Improve Your Algebra Skills with Practice Problems

How to find $li{m}_{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}su{p}_{x>0}{1}_{(0,t_n)}(x)$. It is given that $t_n\ge <!--\; \ge \; -->0,t_n\to <!--\; \to \; -->0$ as $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ and 1 denotes indicator function. It seems that limit should be simply zero but how to justify it? I'm facing this problem while trying to prove a result in banach spaces.

How do you solve ${\displaystyle \frac{128}{-g}<4}$ ?

If I define the integral for ${\int <!--\; \int \; -->}_a^b\mathbf{\text{F}}(t)dt$ as $({\int <!--\; \int \; -->}_a^b{F}_1(t)dt,{\int <!--\; \int \; -->}_a^b{F}_2(t)dt,\dots <!--\; \dots \; -->,{\int <!--\; \int \; -->}_a^b{F}_n(t)dt)$. How do I then show that
$|{\int <!--\; \int \; -->}_a^b\mathbf{\text{F}}(t)dt|\le <!--\; \le \; -->{\int <!--\; \int \; -->}_a^b|\mathbf{\text{F}}(t)|dt?$
If we write out the inequality we have
$\sqrt{{({\int <!--\; \int \; -->}_a^b{F}_1(t)dt)}^2+{({\int <!--\; \int \; -->}_a^b{F}_2(t)dt)}^2+\cdots <!--\; \cdots \; -->+{({\int <!--\; \int \; -->}_a^b{F}_n(t)dt)}^2}\le <!--\; \le \; -->{\int <!--\; \int \; -->}_a^b\sqrt{F_1(t{)}^2+F_2(t{)}^2+\cdots <!--\; \cdots \; -->+F_n(t{)}^2}dt.$
attempt:
If I use Jenssen's inequality I get for the left side:
$$\begin{gathered} \sqrt{{({\int <!--\; \int \; -->}_a^b{F}_1(t)dt)}^2+{({\int <!--\; \int \; -->}_a^b{F}_2(t)dt)}^2+\cdots <!--\; \cdots \; -->+{({\int <!--\; \int \; -->}_a^b{F}_n(t)dt)}^2}\le <!--\; \le \; -->\sqrt{{\int <!--\; \int \; -->}_a^b{F}_1^2(t)dt+{\int <!--\; \int \; -->}_a^b{F}_2^2(t)dt+\cdots <!--\; \cdots \; -->+{\int <!--\; \int \; -->}_a^b{F}_n^2(t)dt} \\ =\sqrt{\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{\int <!--\; \int \; -->}_a^b{F}_i^2(t)dt}=\sqrt{{\int <!--\; \int \; -->}_a^b\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{F}_i^2(t)dt}. \end{gathered}$$
The problem I have now is that the square root function is concave, so if I use Jenssen again, I get the opposite inequality as the one I need. Any idea on how to solve this?

State the number of possible triangles that can be formed using the given measurements, then sketch and solve the triangles, if possible.
$m\mathrm{\angle <!--\; \angle \; -->}A={48}^{\circ <!--\; \circ \; -->},c=29,a=4$

Suppose you are living on a straight line. There are several satellites at height 20,000 kilometers and you get readings saying that satellite 1 is directly above the point $x_1\pm {10}^{-10}$ and is at a distance $h_1=21,000\pm {10}^{-2}$ from you, satellite 2 is directly above $x_2\pm {10}^{-10}$ and at a distance $h_2=52,000\pm {10}^{-2}$. Where are you ($x_0$) and to what accuracy? Hint: Consider separately the cases $x_1<x_2$ and $x_2>x_1$.
My results:
Actually we have two equations:
$(x_1-<!--\; -\; -->x_0{)}^2+(2\cdot <!--\; \cdot \; -->{10}^4{)}^2=h_1^2$
$(x_2-<!--\; -\; -->x_0{)}^2+(2\cdot <!--\; \cdot \; -->{10}^4{)}^2=h_2^2$
We can express $x_0$ from them as:
$x_0=\frac{x_1+x_2}{2}+\frac{(h_1-<!--\; -\; -->h_2)(h_1+h_2)}{2(x_2-<!--\; -\; -->x_1)}$
But how to compute accuracy of measurement? I'm troubling with $x_2-<!--\; -\; -->x_1$ in denominator.

