New answers tagged algebra-precalculus
3
$$\frac {e^{2x}}{e^x+1} = \frac {e^{2x}+e^x-e^x}{e^x+1}=$$
$$\frac {e^{x}(e^x+1)-e^x}{e^x+1}=e^x-\frac {e^x}{e^x+1}$$
0
$$I=\int \frac{e^{2x}}{1+e^x}dx=\int \frac{e^{2x}-1+1}{1+e^x}dx\\
\\
=\int \frac{e^{2x}-1}{1+e^x}dx+\int \frac{1}{e^x+1}dx\\
\\
=\int (e^x-1)dx+\int \frac{e^{-x}}{1+e^{-x}}dx\\
\\
=e^x-x-\ln(1+e^{-x})+c$$
2
To get the result in your book, note that
$$e^{2x}=e^x e^x=e^x(e^x+1-1)=e^x(e^x+1)-e^x$$
3
I think it is simpler to first use the substitution
$$u=e^x$$
to obtain the integral of a rational function:
$$\int\dfrac{u}{1+u}\,du$$
and then use division of polynomials.
0
Just think of $e^{2x}$ as $(e^x)^2$. Now your integrand takes the form $\frac{u^2}{1+u}$, so you can long divide those polynomials in $u$ as usual. That's how they got that decomposition. This does not handle the differential terms as in substitution. That would have to happen after the algebraic moves.
0
We have
$$\begin{cases}
\dfrac{2}{x-3} \gt \dfrac{5}{x+6} \\
\dfrac{1}{3} \lt \dfrac{1}{x-2}
\end{cases}$$
We start by solving each inequality by itself. I will assume you can already do this, so we now have
$$\begin{cases}x<-6,\ 3<x<9\\
2<x<5\end{cases}$$
If we imagine drawing these on a number line then we get the following:
-6 -5 -...
3
Let us consider the inequalties
$(1) \quad \dfrac{2}{x-3} \gt \dfrac{5}{x+6} $
and
$(2) \quad \dfrac{1}{3} \lt \dfrac{1}{x-2} $.
Then compute the set $L_1$ of the solutions of $(1)$ and the set $L_2$ of the solutions of $(2)$.
The set of solutions of
$\left\{
\begin{aligned}
\dfrac{2}{x-3} \gt \dfrac{5}{x+6} \\
\dfrac{1}{3} \lt \dfrac{1}{x-2}
\end{...
3
One solution to this can be found when you input the equation in wolfram alpha:
It appears that wolfram alpha considers negative values outside of the domain of the principal root. If you click on "Use the real-valued root" you will get the expected answer.
2
Let us consider the second inequality. There are two cases here:
$x>2$: then$$\frac13<\frac1{x-2}\iff\frac{x-2}3<1\iff x-2<\frac13;$$
$x<2$: then$$\frac13<\frac1{x-2}\iff\frac{x-2}3>1\iff x-2>\frac13.$$
Now, you do the same thing with the other inequality. Finally, you take the intersection of your answers.
2
The first it's $$\frac{2}{x-3}-\frac{5}{x+6}>0$$ or
$$\frac{x-9}{(x-3)(x+6)}<0,$$
which gives $$(-\infty,-6)\cup(3,9).$$
Solve the second inequality by the same way.
Can you end it now?
I got the following answer:
$$(3,5)$$
1
WA always thinks that $x^{\frac{1}{3}}=\sqrt[3]{x},$ but it's wrong.
By definition, the domain of $x^r$, where $r\in\mathbb Q$, is $(0,+\infty)$.
The domain of $\sqrt[n]{x}$, where $n\in\mathbb N$, $n$ is even and $n\geq2$, is $[0,+\infty)$;
The domain of $\sqrt[n]{x}$, where $n\in\mathbb N$, $n$ is odd and $n\geq3$, is $\mathbb R$.
0
Thr answer of Wolfram is not correct: in fact $\sqrt[3]{x}$ is defined $\forall x \in R$ (also negative or $0$). So in your case $\sqrt[3]{(x-1)^2(x+2)}$ is defined $\forall x \in R$
1
If you know an approximate of the solution, you can use the fixed-point method to find a less approximate (!) solution. Define $$\alpha_{n+1}=\sqrt[k+1]{{4.68-4.50\cos \alpha_{n}\over 1.23}\cdot \alpha_n^{k}}$$and find the limit. I tried $k=3$ and obtained $\approx 4.5166$. All you have to do is that using a scientific calculator press an approximate of ...
0
You can observe that
$$
\lvert\sin x\rvert=\sqrt{1-\cos^2x}=\sqrt{1-\frac{24^2}{25^2}}=\frac{7}{25}
$$
Since $\pi/2<x<\pi$, we conclude that $\sin x=7/25$.
