Improve Your Algebra Skills with Practice Problems

Why does $\sqrt{x}/y=\sqrt{x/y/y}$
Sorry for the awkward title, hard to to sum a mathematical problem with words alone.
Having said that, I recently learned that the root of any value, x, and then that over value y, is identical to the root of $x/y/y$ , as in $\sqrt{x/y/y}$
For e.g, $\sqrt{35}/7=\sqrt{5/7}$ and I cannot logically deduce why. So far, I know that:
$$\begin{gathered} 35=7\times <!--\; \times \; -->5 \\ \frac{\sqrt{35}}{7}=\frac{\sqrt{7}\times <!--\; \times \; -->\sqrt{5}}{7} \end{gathered}$$
And somehow this being equal to:
√(5/7)
So it seems that we divided the numerator, √35 by √5 to give √7. However, 7/√7 is not √7! Yet, apparently it is?
So, a fraction is of course equal to itself if scaled up or down by the same amount, in that 10/2 is the same as 5/1, or 5. So scaling down √35 by √5 makes sense, as it only leaves √7, but it seems that the same cannot be said for the denominator, 7. 7/√7 = √7!? Yet, the two expressions really are the same, √35/7 and √(5/7).
Thanks for any help in advance.
P.S This works for any values, so it's no coincidence.

Convert fraction into decimal
Trying to help my daughter with a question for her maths that has got me stuck...wish I was better at maths!
Convert the fraction 27/50 into decimal
Could someone please help me with this question.

Let $A\subset <!--\; \subset \; -->\mathbb{R}$ be Lebesgue measurable such that $0<\mathrm{\lambda <!--\; \lambda \; -->}(A)<\mathrm{\infty <!--\; \infty \; -->}$ and consider the function $f:\mathbb{R}\to <!--\; \to \; -->[0,\mathrm{\infty <!--\; \infty \; -->})$, $f(x)=\mathrm{\lambda <!--\; \lambda \; -->}(A\cap <!--\; \cap \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},x))$. I would like to compute $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(x)$.
I have already proved that f is continuous (I have in fact showed that it is Lipschitz continuous), but I am not really sure if this allows me to say that $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(x)=\mathrm{\lambda <!--\; \lambda \; -->}(A\cap <!--\; \cap \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},\mathrm{\infty <!--\; \infty \; -->}))=\mathrm{\lambda <!--\; \lambda \; -->}(A)$. Intuitively, I am sure that this is right, but I am not sure whether the fact that $f$ is continuous is enough to write this. I feel that I am somehow assuming that the Lebesgue measure $\mathrm{\lambda <!--\; \lambda \; -->}$ is continuous in some sense. Could you please tell me if my conclusion is right and if the mere continuity of $f$ is enough for it?

How do you solve and graph ${\displaystyle p-6\ge 3}$ ?

How do you solve and graph ${\displaystyle 3x\ge 15}$ ?

Find the limit of fraction involving logarithms
I am looking for a way to prove the following limit for integer $x$s:
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\log <!--\; \; -->(x+2)-<!--\; -\; -->\log <!--\; \; -->(x+1)}{\log <!--\; \; -->(x+2)-<!--\; -\; -->\log <!--\; \; -->(x)}=\frac{1}{2}$
I could find the result by using a computer program but I cannot formally establish the above equality.

I have a prediction $f(x)$ of some continuous process variable, based on an input variable $x$ (think: location). The prediction is incorrect, with the error being normal distributed with expected value $\mathrm{\mu <!--\; \mu \; -->}$ and standard deviation $\mathrm{\sigma <!--\; \sigma \; -->}$.
Hence, the probability density function of $f(x)$ should be
$\frac{1}{\mathrm{\sigma <!--\; \sigma \; -->}\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}{e}^{-<!--\; -\; -->\frac{((f(x)-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->})/\mathrm{\sigma <!--\; \sigma \; -->}{)}^2}{2}}$
Is this correct? (No it is not, see answer below)
Now, I have a measurement $m$ of the process variable, based on an unknown input variable $x_m\cdot <!--\; \cdot \; -->m$ is assumed to be correct, but quantized to integral numbers.
Given a set of $x_i$ together with their predictions $f(x_i)$, how can I compute a probability that $x_m$ was in the vicinity of $x_i$?
I apologize if the question makes no sense.

ow do you write the interval ${\displaystyle (11,\infty )}$ as an inequality?