Hidden Cauchy-Schwarz inequality
If $x_i><!--\; >\; -->0$ and $x_i{y}_i-<!--\; -\; -->{z_i}^2><!--\; >\; -->0$ for $i\le <!--\; \le \; -->n$ , then prove that
$\frac{n^3}{(\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{x}_i)(\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{y}_i)-<!--\; -\; -->(\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{z_i}^2)}\le <!--\; \le \; -->\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{1}{x_i{y}_i-<!--\; -\; -->z_i^2}$
This is a question from the IMO 1969. At first it seems that we can assume that $a_i=\sqrt{x_i{y}_i}-<!--\; -\; -->z_i$ and $b_i=\sqrt{x_i{y}_i}+z_i$. Hence RHS becomes $\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{1}{a_i{b}_i}$.Now how do we simplify the LHS so that we can use the Cauchy-Schwarz inequality?

How do you solve and write the following in interval notation: ${\displaystyle x-3>8}$ and ${\displaystyle x+2>}$ ?

Prove or disprove that $|a_1|+|a_2|+\dots <!--\; \dots \; -->+|a_n|\le <!--\; \le \; -->n\sqrt{a_1^2+\dots <!--\; \dots \; -->+a_n^2}$ by showing that $RHS-<!--\; -\; -->LHS\ge <!--\; \ge \; -->0$if possible.

4.6 The temperature of seawater at different depths and latitudes. Just as the atmosphere may be divided into layers characterized by how the temperature changes as altitude increases, he oceans may be divided into zones characterized by how the temperature changes as depth increases. We shall divide the oceans into three zones: the surface zone comprises the water at depths between 0 m and 200 m; the thermocline comprises the water at depths between 200 m and 1000 m; and the deep zone comprises water at depths exceeding 1000 m.
We shall assume that temperature is unaffected by longitude.
We shall assume that at all latitudes, temperature is a continuous function of depth. Informally, this means that you can sketch the graph of temperature versus depth (at a particular latitude and longitude) without lifting your pencil from thepage.
We shall assume that the wind and waves serve to mix water in the surface zone so effectively that the temperatureremains constant with depth. it does change with latitude, though, as the temperature in the surface zone is greatly affected by solar radiation [4]. We shall assume that the summer temperature in the surface zone is a linear function of latitude. Further, we shall assume that the summer temperature in the surface zone is 2 °C at the poles, and 24 °C at the equator.
We shall also assume that, because solar energy never makes it to the deepest water, the temperature remains constant with depth in the deep zone. In fact, we shall assume that water in the deep zone, no matter the latitude, remains at 2 °C throughout the year.
Since water in the surface zone can be warm, water in the deep zone is always cold, and the temperature at a given latitude is a continuous function of depth, the temperature must change with depth throughout the thermocline. We shall assume that, at each latitude and longitude, temperature is a linear function of depth in this zone.
Context: University coding assignment
So far from this information, I've managed to develop two models:
For the surface zone, I have $T(l)=\frac{-<!--\; -\; -->11}{45}l+24$ and for the deep zone, I have T(l)=2 where T is the summer temperature in degrees Celsius and l is the latitude in degrees (where 0 is the equator and 90 are the poles).
I'm having trouble interpreting this information to develop a linear model for the summer temperature of the seawater (T) at the thermocline zone. I know this model will differ from the last two as it will be a function of depth (rather than latitude), but the lack of data is really throwing me.
Any guidance would be greatly appreciated.

What number is midway between 2/4 and 3/4?
Sorry for posting this question but appreciate if you could help me.
What number is midway between 2/4 and 3/4?

