New answers tagged algebra-precalculus
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The fundamental theorem of calculus "says" that
$$g(x)=\int_0^x f(t) \, dt \implies g'(x)=f(x)$$ So, you first need to find the zero('s) of
$$g'(x)=-\frac{3}{4}+x+\cos\left(\frac{\pi}{4}x^2+x\right)$$ This is a transcendental equation which, by definition, will not show analytical solutions. As a consequence, some numerical methods will be required ...
0
These problems don't happen if you'll use the following identity.
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
Since, $$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=\frac{1}{2}\sum_{cyc}(a^2+b^2-2ab)=\frac{1}{2}\sum_{cyc}(a-b)^2,$$
we see that $\sum\limits_{cyc}(a-b)^2=0$ for $a=b=c$ only and it can give a extraneous root of the equation.
Now, ...
0
Let $u=x^2-1$ then $du=2xdx$. Evaluating u on the limit points the integral becomes
$$
\frac{1}{2}\int_3^8 f'(u)du.
$$
What can you say now using the graph given?
1
$$\log\left[\frac{e^{-x}}{1-e^{-x}}\right]=\log\left[\frac1{e^x-1}\right]=-\log[e^x-1]=(\mu-y)/\sigma$$which gives$$x=\ln[1+e^{(y-\mu)/\sigma}]$$
2
Yes, "tangent" takes its name from its association with a tangent line of the unit circle. Likewise, "secant" is named for a secant line. And "sine" for (a mistranslation of) semi-chord.
Curiously, according to the Earliest Known Uses of Some of the Words in Mathematics site (which has entries for all of the terms), "Vieta ...
1
Very good question! The first example is misleading in writing equivalences everywhere; the second equivalence
$$3-x+ 9\sqrt[3]{(3-x)(6+x)}(\sqrt[3]{3-x} + \sqrt[3]{6+x})+6+x$$
$$\iff$$
$$3-x+ 9\sqrt[3]{(3-x)(6+x)} +6+x=27$$
should (at least conceptually) be an implication $(\implies)$. It is true that it is in fact an equivalence, but this is not yet clear ...
1
This should be obvious.
Every positive integer has a prime factorization. So it may be written as $n = p_1^{a_1}p_2^{a_2}......p_t^{a_t}$ where $p_i$ are the distinct prime factors and $a_i$ are the power of the prime factors.
Now $p_i^{a_i} \ge p_i^1=p_i$ with equality holding if and only if $a_i = 1$. So $n = p_1^{a_1}p_2^{a_2}......p_t^{a_t} \ge p_1p_2.....
2
This is a good reason and it is consistent also with the other equivalent definition for the tangent
(credit)
Refer also to The Etymology of Trig Functions.
0
Here is how the calculation works.
Say the loan amount is $P$, APR is $x$ in $\%$, number of months is $n$ and monthly payment is $M$.
Say monthly rate of interest is $r$. Then $r = \frac{x}{12}$.
Interest at the first monthly payment $ = Pr$
Principal part of monthly payment $ = M - Pr$.
So Principal left after first payment $ = P - (M - Pr) = P(1+r)-M$
...
0
The APR assumes a monthly payment, and is technically a "nominal annual rate, compounded monthly." What this means is that you divide the interest rate by $12$, and that is the monthly interest rate. If the loan amount is $P$, then the interest for the first month is $P\frac i{12}$ so at the end of one month, the borrower owes $P\left(1+\frac i{...
0
My answer is incorrect according to saulspatz, but I'll leave it up so people who thought it is the correct answer (I think a few people would think this) can see that it is not the correct answer...
This is how I think it works, but I could be wrong, so correct me if I'm wrong.
Initial loan £$2100$ with $2$% APR means: $2$% of £$2100 =$ £$42$ interest in ...
0
Let
$\begin{align}y = -2\arcsin\sqrt{1-x} &\Rightarrow \arcsin\sqrt{1-x} = \frac{-y}{2} \\&\Rightarrow 1-x = \sin^2\left(\frac{-y}{2}\right) = \frac{1-\cos(\pm y)}{2} ~~~(\text{ as cos is even })\\&\Rightarrow \cos(\pm y) = 2x-1 \\& \Rightarrow\sin(\frac{\pi}2\pm y) = 2x-1 \\ &\Rightarrow \pm y = \arcsin(2x-1) -\frac{\pi}{2} \\ &\...
2
A comment that got too long with some observations:
Assume wlog $abmn \ne 0$ as those cases are easy to consider and let $f(x)=a\sin mx +b \cos nx$; the range of $f$ is always an interval by continuity.
When $m/n$ is irrational the solution is straightforward as we can use Kronecker theorem to find $x$ st $\sin mx =\pm \frac{a}{|a|}(1-\epsilon) , \cos nx =\...
