Improve Your Algebra Skills with Practice Problems

How do you solve and graph ${\displaystyle \frac{k}{5}\le -17}$ ?

Keith has $500 in a savings account at the beginning of the summer. He wants to have at least $200 at the end of the summer. He withdraws $25 per week for food, clothing, and movie tickets. How many weeks can Keith withdraw money from his account?

Consider a complete, finite measure space $(X,\mathcal{M},\mathrm{\mu <!--\; \mu \; -->})$. Let $\{f_n{\}}_{n=1}^{\mathrm{\infty <!--\; \infty \; -->}}\subset <!--\; \subset \; -->L^1(\mathrm{\mu <!--\; \mu \; -->})$ be a sequence such that $\underset{n}{sup}\int <!--\; \int \; -->|f_n|d\mathrm{\mu <!--\; \mu \; -->}<\mathrm{\infty <!--\; \infty \; -->}$. Furthermore, assume that for every $\mathrm{\epsilon <!--\; \epsilon \; -->}>0$ there exists $\mathrm{\delta <!--\; \delta \; -->}>0$ such that ${\int <!--\; \int \; -->}_E|f_n|d\mathrm{\mu <!--\; \mu \; -->}<\mathrm{\epsilon <!--\; \epsilon \; -->}$ whenever $\mathrm{\mu <!--\; \mu \; -->}(E)<\mathrm{\delta <!--\; \delta \; -->}$. I want to show the followings:
$(a)\underset{n}{sup}{\int <!--\; \int \; -->}_{\{|f_n|\ge <!--\; \ge \; -->M\}}|f_n|d\mathrm{\mu <!--\; \mu \; -->}\to <!--\; \to \; -->0\text{ as }M\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}.$
$(b)\underset{n}{sup}{\int <!--\; \int \; -->}_{\{|f_n|\le <!--\; \le \; -->\mathrm{\delta <!--\; \delta \; -->}\}}|f_n|d\mathrm{\mu <!--\; \mu \; -->}\to <!--\; \to \; -->0\text{ as }\mathrm{\delta <!--\; \delta \; -->}\to <!--\; \to \; -->0.$
I have the following proof for (a):
Let $E=\{x\in <!--\; \in \; -->X:|f_n(x)|\ge <!--\; \ge \; -->M\}$. By Chebyshev's inequality,
$\mathrm{\mu <!--\; \mu \; -->}(E)\le <!--\; \le \; -->\frac{1}{M}{\int <!--\; \int \; -->}_X|f_n|d\mathrm{\mu <!--\; \mu \; -->}\mathrm{\forall <!--\; \forall \; -->}n\in <!--\; \in \; -->\mathbb{N}.$
Since $\mathrm{\mu <!--\; \mu \; -->}(E)\to <!--\; \to \; -->0$ as $M\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, $\mathrm{\forall <!--\; \forall \; -->}\mathrm{\delta <!--\; \delta \; -->}>0$, $\mathrm{\mu <!--\; \mu \; -->}(E)<\mathrm{\delta <!--\; \delta \; -->}$ and therefore,
${\int <!--\; \int \; -->}_E|f_n|d\mathrm{\mu <!--\; \mu \; -->}<\mathrm{\epsilon <!--\; \epsilon \; -->}.$
Since this is true for all $\mathrm{\epsilon <!--\; \epsilon \; -->}>0$ and $n\in <!--\; \in \; -->\mathbb{N}$,
${\int <!--\; \int \; -->}_E|f_n|d\mathrm{\mu <!--\; \mu \; -->}={\int <!--\; \int \; -->}_{\{|f_n|\ge <!--\; \ge \; -->M\}}|f_n|d\mathrm{\mu <!--\; \mu \; -->}\to <!--\; \to \; -->0\text{ as }M\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}.$
Taking the supremum on the left-hand side, we get
$\underset{n}{sup}{\int <!--\; \int \; -->}_{\{|f_n|\ge <!--\; \ge \; -->M\}}|f_n|d\mathrm{\mu <!--\; \mu \; -->}\to <!--\; \to \; -->0\text{ as }M\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}.$
I wonder if my proof is coorect, and for (b), I wonder how to show that the measure of the set $\{|f_n|\le <!--\; \le \; -->\mathrm{\delta <!--\; \delta \; -->}\}$ is small so that a similar argument can be stated as in (a)?