Also
$$
\lvert\cos y\rvert=\sqrt{\frac{1}{1+\tan^2y}}=\frac{40}{41}
$$
and from $\pi<y<3\pi/2$ we conclude that $\cos y=-40/41$. Thus
$$
\sin y=\tan y\cos y=-9/41
$$
1
Your method is correct. Indeed, refer to the graphs:
$\hspace{1cm}$
$$\cos x=-\frac{24}{25}=\frac{\overbrace{-24}^{adjacent}}{25}, x\in \left(\frac{\pi}{2},\pi\right); \quad \tan y=\frac{9}{40}=\frac{\overbrace{-9}^{front}}{\underbrace{-40}_{adjacent}},y\in \left(\pi,\frac{3\pi}{2}\right).$$
Also note:
$$\cos x=-\cos(\pi -x)=-\frac{24}{25} \Rightarrow \cos ...
0
Christiane
$$C_x=1200\cos{\frac{\pi}{6}}=600\sqrt{3}N$$
$$C_y=1200\sin{\frac{\pi}{6}}=600N$$
Hayley
$$H_x=200\cos{\frac{7\pi}{6}}=-100\sqrt{3}N$$
$$H_x=200\sin{\frac{7\pi}{6}}=-100N$$
Since we want the ring not moving at all, then the sum of vectors in $x$-axis and $y$-axis should be equal to 0
$$B_x=-C_x-H_x=-500\sqrt{3}N$$
$$B_y=-C_y-H_y=-500N$$
Because ...
2
Let $\sqrt{x}=a$
$$a^2-8a+7=0$$
Find the factor
$$(a-7)(a-1)=0$$
Then you have
$$a=7\text{ and }a=1$$
Remember that $a=\sqrt{x}$, then
$$x=1 \text{ or } x=49$$
0
Hint: This follows from Gauß's lemma .
0
Assume $r=\frac{q}{p}$ is rational and f is monic and $f(x)=x^n+a_{n-1}x^{n-1}+...+a_1 x+a_0$
so we have $0=p^{n-1} f(r) = \frac{q^n}{p}+a_{n-1} q^{n-1}+a_{n-2}q^{n-2}p+...+a_1 qp^{n-2}+a_0 p^{n-1}=0$
So fraction $\frac{q^n}{p}$ should be equal to integer $-(a_{n-1} q^{n-1}+a_{n-2}q^{n-2}p+...+a_1 qp^{n-2}+a_0 p^{n-1})$ which means $p=1$
1
hint
It can be written as
$$8x-7x-8\sqrt{x}+7=$$
$$8\sqrt{x}(\sqrt{x}-1)-7(\sqrt{x}-1)(\sqrt{x}+1)=$$
$$(\sqrt{x}-1)\Bigl(8\sqrt{x}-7(\sqrt{x}+1)\Bigr)=$$
$$(\sqrt{x}-1)(\sqrt{x}-7)=0$$
thus...
2
I feel like it's possible for $r$ to be a rational, because a polynomial with integer coefficients can have rational roots.
Yes, you're right. The statement is false. For clearly, for example, $$(x-1)(2x-1)$$ belongs in your ring and has the factor $(x-1/2).$
1
$\displaystyle x-8\sqrt{x}+7=0$
or, $\displaystyle (\sqrt{x} -1)(\sqrt{x} -7)=0$
Hence, $x = 1$ and $49$
1
Why did things get compilicated?
If you replace $x$ with $y^2$ and $\sqrt x$ with $y$ you should get
$y^2 - 8y +7 =0$ and that should not be complicated.
========
Factor or use the quadratic formula so
$(y - 7)(y-1) = 0$ so $y = 7$ or $y=1$.
Or $y = \frac {-(-8) \pm \sqrt {64-4*7*1}}2 = \frac {8\pm {36}}2 = \frac {8\pm 6}2=4\pm 3=1,7$; so $y=7$ or $y =...
4
If
$x - 8\sqrt x + 7 = 0, \tag 1$
then
$x + 7 = 8\sqrt x; \tag 2$
then
$x^2 + 14 x + 49 = (x + 7)^2 = (8\sqrt x)^2 = 64x; \tag 3$
thus
$x^2 - 50x + 49 = 0, \tag 4$
which factors as
$(x - 1)(x - 49) =
x^2 - 50x + 49 = 0; \tag 5$
thus
$x = 1 \; \text{or} \; x = 49; \tag 6$
it is now a simple matter to check that $1$, $49$ obey (1).
4
Isolate the square root, then square both sides:
\begin{align*}
x-8\sqrt x+7&=0\\
\implies x+7&=8\sqrt x\\
\implies (x+7)^2 &= 64 x\\
\implies x^2 -50x + 49&=0\\
\implies (x-49)(x-1) &=0\\
\implies x = 49 \text{ or } x&=1,
\end{align*}
which are the two solutions.
