Improve Your Algebra Skills with Practice Problems

Simple algebraic manipulation with 2 equations
My first equations is this: $d_2=d-<!--\; -\; -->30.$
My second equations is this: $\frac{1}{d_2}=\frac{1}{12}-<!--\; -\; -->\frac{1}{1.066(d-<!--\; -\; -->30)}$
I am trying to solve for $d_2$ in the second equation and then set the two equations equal to each other to solve for d.
I am having trouble solving for $d_2$ in the second equation. I think the fractions are giving me a hard time. Any help is much appreciated, thanks!

Why do we miss 8 in the decimal expansion of 1/81, and 98 in the decimal expansion of 1/9801?
I've seen this happen that when you divide in a fraction using the square of any number with only nines in the denominator. Like in
$\frac{1}{9^2}=\frac{1}{81}=0.01234567\underset{\uparrow <!--\; \uparrow \; -->}{}901234567\underset{\uparrow <!--\; \uparrow \; -->}{}901234567\dots <!--\; \dots \; -->,$
and in
$\frac{1}{{99}^2}=\frac{1}{9801}=0.0001020304050607080910111213\dots <!--\; \dots \; -->9697\underset{\uparrow <!--\; \uparrow \; -->}{}99000102\dots <!--\; \dots \; -->,$
the decimals go on predictably when suddenly in the first one you miss 8, and in the second you miss 98 and it keeps going on forever. How does this happen? Why do we miss numbers like 8 in the decimal representation of $\frac{1}{9^2},$ , or like 98 in the decimal of $\frac{1}{{99}^2},$ or 998 in $\frac{1}{{999}^2},$ , or $9998$ in $\frac{1}{{9999}^2}$ ?

How was the conditional probability obtained in this equation?
I'm reading about inverse problems and the Bayesian approach and I'm having troubles understanding how the following equations were obtained.
We consider the following equation initially.
y=G(u)+$\mathrm{\eta <!--\; \eta \; -->}$
y is a set of measured data. G is a mathematical model and u are the parameters of the model. The noise that is present in the observed data is given by $\mathrm{\eta <!--\; \eta \; -->}$ (0 mean noise).
The authors then go on to describe the Bayesian approach and they say that the likelihood function, that is the probability of y given u is given by:
$\mathrm{\rho <!--\; \rho \; -->}(y|u)=\mathrm{\rho <!--\; \rho \; -->}(y-<!--\; -\; -->G(u))$
How did they derive this relation above?

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Let $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be a continuous bounded function, $(X_k{)}_k$ a sequence of i.i.d random variables such that

Let $x\in <!--\; \in \; -->\mathbb{R},Y_k=f(x+\underset{q=1}{\overset{k}{\sum <!--\; \sum \; -->}}{X}_q),{\mathcal{F}}_k=\mathrm{\sigma <!--\; \sigma \; -->}(X_1,...,X_k).$

Verify that:

and deduce that for every $y\in <!--\; \in \; -->\text{supp}{P}_{X_1}:=\{u,\mathrm{\forall <!--\; \forall \; -->}r>0,P_{X_1}([y-<!--\; -\; -->r,y+r])>0\},f(x+y)=f(x).$

One way to prove the first question is to note that

For part 2 we can deduce that for $P_{X_1}$-almost every $y\in <!--\; \in \; -->\mathbb{R},f(x+y)=f(x),$, how to conclude using continuity that $f(x+y)=f(x)$ for every $y\in <!--\; \in \; -->\text{supp}{P}_{X_1}$ ?
How to deduce part 2 ?

I have this (I thought easy) problem:
Let $K$ be a compact Hausdorff space and $\mathrm{\mu <!--\; \mu \; -->}$ a regular Borel probability measure on $K$. Show that:
$\mathrm{supp}<!--\; \; -->\mathrm{\mu <!--\; \mu \; -->}=\bigcap <!--\; \bigcap \; -->\{A\subseteq <!--\; \subseteq \; -->K:A\text{ is closed and }\mathrm{\mu <!--\; \mu \; -->}(A)=1\}$
Here, $\mathrm{supp}<!--\; \; -->\mathrm{\mu <!--\; \mu \; -->}=\{x\in <!--\; \in \; -->K:\mathrm{\forall <!--\; \forall \; -->}\text{ open }U\ni <!--\; \ni \; -->x,\mathrm{\mu <!--\; \mu \; -->}(U)><!--\; >\; -->0\}$, the so-called topological support of $\mathrm{\mu <!--\; \mu \; -->}$.
Clearly, calling the RHS of the above $L$, if $x\in <!--\; \in \; -->K\setminus <!--\; \setminus \; -->L$ then there is an open neighbourhood of $x$ disjoint from $L$, which by measure monotonicity implies the measure of that neighbourhood is 0 as $\mathrm{\mu <!--\; \mu \; -->}(L)=1,\mathrm{\mu <!--\; \mu \; -->}(K\setminus <!--\; \setminus \; -->L)=0$, so $x\notin <!--\; \notin \; -->\mathrm{supp}<!--\; \; -->\mathrm{\mu <!--\; \mu \; -->}$. Likewise, if $x\notin <!--\; \notin \; -->\mathrm{supp}<!--\; \; -->\mathrm{\mu <!--\; \mu \; -->}$, then one neighbourhood of $x$ is null, so there is at least one closed set disjoint from $x$ with measure 1. Thus $x$ is not in the intersection of such closed sets, $x\notin <!--\; \notin \; -->L$. Easily we have $\mathrm{supp}<!--\; \; -->\mathrm{\mu <!--\; \mu \; -->}=L$.
Why did the textbook assume regularity?