1
The problem is just after $$1.48595=\left(1+\frac{r}{2}\right)^{20}$$ You have now
$$\sqrt[20]{1.48595}=1+\frac{r}{2}\implies r=0.04$$
1
The wrong step is:
$$2(1.48595) = 2(1+\frac r2)^{20}$$
$$2.972 = (2+r)^{20}$$
Since $$(2+r)^{20} = (2(1+\frac r2))^{20} = 2^{20}(1+\frac r2)^{20} \ne 2(1+\frac r2)^{20}$$
Multiplying by $2$ in the previous step is unnecessary; we can take the root right then and there.
2
What you quote is not the full solution, is it?
The point $(u,v)$ on the circle satisfies
$$u^2+v^2=1$$
which we just rewrite in equivalent form:
$$\frac{(5u)^2}{25}+\frac{(3v)^2}{9}=1$$
and from this we see that the point $(x,y)=(5u,3v)$, which belongs to the ellipse, satisfies
$$\frac{x^2}{25}+\frac{y^2}{9}=1$$
and so this is the equation of the ellipse.
...
1
Too long for a comment;
Very interested by @am301's approach, I noticed that the $[0,2]$ Padé approximant built around $x=2n\pi$ is
$$e^{e^{\cos (x)}}=\frac{2 e^e}{e (x-2n \pi )^2+2}$$ which is close to the Lorentzian-shape function proposed in @am301's answer.
This approximation would lead to the estimates
$$\{2.12,5.07,7.19,12.14,12.92\}$$ which are not ...
0
Is this a joke? Whatever value you plug in $p$, the result of $$f(a,b,c)=\frac{1}{12} \left(a b c+\frac{1}{a b c}\right)$$
is always $$p \left(5 p^2-1\right)$$
Prime or not prime, it is not a "minimum" or a "maximum". It is the constant value of $f(a,b,c)$, if $a,b,c$ satisfy $$\frac{a b+1}{3 b}=p,\frac{b c+1}{4 c}=p,\frac{a c+1}{5 a}=p$$
0
The units digit of $383^1$ is $3$.
The units digit of $383^2$ is $9$ ($3 \times 3 = 9$).
The units digit of $383^3$ is $7$ ($3 \times 9 = 27$).
The units digit of $383^4$ is $1$ ($3 \times 7 = 21$).
Calculating,
$\; 383^4 \equiv 721 \pmod{1000}$
Calculating,
$\; 383^{101} \equiv 383 \cdot 721^{25} \equiv 383 \cdot (1 + 720)^{25} \equiv 383 \cdot (1 + 2^4 \...
4
To bound an area of $3$ square metres the least possible curve length is such that the curve is a circle, by the isoperimetric inequality. This circle will have a radius of $\sqrt{\frac3\pi}$ metres and a length of $2\pi\sqrt{\frac3\pi}=6.1399\dots$ metres. Since we only have $6$ metres to work with, the task is impossible.
For the same reason, we cannot ...
0
The units digit of $17^1$ is $7$.
The units digit of $17^2$ is $9$ ($7 \times 7 = 49$).
The units digit of $17^3$ is $3$ ($7 \times 9 = 63$).
The units digit of $17^4$ is $1$ ($7 \times 3 = 21$).
Calculating,
$\; 17^4 = 289 \cdot 289 = (280+9)^2 = 280^2 + 18 \cdot 280 + 81 \equiv$
$\quad\quad\quad 400 + 800 + 600 + 640 + 81 \equiv 521 \pmod{1000}$
...
0
Let $\;x=e^{-0.0001n}\;$.
The equation
$e^{0.0002n} + e^{0.0003n} = 0.01$
turns into
$x^3-100x-100=0$.
The only positive solution is
$x=\dfrac{20}{3}\sqrt{3}\cos\left(\dfrac{\theta}{3}\right)\simeq10.46680532$
where $\;\theta=\cos^{-1}\left(\dfrac{3\sqrt{3}}{20}\right)\simeq74.94135263^\text{o}.$
Hence
$n=-10000\ln\left[\dfrac{20}{3}\sqrt{3}\cos\left(\dfrac{\...
1
Adding on @DatBoi's answer, if $t=0.1$, then
$$t^2+t^3=0.011$$ and we have a good approximation. Now using a Newton's iteration,
$$t'=t-\frac{t^2+t^3-0.01}{2t+3t^3}=0.1-\frac{0.001}{0.23}=0.09565\cdots$$
You can use a few more iterations, or use the general formula for the cubic equation. $t$ gives $n$.
2
HINT
Let $e^{0.0001n}=t$
$$t^2+t^3 = 0.01$$
1
The equation of a sine curve is
$$ y = A \sin \dfrac{2 \pi x}{\lambda}$$
So in your particular case amplitude and wavelengths are
$$A=-3, \lambda= 8$$
Your graph is negative in start amplitude compared to standard sine of positive initial slope, but is not an inverse sine graph. Inverse graphs have $(x,y)$ axes interchanged.
