# Make the internet a better place to learn

1

## How do you write the prime factorization of 20st?

EZ as pi
Featured 2 days ago · Prealgebra

$20 s t = 2 \times 2 \times 5 \times s \times t$

#### Explanation:

You need to write the given term or expression as the product of its prime factors. This is sometimes also called expanded form.

$20 s t = 2 \times 2 \times 5 \times s \times t \text{ } \leftarrow$ these are all prime factors.

The variables are really easy to do because even if there are indices, you can see exactly how many of each there are.

So $36 {s}^{3} {t}^{2}$ would be expanded as:

$2 \times 2 \times 3 \times 3 \times s \times s \times s \times t \times t$

3

## What is lim_(xtooo)(5x^3+2x-3)/(4-x^2-2x^3)? I'm completely blank on this one please help

Jim H
Featured 14 hours ago · Calculus

$- \frac{5}{2}$

#### Explanation:

Factor out of the numerator and denominator the greatest power of $x$ in the denominator, and then reduce. (Or, divide numerator and denominator by the greatest power of $x$ in the denominator.)

Use ${\lim}_{x \rightarrow \infty} \frac{c}{x} ^ n = 0$ for all $c$ and all positive $n$.

${\lim}_{x \rightarrow \infty} \frac{5 {x}^{3} + 2 x - 1}{4 - {x}^{2} - 2 {x}^{3}} = {\lim}_{x \rightarrow \infty} \frac{{x}^{3} \left(5 + \frac{2}{x} ^ 2 - \frac{1}{x} ^ 3\right)}{{x}^{3} \left(\frac{4}{x} ^ 3 - \frac{1}{x} - 2\right)}$

$= {\lim}_{x \rightarrow \infty} \frac{5 + \frac{2}{x} ^ 2 - \frac{1}{x} ^ 3}{\frac{4}{x} ^ 3 - \frac{1}{x} - 2}$

$= \frac{5 + 0 - 0}{0 - 0 - 2} = - \frac{5}{2}$

Note

The above is the long explanation. The short one is look at just the dominating terms -- the greatest powers of $x$-- in the numerator and denominator. At infinity, the quotient behaves like the ratio of those greatest powers.

Examples

${\lim}_{x \rightarrow \infty} \frac{3 {x}^{5} + 7 {x}^{2} - 19}{4 {x}^{5} - 2 {x}^{3} + 3 x - 2} = {\lim}_{x \rightarrow \infty} \frac{3 {x}^{5}}{4 {x}^{5}} = {\lim}_{x \rightarrow \infty} \frac{3}{4} = \frac{3}{4}$

${\lim}_{x \rightarrow \infty} \frac{7 {x}^{2} - 5 x + 7}{2 {x}^{3} + 5 x + 8} = {\lim}_{x \rightarrow \infty} \frac{7 {x}^{2}}{2 {x}^{3}} = {\lim}_{x \rightarrow \infty} \frac{7}{2 x} = 0$

${\lim}_{x \rightarrow \infty} \frac{{x}^{4} + 3 x - 9}{6 x + 2} = {\lim}_{x \rightarrow \infty} \frac{{x}^{4}}{6 x} = {\lim}_{x \rightarrow \infty} {x}^{3} / 6 = \infty$

2

## How do you find the arc length of the curve f(x)=x^3/6+1/(2x) over the interval [1,3]?

HSBC244
Featured 3 days ago · Calculus

The arc length is $\frac{14}{3}$ units.

#### Explanation:

The arc length of a curve on the interval $\left[a , b\right]$ is given by evaluating ${\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$.

The derivative of $f ' \left(x\right)$, given by the power rule, is

$f ' \left(x\right) = \frac{1}{2} {x}^{2} - \frac{1}{2 {x}^{2}} = \frac{{x}^{4} - 1}{2 {x}^{2}}$

Substitute this into the above formula.

${\int}_{1}^{3} \sqrt{1 + {\left(\frac{{x}^{4} - 1}{2 {x}^{2}}\right)}^{2}} \mathrm{dx}$

Expand.

${\int}_{1}^{3} \sqrt{1 + \frac{{x}^{8} - 2 {x}^{4} + 1}{4 {x}^{4}}} \mathrm{dx}$

Put on a common denominator.

int_1^3sqrt((x^8 + 2x^4 + 1)/(4x^4)dx

Factor the numerator as the perfect square trinomial, and recognize the denominator can be written of the form ${\left(a x\right)}^{2}$.