If $\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}=1$, prove $\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\le <!--\; \le \; -->3$

For a filtration ${\mathcal{F}}_0\subset <!--\; \subset \; -->\cdots <!--\; \cdots \; -->\subset <!--\; \subset \; -->{\mathcal{F}}_t\subset <!--\; \subset \; -->\cdots <!--\; \cdots \; -->\subset <!--\; \subset \; -->{\mathcal{F}}_{\mathrm{\infty <!--\; \infty \; -->}}$. We know that if $X$ is ${\mathcal{F}}_0$-measurable, then $X$ is also ${\mathcal{F}}_t$-measurable for $t\ge <!--\; \ge \; -->0$.
I have a question about the martingale $X_t$ (which ${\mathcal{F}}_t$ measurable) is that given a stopping time $T$,
why $1_{T=\mathrm{\infty <!--\; \infty \; -->}}{X}_{\mathrm{\infty <!--\; \infty \; -->}}$ is ${\mathcal{F}}_T$-measurable?
It is clear that $1_{T=\mathrm{\infty <!--\; \infty \; -->}}$ is ${\mathcal{F}}_T$-measurable. But because $X_{\mathrm{\infty <!--\; \infty \; -->}}$ is ${\mathcal{F}}_{\mathrm{\infty <!--\; \infty \; -->}}$-measurable. How can we say $X_{\mathrm{\infty <!--\; \infty \; -->}}$ is also ${\mathcal{F}}_T$-measurable?

Conditions present after solving for x
The question asked me to solve
${\displaystyle \frac{x^2+2x-<!--\; -\; -->8}{x^2-<!--\; -\; -->x-<!--\; -\; -->2}}=3$
The answer is $x=0.5$, which I worked out, but in the answers it says that I must state the condition that $x\ne 2$ to get full marks. Why is this condition present? How can I recognise when a condition exists, when I encounter similar problems?

Decompose a fraction in a sum of two
Let's say that I have this fraction:
$\frac{2x}{x^2+4x+3}$
I would like to decompose in two fraction:
$\frac{A}{x+3}+\frac{B}{x+1}$
Which is the procedure for that? :)

Without using the Second Derivative Test, can the convexity of the natural exponential function be shown directly from the definition of convexity? The expression
$e^t=\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{t^n}{n!}$
can be used.

Is there any value of x with the property that $x<x-<!--\; -\; -->1$ ?

How do you graph the inequality ${\displaystyle x-1>-10}$ on a number line?

How do you solve ${\displaystyle \frac{h}{2}<-5}$ ?

How do you solve and graph ${\displaystyle a+2>-3}$ ?

How do you solve ${\displaystyle \frac{3}{4}t\le 12}$ ?

Advanced GCSE Indices
At school, we've been asked to simplify the following:
${\left(\frac{S^{9/10}}{S^{3/5}}\right)}^{-<!--\; -\; -->4}$
I've given it a go but haven't been taught this yet. Could someone please explain the process which one would use?

How do you solve ${\displaystyle -4t>-20}$ ?

I have an exercise in my measure and integration theory course and I'm a little bit stuck with the $\mathrm{\sigma <!--\; \sigma \; -->}$-rings and $\mathrm{\sigma <!--\; \sigma \; -->}$-algebras. The exercise goes as follows:
Let $X$ be a set and $\mathrm{\epsilon <!--\; \epsilon \; -->}=\{\{x\}|x\in <!--\; \in \; -->X\}$. Determine $\mathrm{\sigma <!--\; \sigma \; -->}(\mathrm{\epsilon <!--\; \epsilon \; -->})$, the smallest $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra that contains epsilon.
I know that $\mathrm{\epsilon <!--\; \epsilon \; -->}=\{\{x\}|x\in <!--\; \in \; -->X\}$ is the set that contains the singeltons as elements.
My first problem when defining the $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra is whether it has to contain the empty set AND the set $X$, OR the empty set and the set $\mathrm{\epsilon <!--\; \epsilon \; -->}$? Because this is one of three conditions for forming a $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra.
The second condition, that $\mathrm{\sigma <!--\; \sigma \; -->}(\mathrm{\epsilon <!--\; \epsilon \; -->})$ has to contain the complement of a subset $A\in <!--\; \in \; -->\mathrm{\sigma <!--\; \sigma \; -->}(\mathrm{\epsilon <!--\; \epsilon \; -->})$, is (I think) clear for me. But then the third condition, that it has to contain the the union of all subsets of $\mathrm{\sigma <!--\; \sigma \; -->}(\mathrm{\epsilon <!--\; \epsilon \; -->})$, is also unclear. I am not sure how to "build" this union.