5.30 converted to a fraction or mixed number in lowest terms
The correct answer they got on my worksheet is $5\frac{3}{50}$, but I get $5\frac{15}{50}$

How do you graph ${\displaystyle -3\ge n}$ ?

I have a bunch of points where the best circle is fitted into them. The algorithm is based on the least squares approach to fit a circle.
My question is, how would error analysis be performed. I am thinking of something like:
$\underset{i}{\sum <!--\; \sum \; -->}(r_i-<!--\; -\; -->r_{(\text{actual})}{)}^2$ where $r_i$ are the distances from the center of the fitted circle to the point, and $r_{actual}$ is the actual radius of the circle.
Any suggestions or insights on how to proceed?

Prove inequality by induction $\mathrm{\forall <!--\; \forall \; -->}n\in <!--\; \in \; -->\mathbb{N},n\ge <!--\; \ge \; -->2:(1+\frac{1}{n}{)}^n<\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{1}{k!}<3$
I have the following statement to prove:
$\mathrm{\forall <!--\; \forall \; -->}n\in <!--\; \in \; -->\mathbb{N},n\ge <!--\; \ge \; -->2:(1+\frac{1}{n}{)}^n<\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{1}{k!}<3$
What I have already proven is that
$(1+\frac{1}{n}{)}^n<(1+\frac{1}{n+1}{)}^{n+1}$
I have started, as usual by induction with $n=2$. Then I went on to say
$(1+\frac{1}{n+1})\ast <!--\; \ast \; -->(1+\frac{1}{n+1}{)}^n<\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{1}{k!}+\frac{1}{(n+1)!}<3$
But this is where I seem to get stuck. I cannot use the induction hypothesis, since the denominator on the left side is now $n+1$ instead of $n$. I assume, I have to use the other inequality, which I have already proven to be true, but I do not know how. Any hints?

Why does finding the union of these three sets yield a negative number?
I've been working on a homework problem that I can't seem to be able to solve.
The question states:
Suppose 25 people attended a conference that contains 3 sessions. 15 people attended session #1; 18 people attended session #2; 12 people attended session #3. At least how many people attended all 3?
$u,\sin <!--\; \; -->2u=-<!--\; -\; -->2\sin <!--\; \; -->u\cos <!--\; \; -->u$
However, when I'm applying this problem to 3 sets, I obtained a negative number.
I computed the result: |A∪B∪C| ≤ 5 where |A∩B| ≥ 8, |A∩C| ≥ 2, |B∩C| ≥ 5, however, I don't quite understand the negative part where 25 ≥ |(A∩B)∪C| = |A∩B| + |C| - |A∩B∩C| (shorter way to find the minimum). The result just ends up as |A∩B∩C| + 25 ≥ 20 which would result in a negative unless I can somehow divide both sides by -1 which would change the equality?

Consider the wave equation with initial data:
$u_{tt}(t,x)+u_{xx}(t,x)=0$
$u(0,x)=u_0(x)$
$u_t(0,x)=u_1(x)$
Hadamard showed that this problem is ill-posed: there exist large solutions with arbitrarily small initial data. For instance, if we take $u(t,x)=a_{\mathrm{\omega <!--\; \omega \; -->}}\sinh <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t)\sin <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}x)$, then $u_0(x)=0$ and $u_1(x)=a_{\mathrm{\omega <!--\; \omega \; -->}}\mathrm{\omega <!--\; \omega \; -->}\sin <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}x)$, then we can make u(t,x) grow arbitrarily fast while keeping $u_0$ and $u_1$ small.
Tweaking this construction, it is not hard to see that for any given k and $\mathrm{ϵ<!--\; ϵ\; -->}>0$, we can construct initial data such that
$||u_0|{|}_{\mathrm{\infty <!--\; \infty \; -->}}+||u_0^{(1)}|{|}_{\mathrm{\infty <!--\; \infty \; -->}}+\dots <!--\; \dots \; -->+||u_0^{(k)}|{|}_{\mathrm{\infty <!--\; \infty \; -->}}+||u_1|{|}_{\mathrm{\infty <!--\; \infty \; -->}}+||u_1^{(1)}|{|}_{\mathrm{\infty <!--\; \infty \; -->}}+\dots <!--\; \dots \; -->+||u_1^{(k)}|{|}_{\mathrm{\infty <!--\; \infty \; -->}}<\mathrm{ϵ<!--\; ϵ\; -->}$
and $||u(\mathrm{ϵ<!--\; ϵ\; -->},x)|{|}_{\mathrm{\infty <!--\; \infty \; -->}}>\frac{1}{\mathrm{ϵ<!--\; ϵ\; -->}}$This can be interpreted as saying that the problem is ill-posed even in a Holder sense.
My question is: can one construct an example of a solution u(t,x) with initial data $u_0(x)$ and $u_1(x)$ such that
$\underset{i=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}||u_0^{(i)}|{|}_{\mathrm{\infty <!--\; \infty \; -->}}+||u_1^{(i)}|{|}_{\mathrm{\infty <!--\; \infty \; -->}}<\mathrm{ϵ<!--\; ϵ\; -->}$