${\int}_{1}^{3} \sqrt{{\left({x}^{4} + 1\right)}^{2} / {\left(2 {x}^{2}\right)}^{2}} \mathrm{dx}$

Eliminate the square root using ${\left({a}^{2}\right)}^{\frac{1}{2}} = a$

${\int}_{1}^{3} \frac{{x}^{4} + 1}{2 {x}^{2}} \mathrm{dx}$

Factor out a $\frac{1}{2}$ and put it in front of the integral.

$\frac{1}{2} {\int}_{1}^{3} \frac{{x}^{4} + 1}{x} ^ 2 \mathrm{dx}$

Separate into different fractions.

$\frac{1}{2} {\int}_{1}^{3} {x}^{4} / {x}^{2} + \frac{1}{x} ^ 2 \mathrm{dx}$

Simplify using ${a}^{n} / {a}^{m} = {a}^{n - m}$ and $\frac{1}{a} ^ n = {a}^{-} n$.

$\frac{1}{2} {\int}_{1}^{3} {x}^{2} + {x}^{-} 2 \mathrm{dx}$

Integrate using $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right)$, with $n \in \mathbb{R} , n \ne - 1$.

$\frac{1}{2} {\left[\frac{1}{3} {x}^{3} - \frac{1}{x}\right]}_{1}^{3}$

Evaluate using the second fundamental theorem of calculus, which states that for ${\int}_{a}^{b} F \left(x\right) = f \left(b\right) - f \left(a\right)$, if $f \left(x\right)$ is continuous on $\left[a , b\right]$ and where $f ' \left(x\right) = F \left(x\right)$.

$\frac{1}{2} \left(\frac{1}{3} {\left(3\right)}^{3} - \frac{1}{3} - \left(\frac{1}{3} {\left(1\right)}^{3} - \frac{1}{1}\right)\right)$

Combine fractions and simplify.

$\frac{1}{2} \left(9 - \frac{1}{3} - \frac{1}{3} + 1\right)$

$\frac{1}{2} \left(10 - \frac{2}{3}\right)$

$5 - \frac{1}{3}$

$\frac{14}{3}$

Hopefully this helps!

2

## Why would frameshift mutation have a larger impact than a substitution mutation on the organism in which the mutation occurred?

Elijah F.
Featured 3 days ago · Biology

Because it messes up the whole nucleotide sequence.

#### Explanation:

The addition or loss of DNA bases changes a gene's reading frame. A reading frame consists of groups of 3 bases that each code for one amino acid.

A frameshift mutation shifts the grouping of these bases, changing the code for an amino acids.

The resulting protein is usually not functional. Premature chain termination mostly takes place.

Insertions, deletions, and duplications can all be frameshift mutations.

Example:
Imagine each letter as a nucleotide, and groups of three letters as a codons.

Normal sequence, no mutations:
THE FAT CAT ATE THE FAT RAT

Frame-shift mutation, Insertion of an extra A at the A in CAT:
THE FAT CAA TAT ETH EFA TRA T

Frame-shift mutation, Deletion of an A at the A in CAT:
THE FAT CTA TET HEF ATR AT

Frame-shift mutation, Duplication of an A at the A in CAT:

THE FAT CAA TAT ETH EFA TRA T

The result of frame-shift mutations is that the wrong codons will be synthesized.

Another more scientific example:
Nucleotide sequence is on the top grouped into 3's (the codons), the amino acid it codes for is abbreviated underneath.

Original Sequence:
CAT--TCA--CAC--GTA--CTC--ATG
[his]--[ser]--[his]--[val--[leu]--[met]

Frameshift one base to the right:
C--ATT--CAC--ACG--TAC--TCA--TGC
[x]--[ile]--[his]--[thr]--[tyr]--[ser]--[cys]

1

## What are the factors that limit dispersal of a species?

Elijah F.
Featured 3 days ago · Biology

Geographic Distance, Environmental Resistance, Adaptive Constraints and Natural Selection

#### Explanation:

Geographical Distance:
Geographical isolation (on a far away island for example) would limit species from dispersing because the distance between a new environment is too far away.

Environmental Resistance:
Environmental factors , and climatic conditions may limit an organism's geographical distribution.