7
Let $y=\sqrt{x}$, therefore $y^2=x$
$y^2-8y+7=0$ therefore $(y-7)(y-1)=0$ hence $y=1,7$.
Therefore $x=1,49$. These are indeed the only solutions.
1
You could have nonreal and complex coefficient for A and B. They're just numbers!
$$ \frac{x-3}{x^2-7}=\frac{A}{x-\sqrt{7}}+\frac{B}{x+\sqrt{7}}$$
So by equality we have:
$$ \frac{x-3}{x^2-7}=\frac{A(x+\sqrt{7})+B(x-\sqrt{7})}{(x+\sqrt{7})(x-\sqrt{7})}$$
S0:
$$A+B=1 \; and \; (A-B)\sqrt{(7)}=-3$$
Finally :
$$A=\frac{\sqrt{7}-3}{2\sqrt{7}}$$
$$B=\frac{\...
1
Yes, you can do it, provided you don't mind using irrational real numbers. That is, if $a$ is positive, then you have that $$x^2-a=(x-\sqrt a)(x+\sqrt a).$$
3
There are several ways to define $x^y$ for real $y$ and $x>0$. Here are three popular variations:
$x^y=\exp(y\ln x)$, where $\exp$ is defined through, say, a power series
$x^y=\lim_{n\to\infty}x^{y_n}$ where $y_n$ is a sequence of rationals converging to $y$, and $x^{y_n}$ is defined through roots and repeated multiplication
The function $y\mapsto x^y$ ...
3
One can also define $2^{\sqrt{3}}$ as follows. First, $2^k$ for integral $k$ is clear. If $k>0$ is an integer, it is clear how to define $2^{1/k}$: as the unique positive real $k^{\mathrm{th}}$ root of $2$. Then, for $r$ any positive real, choose a sequence $a_n = \frac{p_n}{q_n}$, where $p_n, q_n$ are integers, converging to $r$. Then $2^r:=\lim_{n\to\...
0
The meaning is $e^{\ln(2)\sqrt 3}=\exp(\ln(2)\sqrt 3)$. The exponential is a well defined function.
0
$|x+y-1| + |2x + y + 1|=1$
As usual, you can do this in cases depending on whether $x+y-1$ and $2x+y+1$ are great or equal to zero, or less than zero.
The trick is to realize that $x + y -1 = 0$ and $2x+y+1$ equal zero are two intersecting lines that cut the plane into four regions and each region will represent one of the four cases.
For instance. The ...
0
If you know already a bit of linear algebra you may proceed as follows:
You surely can draw $|u|+|v|=1$.
Now, consider the mapping
$$\begin{pmatrix}u \\ v\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 2 & 1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}+\begin{pmatrix}-1 \\ 1\end{pmatrix}$$
The inverse mapping (just solve for $\begin{pmatrix}x \\ y\end{...
1
From the denominator of the fraction $\frac{\frac{2-b}{2b}}{b-2}$ pick up $-1$ and obtain: $ \frac{\frac{2-b}{2b}}{-(2-b)}$. Now rewrite the fraction as: $\frac{2-b}{2b}\cdot \frac{-1}{2-b}$. $2-b$ simplifies and you have: $-\frac{1}{2b}$.
2
$\displaystyle \frac{\frac{2-b}{2b}}{b-2}$
$\displaystyle = \left(\frac{2-b}{2b} \right)\left(\frac{1}{b-2}\right)$
$\displaystyle = \left(\frac{(-1)(b-2)}{2b} \right)\frac{1}{(b-2)}$
Now cancel $(b-2)$ from the numerator and the denominator to get
$\displaystyle -\frac{1}{2b} $
Is it clear now? Please let me know.
1
When you have a fraction, you often simplify it by multiplying it by $1$ in the form of $\frac aa$ for some convenient $a$. Here you have a fraction with $ \frac {2-b}{2b}$ in the numerator and $b-2$ in the denominator. A convenient $a$ is $\frac 1{b-2}$. Then we have
$$\frac {\frac {2-b}{2b}}{b-2}=\frac {\frac {2-b}{2b}}{b-2}\cdot \dfrac{\left(\frac 1{b-...
0
The equation: $$ A |L_1| +B |L_2| =C $$, if $L_1,L_2$ are non-paralal lines, and $A,B,C >0$, represents a parallelogram with $L_1$ and $L_2$ as diagonals and their point of intersection is the center of the 'gram.
If $L_1$ and $L_2$ are perpendicular it is square or rhombus. If $L_1,L_2$ are non-perpendicular it is rectangle or a parallelogram. Where
...
1
You have to distinguish several cases:
Assuming $$x+y-1\geq 0$$ and $$2x+y+1\geq 0$$then we have$$2y+3x=1$$ or $$y=-\frac{3}{2}x+\frac{1}{2}$$
In the case of $$x+y-1<0$$ and $$2x+y+1\geq 0$$ you will get
$$x=-1$$
Can you finish?