Suppose that every point of an interval $X=[0,1]$ lies in at least $k$ subintervals $A_i\subset <!--\; \subset \; -->X$, $i=1,...,n$. Show that
$k\le <!--\; \le \; -->\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}|A_i|$
where $|A_i|$ denotes the length of $A_i$.
My thoughts:
Let's consider $k$ copies of $X$. We will denote j-th copy as $X^j$.
If we would be able to cover them with $\{A_i{\}}_{i=1}^n$ then we're done.
(it is also enough to left uncovered the set of measure zero in every copy).
Let's fix an arbitrary point $x\in <!--\; \in \; -->X^1$. There exists $1\le <!--\; \le \; -->i_1<<!--\; <\; -->...<i_k\le <!--\; \le \; -->n$ such that $x\in <!--\; \in \; -->A_{i_j},j=1,...,k$. Then we will use $A_{i_j}$ to cover everything it can on the copy $X^j$. Now, let's choose another point $x\in <!--\; \in \; -->X^1$ and repeat the process.
There is a hard part in this proof. I can't understand why it can not be the case when we used all our $A_i$ but didn't fully cover every copy $X^j$(or didn't left uncovered the set of measure zero in every copy).

How can we algebraically solve $|x-1|+|x-2|>1?$

What is the least common multiple of 15 and 50?

Are the sums equal to each other?
They are $2$ different results for the integral
$\int <!--\; \int \; -->x{e}^{2x}\sin <!--\; \; -->\left(\frac{x}{3}\right)dx$
1. ${\displaystyle \frac{-<!--\; -\; -->3}{1369}{e}^{2x}(3(35-<!--\; -\; -->74x)\sin <!--\; \; -->\left(\frac{x}{3}\right)+(37x-<!--\; -\; -->36)\cos <!--\; \; -->\left(\frac{x}{3}\right))}$
2. ${\displaystyle \frac{3}{37}{e}^{2x}(x(6\sin <!--\; \; -->\left(\frac{x}{3}\right)-<!--\; -\; -->\cos <!--\; \; -->\left(\frac{x}{3}\right))-<!--\; -\; -->\frac{3}{37}(35\sin <!--\; \; -->\left(\frac{x}{3}\right)-<!--\; -\; -->13\cos <!--\; \; -->\left(\frac{x}{3}\right)))}$
Are the sums equal to each other?

I am currently working on a problem from my statistics class. It is as follows:
$$\begin{gathered} \text{A computer consulting firm presently has bids out on three projects. Let}{A}_1=\{\text{awarded project}i\},\text{for}i=1,2,3,\text{ and suppose that }P(A_1)=.22, \\ P(A_2)=.25, \\ P(A_3)=.23, \\ P(A_1\cap <!--\; \cap \; -->A_2\cap <!--\; \cap \; -->A_3)=.01. \\ \text{Expree in words each of th following events, and compute the probability of each event}: \\ a.A_1\cup <!--\; \cup \; -->A_2 \\ b.A_1\mathrm{\prime <!--\; \prime \; -->}\cap <!--\; \cap \; -->A_2\mathrm{\prime <!--\; \prime \; -->} \\ c.A_1\cup <!--\; \cup \; -->A_2\cup <!--\; \cup \; -->A_3 \\ e.A_1\mathrm{\prime <!--\; \prime \; -->}\cap <!--\; \cap \; -->A_2\mathrm{\prime <!--\; \prime \; -->}\cap <!--\; \cap \; -->A_3 \\ d.A_1\mathrm{\prime <!--\; \prime \; -->}\cap <!--\; \cap \; -->A_2\mathrm{\prime <!--\; \prime \; -->}\cap <!--\; \cap \; -->A_3\mathrm{\prime <!--\; \prime \; -->} \\ f.(A_1\mathrm{\prime <!--\; \prime \; -->}\cap <!--\; \cap \; -->A_2\mathrm{\prime <!--\; \prime \; -->})\cup <!--\; \cup \; -->A_3 \end{gathered}$$
The only one I had difficulty quantifying with words was problem f). How would I do that?
Also, I had originally assumed the three events were disjoint; but, by looking at the question again, I found that they aren't, because of the last piece of information given--the intersection of all of the events. I am having a hard time understanding how they are not disjoint. How can these particular three events have something in common? It just seems odd.

Summation and Patterns Question
I have a really urgent question. Today i was looking at this Pattern (Using fractions) :
$\frac{1}{1\ast <!--\; \ast \; -->3}+\frac{1}{3\ast <!--\; \ast \; -->5}+\frac{1}{5\ast <!--\; \ast \; -->7}+\frac{1}{7\ast <!--\; \ast \; -->9}+...$=? The pattern is $\frac{n}{2n+1}$. I was wondering if you could do a summation formula, in which the end term is infinite. Thanks for the help!