Some organisms are better adapted for dispersal. There are numerous examples of flora and fauna which are distributed all over the world, but most are restricted to certain locations/climatic areas.

Natural Selection:
Competition for resources could be too high in a new niche and thus settlement in a new area cannot occur easily. Newly dispersed species may face fierce predators. On the other hand dispersal of a species to islands are often successful.

4

## What was the U.S. Policy of Imperialism in the early 1900's?

Roberto Giammattei
Featured 2 days ago · U.S. History

Kind of Shaky

#### Explanation:

When the United States entered this new dawn, they were met with lots of opportunities. When it comes to the precedents, the U.S. war just emerging from the Spanish-American war, wherein the Americans made the Philippines a protectorate.

During this time, they also did annex Hawaii (they were reluctant before, when they sensed vulnerability as they were attacking the Philippines, fearing that say, Japan would take them, were annexed).

1900's: When Theodore Roosevelt comes into power, he is quintessentially recognized for his "Big Stick" diplomacy. "speak softly, and carry a big stick." -Theodore Roosevelt. What this meant, was that basically the U.S.A. would be meddling in other nation's affairs by arguing with things such as the Platt Amendment (the Monroe Doctrine being the predecessor to all of these justifications); with this one, the U.S. was able to "allow the United States to intervene unilaterally in Cuban affairs, and a pledge to lease land to the United States for naval bases on the island."

Then there's the famous Panamanian rebels, which Teddy helped by blockading Colombia from being able to deal with them (by using the Navy), which then allowed him to construct the Canal.

After Teddy's tenure was over, and Thomas Woodrow Wilson took command, it can be summed up like this : "Tell me what is right and I will fight for it.". Wilson saw Imperialism as something so unjust, and unfair, that he vehemently opposed it. Ironically, Wilson ended up also intervening in other affairs, such as by sending U.S. marines to Haiti (https://en.wikipedia.org/wiki/United_States_occupation_of_Haiti), making Haiti a protectorate of the United States. Also, since Wilson loathed Victoriano Huerta (Mexican Dictator), he took over the port of Veracruz (https://en.wikipedia.org/wiki/United_States_occupation_of_Veracruz), in order to prevent German munitions from arriving there. He also launched an expedition (The "Mexican Expedition") in retaliation for an attack Pancho Villa (Mexican Revolutionary), made on Columbus, New Mexico. (https://en.wikipedia.org/wiki/Pancho_Villa_Expedition)

He did however, sign an agreement, wherein the Philippines would be released once there was a "stable government" in the Pujo Committee. The nation would eventually gain their independence from the U.S.A. by 1946. Therefore, they began in an imperialist manner, slowly turned more towards Democratism. Woodrow also purchased the Virgin Islands from Denmark in 1917.

3

## Prove that if u is an odd integer, then the equation x^2+x-u=0 has no solution that is an integer?

Steve
Featured 2 days ago · Algebra

Proposition
If $u$ is an odd integer, then the equation ${x}^{2} + x - u = 0$ has no solution that is an integer.

Proof
Suppose that there exists a integer solution $m$ of the equation:

${x}^{2} + x - u = 0$

where $u$ is an odd integer. We must examine the two possible cases:

$m$ is odd; or
$m$ is even.

First, let us consider the case where $m$ is odd, then there exists an integer $k$ such that:

$m = 2 k + 1$

Now, since $m$ is a root of our equation, it must be that:

${m}^{2} + m - u = 0$
 :. (2k + 1)^2 + (2k + 1) − u = 0
 :. (4k^2 + 4k + 1) + (2k + 1) − u = 0
 :. 4k^2 + 6k + 2 − u = 0
$\therefore u = 4 {k}^{2} + 6 k + 2$
$\therefore u = 2 \left(2 {k}^{2} + 3 k + 1\right)$

And we have a contradiction, as $2 \left(2 {k}^{2} + 3 k + 1\right)$ is even, but $u$ is odd.

Next, let us consider the case where $m$ is even, then there exists an integer $k$ such that:

$m = 2 k$

Similarly, since $m$ is a root of our equation, it must be that:

${m}^{2} + m - u = 0$
 :. (2k)^2 + (2k) − u = 0
 :. 4k^2 + 2k − u = 0
$\therefore u = 4 {k}^{2} + 2 k$
$\therefore u = 2 \left(2 {k}^{2} + k\right)$

And, again, we have a contradiction, as $2 \left(2 {k}^{2} + k\right)$ is even, but $u$ is odd.