1
If $b\ne 0$ then $b^2>0$ so $(b^2C\le b^2D\iff C\le D)$ for any $C,D.$
So with $C=1$ and $D=2a^2+b^2,$ if $b\ne 0$ then $$a^4+b^2\le (a^2+b^2)^2 \iff a^4+b^2\le a^4+2a^2b^2+b^4 \iff$$ $$\iff b^2\le 2a^2b^2+b^4\iff$$ $$\iff b^2(1)\le b^2(2a^2+b^2)\iff$$ $$\iff 1\le 2a^2+b^2.$$ The last inequality above cannot be true for every $a,b>0.$ E.g. if $a=b=1/...
1
$y = \sin(x)$ gets a value $1$ at $x=\frac{\pi}{2}$
$ \Rightarrow y = -3\sin(x)$ gets a value $-3$ at $x=\frac{\pi}{2}$
$ \Rightarrow y = -3\sin(ax)$ gets a value $-3$ at $ax=\frac{\pi}{2}$
We have $y=-3$ at $x = 2$.
So, $ax = \frac{\pi}{2} \Rightarrow 2a = \frac{\pi}{2} \Rightarrow\boxed{a = \frac{1}{4}\pi}$
The equation is $\boxed{ y = -3\sin(ax)=-3\sin\...
0
Imagine there is a solution $(a,x)$. If $x\neq0$, the equation is equivalent to $\frac{x}{x+1}=a$. Now if $x\neq-1$, the equation is equivalent to $x=a(x+1)=ax+a$. From which, $$x(1-a)=a\tag{1}$$ And now if $a\neq1$, $$x=\frac{a}{1-a}\tag{2}$$
So as long as $a\neq1$ and also $a$ is not such that equation (2) implies $x=0$ or $x=-1$, there is a solution. Is ...
0
CW answer. It means Country Western.
Note how the column with coefficients of $p^2$ has the integer part, 2, 5, 9, 14, 20, 27, 35, 44, is quadratic in $n$
parisize = 4000000, primelimit = 500000
? f1 = s
%1 = s
? f2 = s^2 - 2 * p
%2 = s^2 - 2*p
? f3 = s * f2 - p * f1; f1 = f2; f2 = f3
%3 = s^3 - 3*p*s
? f3 = s * f2 - p * f1; f1 = f2; f2 = f3
%4 = s^4 - 4*p*...
2
$$\bigg( \left(\frac{1}{2} \right)^2+ \left(\frac{1}{2} \right)^2 \bigg)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}$$
but
$$\bigg( \left(\frac{1}{2}\right)^4+ \left(\frac{1}{2}\right)\bigg)^2= \left(\frac{1}{2}\right)^4+\frac{1}{4} > \frac{1}{4}= \bigg( \left(\frac{1}{2} \right)^2+ \left(\frac{1}{2}\right)^2 \bigg)^2$$
1
$a=0, b=\frac 1 2 $ is a counter-example.
If $0$ is not allowed just take $a>0$ but small enough. Eg: $a=\frac 1 {10}$ will do.
1
HINT
You can simply expand the RHS and subtract the LHS:
\begin{align*}
(a^{2} + b^{2})^{2} = a^{4} + 2a^{2}b^{2} + b^{4} \Rightarrow (a^{2} + b^{2})^{2} - a^{4} - b^{2} = 2a^{2}b^{2} + b^{4} - b^{2} = b^{2}(2a^{2} + b^{2} - 1)
\end{align*}
When the last expression is nonnegative?
2
Setting $p=-1$, these are Lucas polynomials:
$$L_n(x) = 2^{-n} \left((x-\sqrt{x^2+4})^n + (x+\sqrt{x^2+4})^n \right)$$
Your coefficients correspond to OEIS A034807 and may be expressed in terms of Lucas numbers.
Thanks to Steven Stadnicki for him comment below pointing out that the $p$ powers can be inferred, so setting $p=-1$ is WLOG.
For completion:
By ...
0
You can find recursively as:
$$
f_{n+1}(s,p)=s f_n(s,p)-p f_{n-1}(s,p)
$$
0
When you add $1/2 \,\mathrm{L}$ bleach solution to $12 \,\mathrm{L}$ of water (and mix thoroughly), you have a new solution of $25/2 \,\mathrm{L}$. The concentration of bleach is that solution is $\frac{1/2}{25/2} = 1/25$ of the original concentration and that is the concentration carried by the nappies from the first rinse to the second rinse. You should ...