Property of fractions
Given two fractions $\frac{h}{k}$ and $\frac{h^{{}^{\prime }}}{k^{{}^{\prime }}}$ both in reduced form. I am unable to find a case when $\frac{h+h^{{}^{\prime }}}{k+k^{{}^{\prime }}}$ does not lie in the interval $[\frac{h}{k},\frac{h^{{}^{\prime }}}{k^{{}^{\prime }}}]$. Is there such a case ?
PS: I was able two prove no such case exists for consecutive terms of Farey series. But can't prove in general.

I would like to run a Kalman filter over a set of measurements which may (will) be correlated. Essentially, each new measurement contains (say) 90% of the same information from the previous measurement. The measurements are made by integrating over time, and subsequent measurements are taken by overlapping integrations.
For various reasons, I can't necessarily decimate the measurements so that they are independent of each other, but if I apply each measurement at its full weighting, the filter will become over confident and its covariance will shrink too quickly.
My first thought is that there must be someway to increase the covariance from each measurement so that it doesn't overly affect the filter state, but I would like a rigorously correct solution, and I welcome any pointers.
A least squares approach which allowed expressing the correlated measurements would also be welcome, but I kind of prefer the Kalman filter because it gives me a running covariance as well as the filtered result.
Thank you in advance for any pointers! Is there a better stack exchange for this question?

How do you solve ${\displaystyle 5x\le 21+2x}$ ?

How do you solve and graph ${\displaystyle 8\le r-14}$ ?

How do you solve ${\displaystyle 40\ge -5z}$ ?

So my task is the following: Consider a sequence of random variables $(X_n{)}_{n\in <!--\; \in \; -->\mathbb{N}}$ with $X_n\stackrel{a.s.}{\to <!--\; \to \; -->}X$ and $X_n\le <!--\; \le \; -->B$ a.s. for $B\in <!--\; \in \; -->\mathbb{R}$. Show that $X_n\stackrel{{\mathcal{L}}^p}{\to <!--\; \to \; -->}X$ for all $p\in <!--\; \in \; -->\mathbb{N}$.

My ideas till now:
1. In a first step i tried to show that the expected value is bounded by both sides from zero: $0\le <!--\; \le \; -->\mathbb{E}(|X_n-<!--\; -\; -->X{|}^p)\le <!--\; \le \; -->\dots <!--\; \dots \; -->0$ Problem here is that i would need that $|X_n|\le <!--\; \le \; -->B$ a.s.
2. My second attempt was to use the dominated convergence theorem. With $Y_n=B$ it follows that if $Y=B$ that $Y_n\stackrel{a.s.}{\to <!--\; \to \; -->}Y$ and $\mathbb{E}(|Y_n-<!--\; -\; -->Y|)$. Now I also need that $|X_n|\le <!--\; \le \; -->Y$ and then it follows that $\mathbb{E}(|X_n-<!--\; -\; -->X|)\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{\to <!--\; \to \; -->}0$.

Probably someone can give me a hint or another way of solving this.

How do you solve ${\displaystyle -7\le 3-2x}$ ?

How do you solve ${\displaystyle -4x+3\le -5+11}$ ?