I have a measure $\mathrm{\mu <!--\; \mu \; -->}$ that is supported on $[-<!--\; -\; -->3,3]\times <!--\; \times \; -->\mathbb{R}$. What we are given is that, if we fix the first component $i$, then $\mathrm{\mu <!--\; \mu \; -->}(i,\cdot <!--\; \cdot \; -->)$ is a probability measure. Formally (maybe it should integrate to $\mathrm{d}i$ or similar, I cannot define this very well):
${\int <!--\; \int \; -->}_{x\in <!--\; \in \; -->\mathbb{R}}\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}(i,x)=1\mathrm{\forall <!--\; \forall \; -->}i\in <!--\; \in \; -->[-<!--\; -\; -->3,3].$
I find it very hard to understand this tuple-indexing notation. My question is the following: What kind of assumptions do we need to have a result similar to:
${\int <!--\; \int \; -->}_{(i,x)\in <!--\; \in \; -->[-<!--\; -\; -->3,3]\times <!--\; \times \; -->\mathbb{R}}\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}(i,x)={\int <!--\; \int \; -->}_{i\in <!--\; \in \; -->[-<!--\; -\; -->3,3]}\mathrm{d}i$

I just think that since for any fixed $i$ the measure $\mathrm{\mu <!--\; \mu \; -->}$ integrates to 1 (on the second dimension -- apologies for my poor terminology), then integrating over all $(i,x)\in <!--\; \in \; -->[-<!--\; -\; -->3,3]\times <!--\; \times \; -->\mathbb{R}$ should also give simply an 'iterated integral' where we first integrate wrt $x\in <!--\; \in \; -->\mathbb{R}$ for a fixed $i$, which will integrate to 1 , and then integrate over $i\in <!--\; \in \; -->[-<!--\; -\; -->3,3]$. But of course, we cannot define
${\int <!--\; \int \; -->}_{(i,x)\in <!--\; \in \; -->[-<!--\; -\; -->3,3]\times <!--\; \times \; -->\mathbb{R}}\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}(i,x)={\int <!--\; \int \; -->}_{i\in <!--\; \in \; -->[-<!--\; -\; -->3,3]}{\int <!--\; \int \; -->}_{x\in <!--\; \in \; -->\mathbb{R}}\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}(i,x)={\int <!--\; \int \; -->}_{i\in <!--\; \in \; -->[-<!--\; -\; -->3,3]}\mathrm{d}y$
where in the red parts I make an abuse of integartion rules.

How to calculate
${\int <!--\; \int \; -->}_Q{(x_1+x_2+\dots <!--\; \dots \; -->+x_n)}^{2}d{\mathrm{\lambda <!--\; \lambda \; -->}}_n$
whereas $n\ge <!--\; \ge \; -->2$ and
$Q=\{(x_1,\dots <!--\; \dots \; -->,x_n)\in <!--\; \in \; -->{\mathbb{R}}^n:0\le <!--\; \le \; -->x_i\le <!--\; \le \; -->1,i=1,\dots <!--\; \dots \; -->,n\}$

I know that I have to use induction and Fubini to calculate
$I(n)=\underset{0}{\overset{1}{\int <!--\; \int \; -->}}\underset{0}{\overset{1}{\int <!--\; \int \; -->}}\cdots <!--\; \cdots \; -->\underset{0}{\overset{1}{\int <!--\; \int \; -->}}{(x_1+x_2+\dots <!--\; \dots \; -->+x_n)}^{2}d{x}_n\cdots <!--\; \cdots \; -->d{x}_{2}d{x}_1$
And I already started to calculate the integrals for $n=2,3,4$ but the numbers I got did not enlighten me and I have no idea for an induction hypothesis. I also exchanged $x$ with 1 but I still have no clue. Maybe I am doing something wrong. Thanks for help in advance.

What is the greatest common factor for 78 and 91?

Rationalise the denominator and simplify $\frac{3\sqrt{2}-<!--\; -\; -->4}{3\sqrt{2}+4}$
Does someone have an idea how to work ${\displaystyle \frac{3\sqrt{2}-<!--\; -\; -->4}{3\sqrt{2}+4}}$ by rationalising the denominator method and simplifying?