
5.$y=Alog(x+B)+C$
6.assuming log is log based 10
I tried to deduce the $A,B,C$ constants in order to generate a formula.
Here is what I've done so far:
1.From assumption #1, I can generate $1=Alog(B)+C$
2.From assumption #2, I can generate $0=Alog(1+B)+C$
3.I make both equations equals by transforming equation from 1. to $0=Alog(B)+C-<!--\; -\; -->1$
4.Making the following equality: $Alog(B)+C-<!--\; -\; -->1=Alog(1+B)+C$
5.I can scratch both $C$ constants out (is it safe to assume $C=0$, or $C$ can be equal to any arbitrary value, without changing the equation's plot?): $Alog(B)-<!--\; -\; -->1=Alog(1+B)$
6.Dividing each part by A gets me: ${\displaystyle \frac{Alog(B)}{A}}-<!--\; -\; -->{\displaystyle \frac{1}{A}}={\displaystyle \frac{Alog(1+B)}{A}}$
7.Allowing me to reduce both logs: $\log <!--\; \; -->(B)-<!--\; -\; -->{\displaystyle \frac{1}{A}}=log(1+B)$
8.If I put both logs on the same side: $log(B)-<!--\; -\; -->log(1+B)={\displaystyle \frac{1}{A}}$
9.I can combine them into: $\log <!--\; \; -->({\displaystyle \frac{B}{1+B}})={\displaystyle \frac{1}{A}}$
And I'd say this is where I'm stuck; I have two unknowns ($A$ and $B$) with one equation to solve. I don't know where to go from here. What's the next step?

Adapt the velocity equation $v=v_0+a\cdot <!--\; \cdot \; -->t$ to solve for average acceleration where the ball starts at t = 0.0 seconds and $v_0=0.00$ meters/second