So we have proved that there is no integer solution of the equation ${x}^{2} + x - u = 0$ where $u$ is an odd integer.

Hence the proposition is proved. QED

1

## Which black leader argued that African Americans needed to accommodate themselves to segregation?

Roberto Giammattei
Featured yesterday · U.S. History

Booker T. Washington

#### Explanation:

Booker T. Washington (https://en.wikipedia.org/wiki/Booker_T._Washington), educator, orator, and advisor to U.S. presidents. His views differed greatly to views upheld by other advocators of black rights (such as Dr. W.E.B. Du Bois, founder of the NAACP, who advocated for complete egalitarianism https://en.wikipedia.org/wiki/W._E._B._Du_Bois).

What Mr. Washington argued, was that Blacks should not be equal to Whites (a view that Whites favored at the time), but he did believe in improving their education, earning him the moniker "The Great Accommodator". He was the first Black person to be invited into the White House (By Theodore Roosevelt), a highly controversial issue at the time, in order to advise T.R. on race relations.

1

## In what situation would a CT scan be more useful than an X-ray?

josef
Featured yesterday · Anatomy & Physiology

CT scan is definitely more useful in any situation because it provides a clearer image in multiple angles for the body organs.

#### Explanation:

A computed tomography scan, or CT scan, is similar to an MRI in that it produces detailed, high-quality images of the body. The CT scan is a more sophisticated and powerful X-ray that takes a 360-degree image of internal organs, the spine and vertebrae. Contrast dyes are often injected into the blood to make structures within the body more visible on the CT scan.

A CT scan produces detailed images of organs, bones, soft tissue and blood vessels and can be used to more easily diagnose cancer, heart disease, appendicitis, musculoskeletal disorders, trauma and infectious diseases.

A CT scanner looks like a large box with a tunnel in the center. The patient lies on a table that slides in and out of the tunnel, while the scanner rotates around the patient, producing cross-section images of the body. The technologist performing the scan sits in a separate room with computers on which the images are displayed. The technologist can speak with the patient using speakers and microphones.

A CT scan is more expensive than an X-ray and is not always available at small or rural hospitals.

3

## Let f_n (x) = (Sigma_(n=1))^n (sin^2x)/(cos^2(x/2) - cos^2((2n+1)/2)x) and g_n (x) = (Pi_(n=1))^n f_n (x) Let I_n = int_0^pi (f_n (x))/(g_n (x)) dx and if (Sigma_(n=1))^n I_n = K pi . Then K is??

Steve
Featured yesterday · Calculus

$K = 1$

#### Explanation:

Firstly poor notation for the summation and product. The standard notation is to use a "dummy" variable, usually $i$ or $r$ as the loop counter, as in

${\sum}_{r = 1}^{n} r = \frac{1}{2} n \left(n + 1\right)$

So using the correct notation we have:

${f}_{n} \left(x\right) = {\sum}_{r = 1}^{n} \setminus {\sin}^{2} \frac{x}{{\cos}^{2} \left(\frac{x}{2}\right) - {\cos}^{2} \left(\frac{\left(2 r + 1\right) x}{2}\right)}$

If we focus in the denominator for a moment, which desperately needs simplification, we see that it is the difference of two squares, so we can factorise prior to simplifying, and we can use the identities:

cos(A)+cos(B) = \ \ \ \ 2cos((​A+B)/2​​ )cos((​A−B)/2​​ )
cos(A) - cos(B) = -2 sin((​A+B)/2​​ )sin((​A−B)/2​​ )
$\sin 2 A = 2 \sin A \cos A$

to get;