How can we find out the interval in an inequality?
I have the inequality: $(x+2)(x-<!--\; -\; -->3)<<!--\; <\; -->0$ . In thise case, I'm expected to find out two values by evaluation $(x+2)=0$ and $(x-<!--\; -\; -->3)=0$ and then divide the real number line in 4 intervals and figure out for which intervals the inequality holds by figuring out whether the expression on the LHS is less than 0 or greater than 0.

How to interpret parameter estimates in factor prediction ( in R )
So I have some data set in a .csv file and there are three factor levels, 1 , 2 , 3, (there are fifteen of each) and each has a corresponding score.
Here are some details.
so the data is contained in a simple csv file, the first column is labelled Team, and the second column is labelled Score.
The first column consists of fifteen 1's, followed by fifteen 2's , followed by fifteen 3's.
The R code I used was
data.source<-"http.www.. " ( the data set)
SportScores<-read.csv(file=data.source)
I set x such that x prints 1 1 1.... 1 2 2 2 ... 2 3 3 3 .... 3 Levels 1 2 3
names(sportScores)
y<-SportScores$Scores
So using lm I get parameter estimates in R as
Intercept (35.800)
x2 (0.066)
x3 (12.40)
the t value are very large for intercept and x3, but very small for x2, ie it indicated to me that we cannot reject the null in this case, but what is the null?
${\mathrm{\beta <!--\; \beta \; -->}}_0=35.8$
${\mathrm{\beta <!--\; \beta \; -->}}_1^c=0.06667$
${\mathrm{\beta <!--\; \beta \; -->}}_2^c=12.40$
But how do I interpret this? I want to see any differences in scores between the 3 levels, etc. I mean, what even is the test being conducted? For example
${\mathrm{\beta <!--\; \beta \; -->}}_1^c=0.06667$
has a small t value, so the null hypothesis is not rejected, but what even is the null hypothesis in this case? Moreover, from the code output itself, how can I know the associated individual standard errors of the estimated means?

Find the minimum of $\underset{cyc}{\sum <!--\; \sum \; -->}\sqrt{\frac{a}{2(b+c)}}$ with $a,b,c><!--\; >\; -->0$
As said in the title, I have to find the minimum of the following:
$\underset{cyc}{\sum <!--\; \sum \; -->}\sqrt{\frac{a}{2(b+c)}}$
with $a+b+c>0$
In my very last attempt, I tried to work it out using AM-GM: Since
$\sqrt{\frac{a}{2(b+c)}}=\frac{1}{\sqrt{2}}\frac{a}{\sqrt{a(b+c)}}\ge <!--\; \ge \; -->\frac{\sqrt{2}a}{a+b+c}$
$\Rightarrow <!--\; \Rightarrow \; -->\underset{cyc}{\sum <!--\; \sum \; -->}\sqrt{\frac{a}{2(b+c)}}\ge <!--\; \ge \; -->\frac{\sqrt{2}(a+b+c)}{a+b+c}=\sqrt{2}$
The problem is, to get the minimum value,
$a=b+c$
$b=c+a$
$c+a+b$
or $a=b=c=0$. This is wrong since $a,b,c>0$ And after some calculus I found the minimum value is $\frac{3}{2}$ when $a=b=c$. So what mistake I have made?

The high soccer team can have no more than 26 players. How do you write and solve an inequality to determine how many more players can make the team if the coach has already chosen 17 players?