0
After the wash and wringing, the nappies contain $\frac 12$ litre of strong solution. In the first rinse you add $12$ litres of water, mix that with what was in the nappies to start, and wring them out. That leaves $\frac 12$ litre of the solution in the nappies. What is the concentration of that solution compared to the one they were washed in? The ...
1
Let $x=\frac{p}{q}$ be a solution in lowest terms, i.e., $\gcd(p,q)=1$, with $q>0$. By substituting this into the equation for $x$, we get
$$\left(\frac{p}{q}\right)^2-\frac{p}{q}=1\Leftrightarrow p^2-pq=q^2\Leftrightarrow p^2=q(p+q).$$
That is, $q$ divides $p^2$. But we assumed $\gcd(p,q)=1$ and thus $q$ must be $1$. Hence, we get
$$p^2=p+1\...
0
Hint: The number in base $10$ is $8285$. Remember that this means the number can be expressed as
$$5\cdot 10^0 + 8\cdot 10^1 + 2\cdot 10^2 + 8\cdot 10^3 = 8285$$
You are told that, in base $b$, the number can be expressed as
$$4\cdot b^0 + 0\cdot b^1 + 1\cdot b^2 = 8285$$
The left side is just $b^2 + 4$. So you want to solve
$b^2 + 4 = 8285$ for $b$.
1
Since $(104)_b = 1\cdot b^2 + 0 \cdot b + 4$ you get the equation
$$b^2+4 = 8285,$$
which you should be able to solve.
2
Hint: In the number $104_b$, the final digit represents the units, the middle digit represents the number of "$b$"s, and the leftmost digit represents the number of "$b^2$"s. (Compare to base $10$, where these would be units, tens, and hundreds.) In other words, you have
$$b^2+4 = 8285.$$
Now solve for $b$.
2
Using the quadratic formula on $x^2-x=1$ we have
$$x = \frac {1\pm\sqrt5}2$$
If $x$ can be rational, so can $2x-1 = \pm \sqrt 5$. But the irrationality of $\pm\sqrt 5$ can be proved by infinite descent, for instance.
0
The only solutions for x are GoldenRatio and 1-GoldenRatio, which are both irrational.
GoldenRatio =( 1+Sqrt(5) ) /2
0
Write $x=a/b$ in its lowest terms so$$x=\frac{1}{x-1}=\frac{b}{a-b},$$forcing a contradiction if $|x|>1$. See if you can do the same for $|x|<1$.
2
By Ruffini's Theorem (or Rational root Theorem), all possible rational solutions of $$x^2-x-1=0$$ are those whose numerator divides 1 (the independent term) and whose denominator divides 1 (the leading coefficient).
So the only possible rational solutions are $\pm1$ and, trivially, they aren't.
0
Yes, this is valid. You are just traversing the same set of $(i,j)$ that constitute your index set in a different order. But you should note that
This would be true for any $a_{i,j}$, not just $a_{i,j} = i \cdot j$.
You couldn't automatically do this if your index set was infinite.
As @CameronWilliams' comment points out, while it is valid it may or may not ...
1
Let the bottle have weight $x$ and liquid when full have weight $y$. We get $x+\frac{3}{4}y = \frac{3}{2}$ and $\frac{3}{4} y = x + \frac{3}{4}$. Solving gives us $x = \frac{3}{8}$ and $ y = \frac{3}{2}$. If the bottle is completely filled, we have the weight as $x+y = \frac{15}{8}$.
3
For the most general case in which the base doesn't even need to be a polygon, it is an application of Cavalieri's Principle. Imagine a pyramid with a blob-shaped base and a square pyramid next to it with the same height and whose base has the same area as that of the first pyramid. Since any level cross-section of the two pyramids would have to have the ...
2
Any polygon can be decomposed into triangles, giving $A = A_1 + \cdots + A_n$.
Thus,
$$\begin{align}
V &= V_1 + \cdots + V_n\\
&=\frac13A_1h + \cdots + \frac13A_nh\\
&=\frac13(A_1 + \cdots + A_n)h\\
&=\frac13Ah.
\end{align}$$
For the base case, you can decompose a prism like so. To see that the three pyramids have the same volume, note that ...
4
We observe that $12:8=3:2$
So
$\begin{array}
112x^3 - 8x^2 -3x + 2 \\
&=&4x^2(3x-2) - (3x-2) \\
&=&(3x-2)(4x^2 - 1)\\
&=&(3x-2)((2x)^2 - 1^2)\\
&=&(3x-2)(2x+1)(2x-1) \\
\end{array}$
1
I doubt this is the intended solution but here it is anyway.
The main tool is the discriminant. Wikipedia says:
The discriminant [of a quartic] is zero if and only if two or more roots are equal. If the coefficients are real numbers and the discriminant is negative, then there are two real roots and two complex conjugate roots. Likewise, if the discriminant ...
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