I would like to verify my solution for the following problem:
Let $g:(0,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->\mathbb{R}$ be a nonzero, continuous function and define $G:[0,1]\times <!--\; \times \; -->[1,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->\mathbb{R}$ by
$G(x,y)=g(xy).$
Show that $G$ is not in $L^1([0,1]\times <!--\; \times \; -->[1,\mathrm{\infty <!--\; \infty \; -->}))$.
Solution:
${\int <!--\; \int \; -->}_1^{\mathrm{\infty <!--\; \infty \; -->}}{\int <!--\; \int \; -->}_0^1|G(x,y)|dxdy={\int <!--\; \int \; -->}_1^{\mathrm{\infty <!--\; \infty \; -->}}{\int <!--\; \int \; -->}_0^1|g(xy)|dxdy={\int <!--\; \int \; -->}_1^{\mathrm{\infty <!--\; \infty \; -->}}{\int <!--\; \int \; -->}_0^y\frac{|g(t)|}{y}dtdy\ge <!--\; \ge \; -->{\int <!--\; \int \; -->}_1^{\mathrm{\infty <!--\; \infty \; -->}}\frac{1}{y}\underset{t\in <!--\; \in \; -->(0,y]}{inf}|g(t)|dy\ge <!--\; \ge \; -->\underset{t\in <!--\; \in \; -->(0,\mathrm{\infty <!--\; \infty \; -->})}{inf}|g(t)|{\int <!--\; \int \; -->}_1^{\mathrm{\infty <!--\; \infty \; -->}}\frac{1}{y}dy=\mathrm{\infty <!--\; \infty \; -->}$
I'm a little concerned about taking the infimum over the non compact set (0,y] (g is not defined at 0).
Is this correct? And if so, how do I address my concern above? Thanks for your time.

Why are zeros/roots (real) solutions to an equation of an n-degree polynomial?
I can't really put a proper title on this one, but I seem to be missing one crucial point. Why do roots of a function like $f(x)=a{x}^2+bx+c$ provide the solutions when $f(x)=0$. What does that y=0 mean for the solutions, the intercept at the x axis? Why aren't the solutions at $f(x)=403045$ or some other arbitrary n?
What makes the x-intercept special?

The length of a rectangle is three times its width. If the perimeter is at most 112 centimeters, what is the greatest possible value for the width?

Question:
A prism has an equilateral triangular base with a perimeter of 12cm. If the length of the prism is 24cm, determine the total surface area of the prism.
hint: What is the area of 1 triangle?
I have draw a diagram and tried a few methods to figure out the area of 1 triangle, but it just doesn't seem to work.
Any help would be appreciated.

Given an observable $z$,$r$ and unobservable $x$, I have a process of the form:
$x_{k+1}=x_k+w_k$
$z_k=a_1{z}_{k-<!--\; -\; -->1}+a_2{z}_{k-<!--\; -\; -->2}-<!--\; -\; -->\frac{s}{2}(r_{k-<!--\; -\; -->1}-<!--\; -\; -->x_{k-<!--\; -\; -->1}+r_{k-<!--\; -\; -->2}-<!--\; -\; -->x_{k-<!--\; -\; -->2})+v_k$
which in my understanding are transit and measurement equations.
I am a beginner in Kalman-filter. In all places I researched, the measurement equation is in the form of $z_k=H{x}_k+v_k$, but how should I apply Kalman filter to the above system?

The limit of gaussian distribution on curve manifold
I am reading the paper. In introduction, it said
In many applications, even simple estimation problems involving angular data are often considered as traditional linear or nonlinear estimation problems and handled with classical techniques such as the Kalman Filter [1], the extended Kalman Filter (EKF), or the unscented Kalman Filter (UKF) [2]. In a circular setting, most traditional approaches to filtering suffer from assuming a Gaussian probability density at a certain point. They fail to take into account the periodic nature of the problem and assume a linear vector space instead of a curved manifold. This shortcoming can cause poor results, in particular when the angular uncertainty is large. In certain cases, the filter may even diverge.
My background is engineering, and I understand what Gaussian distribution in N-dimension means and the role of Gaussian distribution in Kalman filter. But I don't understand why Gaussian distribution fail to estimate angular data in its nature. It will be great if there are mathematical explanation and reference. Thank you.

Laurent expansion of $\frac{1}{z^2}$
I need to find a Laurent expansion of $\frac{1}{z^2}$ with centre in $z_0=1$ and $P(1,2014,2015)$
If it was $\frac{1}{z}$, I'd rewrite the fraction like this:
$\frac{1}{(z-<!--\; -\; -->1)+1}$
But with $z^2$ I'm not sure how to proceed...
My attempt:
According to the given recommendation, I've tried to do the expansion of $\frac{1}{z}$:
$\frac{1}{z}=\frac{1}{1+(z-<!--\; -\; -->1)},a_0=1,q=-<!--\; -\; -->(z-<!--\; -\; -->1)=1-<!--\; -\; -->z$
$\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(1-<!--\; -\; -->z{)}^n=1+(1-<!--\; -\; -->z)+(1-<!--\; -\; -->z{)}^2+...$
And after differentiation:
$0+1+2(1-<!--\; -\; -->z)+3(1-<!--\; -\; -->z{)}^2+...=\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}n(1-<!--\; -\; -->z{)}^{n-<!--\; -\; -->1}$