Let $E\subset <!--\; \subset \; -->{\mathbb{R}}^n$ and $O_i=\{x\in <!--\; \in \; -->{\mathbb{R}}^n\mid <!--\; \mid \; -->d(x,E)<\frac{1}{i}\},i\in <!--\; \in \; -->\mathbb{N}$. Show that if $E$ is compact, then $m(E)=\underset{i\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}m(O_i)$. Does the statement hold if $E$ is closed, but not bounded and if $E$ is open, but bounded?
What I got was that since $\underset{i}{\bigcap <!--\; \bigcap \; -->}{O}_i=E$ and $O_1\supset <!--\; \supset \; -->O_2\supset <!--\; \supset \; -->\dots <!--\; \dots \; -->$, then we have that
$m(E)=m(\underset{i}{\bigcap <!--\; \bigcap \; -->}{O}_i)=\underset{i\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}m(O_i)$
by the continuity of the Lebesgue measure. How is the fact needed that $E$ is compact here?

Fraction simplification Rules
I am studying for GRE and One of the practice questions is a division. After converting my Mixed numeral I get $\frac{90}{72}$ now I just have to simplify.
What I understood is that you divide by Least common multiplier so I
1.) Divide $\frac{90}{72}$ by 2: And I got $\frac{45}{36}$
2.) Divide $\frac{45}{36}$ by 3: And I got $\frac{9}{12}$
3.) and so on until I got to 3/4 which is the wrong answer
If in step 1, I would have divide by 3
1.) Divide $\frac{90}{72}$ by 3 and I got $\frac{10}{8}$ 2.) Divide $\frac{10}{8}$by 2 and I got \frac{5}{4} which is the right answer
What is the rule to simplify fractions?

Let $\mathrm{\gamma <!--\; \gamma \; -->}:[0,1]\to <!--\; \to \; -->{\mathbb{S}}^2$ be a smooth curve. Consider the following set
$\underset{t\in <!--\; \in \; -->[0,1]}{\bigcup <!--\; \bigcup \; -->}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}(\mathrm{\gamma <!--\; \gamma \; -->}(t))\subset <!--\; \subset \; -->{\mathbb{R}}^3$
My question is: does this set have Hausdorff dimension at most 2, and as a result it has zero 3-dimensional Lebesgue measure?. Note that if the smoothness condition is removed then the answer is trivially false.

Proof of an inequality involving different exponents
I was asked the following interesting question but I can not come up with a proof.
Let $a,b,c$ be positive real numbers. Prove that
$\begin{array}{}\text{(}\mathrm{♣<!--\; ♣\; -->}\text{)}& \frac{a^3}{b^3+c^3}+\frac{b^3}{a^3+c^3}+\frac{c^3}{a^3+b^3}\ge <!--\; \ge \; -->\frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}\end{array}$
One way to prove $(\mathrm{♣<!--\; ♣\; -->})$ may be to consider the following function
$f(x)=\frac{a^x}{b^x+c^x}+\frac{b^x}{a^x+c^x}+\frac{c^x}{a^x+b^x}$
and prove that $f(x)$ is a non-decreasing function on $[2,3]$. When we take its derivative, the formula looks quite tedious and we just can not continue on. So is there any better solution?

There is a question in a book that I am trying to solve.
"A man usually rides his bike 1 kilometers per hour, yet the wind slows him to 6.76 kilometers for 26 minutes and 5.55 kilometers for 10. How long until he gets home 11.54 kilometers away?"

How do you solve and graph ${\displaystyle p-4<-7}$ ?