${\cos}^{2} \left(\frac{x}{2}\right) - {\cos}^{2} \left(\frac{\left(2 r + 1\right) x}{2}\right)$
$\setminus \setminus \setminus = {\cos}^{2} \left(\frac{x}{2}\right) - {\cos}^{2} \left(r x + \frac{x}{2}\right)$
$\setminus \setminus \setminus = \left(\cos \left(\frac{x}{2}\right) - \cos \left(r x + \frac{x}{2}\right)\right) \left(\cos \left(\frac{x}{2}\right) - \cos \left(r x + \frac{x}{2}\right)\right)$
$\setminus \setminus \setminus = 2 \cos \left(\frac{r x + x}{2}\right) \cos \left(\frac{- r x}{2}\right) \left(- 2\right) \sin \left(\frac{r x + x}{2}\right) \sin \left(\frac{- r x}{2}\right)$
$\setminus \setminus \setminus = 2 \cos \left(\frac{r x + x}{2}\right) \cos \left(\frac{r x}{2}\right) \left(- 2\right) \sin \left(\frac{r x + x}{2}\right) \left(- \sin \left(\frac{r x}{2}\right)\right)$
$\setminus \setminus \setminus = \left\{2 \sin \left(\frac{r x}{2}\right) \cos \left(\frac{r x}{2}\right)\right\} \left\{2 \sin \left(\frac{r x + x}{2}\right) \cos \left(\frac{r x + x}{2}\right)\right\}$
$\setminus \setminus \setminus = \sin \left(\frac{2 r x}{2}\right) \sin \left(\frac{2 \left(r x + x\right)}{2}\right)$
$\setminus \setminus \setminus = \sin \left(r x\right) \sin \left(\left(r + 1\right) x\right)$

So we can therefore write ${f}_{n} \left(x\right)$ as:

${f}_{n} \left(x\right) = {\sum}_{r = 1}^{n} \setminus {\sin}^{2} \frac{x}{\sin \left(r x\right) \sin \left(\left(r + 1\right) x\right)}$

Let us examine the first few expansions of $f$ and $g$

${f}_{1} \left(x\right) = {\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}$
${f}_{2} \left(x\right) = {\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}$
${f}_{2} \left(x\right) = {\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)} + {\sin}^{2} \frac{x}{\sin \left(3 x\right) \sin \left(4 x\right)}$

And we have:

${g}_{n} \left(x\right) = {\prod}_{r = 1}^{n} \setminus {f}_{n} \left(x\right)$

${g}_{1} \left(x\right) = {f}_{1} \left(x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}$

${g}_{2} \left(x\right) = {f}_{1} \left(x\right) \cdot {f}_{2} \left(x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}\right) \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}\right)$

${g}_{3} \left(x\right) = {f}_{1} \left(x\right) \cdot {f}_{2} \left(x\right) \cdot {f}_{3} \left(x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}\right) \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}\right) \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)} + {\sin}^{2} \frac{x}{\sin \left(3 x\right) \sin \left(4 x\right)}\right)$

So using the definition of ${I}_{n}$:

${I}_{n} = {\int}_{0}^{\pi} \setminus {f}_{n} \frac{x}{g} _ n \left(x\right) \setminus \mathrm{dx}$

We have:

${I}_{1} = {\int}_{0}^{\pi} \setminus {f}_{1} \frac{x}{g} _ 1 \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \setminus \frac{{\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}}{{\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\left[x\right]}_{0}^{\pi}$
$\setminus \setminus \setminus = \pi$

So if ${\sum}_{r = 1}^{n} \setminus {I}_{r} = K \pi$ then;

$n = 1 \implies {I}_{1} = K \pi$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies \pi = K \pi$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies K = 1$

Let's see if this holds with $n = 2$, if so then perhaps we can prove the proposition by Induction.

${I}_{2} = {\int}_{0}^{\pi} \setminus {f}_{2} \frac{x}{g} _ 2 \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \setminus \frac{{\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}}{\left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}\right) \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}\right)} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \frac{1}{{\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \frac{\sin \left(x\right) \sin \left(2 x\right)}{\sin} ^ 2 \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \frac{\sin \left(x\right) 2 \sin x \cos x}{\sin} ^ 2 \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} 2 \cos x \setminus \mathrm{dx}$
$\setminus \setminus \setminus = 2 \setminus {\left[\sin x\right]}_{0}^{\pi} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = 2 \setminus \left(\sin \pi - \sin 0\right)$
$\setminus \setminus \setminus = 0$

So if ${\sum}_{r = 1}^{n} \setminus {I}_{r} = K \pi$ then;

$n = 2 \implies {I}_{1} + {I}_{2} = K \pi$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies \pi + 0 = K \pi$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies K = 1$, consistent with the above case $n = 1$

I think it's fairly easy to see that ${I}_{k} = 0 \forall k \ge 2$, and if I have a bit more time later I will attempt to prove that by Induction.