1
Assuming you mean:
Fix positive natural numbers $s,t$. Let $(x_0,y_0)$ be the unique pair such that $0<x<1$ and
$$\begin{align}
y_0 &= x_0^t \\
y_0&= (1-x_0)^s \text{.}
\end{align}$$
Let $(x,y)$ be the unique pair such that $0<x<1$ and
$$\begin{align}
y&=\exp\left(\tfrac{s}{\ln x}\right)\\
y&=\exp\left(\tfrac{t}{\ln (1-x)}\right)\...
1
The axioms used for this derivation don't assume that
$$1 + (1+1) + (1+1+1) + (1+1+1+1) + \dots = 1 + (1 + (1 + (1 + \dots$$
In fact the freedom to arbitrarily regroup terms is not assumed for series in general, let alone tricky things like this.
0
Benjamin's application of force must be equal and opposite to the combined force of Christine and Hayley.
$C = (1200 \cos 30^\circ, 1200\sin 30^\circ) = (600\sqrt 3, 600)\\
H = (200\cos 210^\circ, 200\sin 210^\circ) = (-100\sqrt 3, -100)$
Hayley and Christine are pulling in opposite directions! Ben is pulling in the same direction as Hayley, and pulling ...
1
You have done the problem the right way, but made two crucial arithmetic errors:
$$1200\cos 30^{\circ} - 200 \cos 30^{\circ} = 1000\cos 30^{\circ} = 500\sqrt{3},$$
and similarly for the other set of equations. (You multiplied $\cos 30^{\circ}$ by itself for some reason.)
If you make those changes and rework the problem, you should get an answer of $1000$ N ...
-1
So you are indeed trying to find integer solutions to the equation
$$mn^3 - m^3n = D$$
It looks like you've flipped the sign of $n_1$, presumably because the associated root is always negative (if $m$ and $D$ are both positive, this equation will always have exactly one negative root, with the other two roots either positive or complex conjugates). Since you ...
1
Update: suppose you want to create a curve from $(0,a)$ to $(b,0)$.
Then the following will do:
$$
\varphi_{a,b}(x)
:= \frac{a}{b} \frac{b - x}{1 + ax}
$$
for all $a,b > 0$.
To expand on @J.W. Tanner's answer: For any $a > 0$
$$
f_a(x) := \frac{a(1 - x)}{a + x}
$$
fulfills the condition:
We have
$f_{a}(1) = \frac{a(1 - 1)}{a + 1} = \frac{0}{a + 1} = ...
1
$$y=\dfrac2{x+1}-1=\dfrac{1-x}{1+x}$$ defines a curve that has $y$-intercept $1$ and $x$-intercept $1$ .
(I started with $y=\dfrac1x$ and added $1$ in the denominator to make the $y$-intercept $1$.
But that has no $x$-intercept, so I multiplied by $2$ and subtracted $1,$
to get $x$-intercept $1$ while maintaining $y$-intercept $1.)$
2
If you graph the functions (should be easy to do by hand. If you don't know how to graph the sine function, learn here), a solution is where the graphs intersect.
Where c = 0:
There is one solution between 0 and $\pi$. This proves option 3 and option 2 correct.
Where c = 1:
There are 3 solutions, two between $-\pi$ and 0 and one between 0 and $\pi$. This ...
2
Think of (cos(t), sin(t)) as (x,y) coordinates on the unit circle.
I would visualize the point on Quadrant 1, and see what happened next.
$$\begin{matrix}
t → t+\pi & (x,y) → (-x, -y) & \text{Quadrant 3}\cr
t → -t & (x,y) → (x, -y) & \text{Quadrant 4}\cr
t → t+\frac{\pi}{2} & (x,y) → (-y, x) & \text{Quadrant 2}\cr
t → -t+\frac{\pi}{2}...
1
If one doesn't count multiplicities of roots
$$
f(x)=36x^4-76x^3+42x^2+1
$$
has one point at which $f(x)=1$, two points at which $f(x)=2$, three points at which $f(x)=3$, and four points at which $f(x)=4$.
4
Hints:
For (a), use the fact that $\sin(t+π)=-\sin t$ and $\cos(t+π)=-\cos t$ as well.
For (b), you need to know that $\sin$ is odd and $\cos$ is even. Thus, $\sin(-t)=-\sin t$ and $\cos(-t)=\cos t.$
For (c), use the facts that $\cos(t+π/2)=-\sin t$ and $\sin(t+π/2)=\cos t.$
You should now be able to do (d) by applying the facts I suggested in (c) and (b)...
10
Note that $(2c)!\sigma(c)=\sum_{n=0}^{c-1}\binom{2c}{2n+1}$ while $(2c)!\gamma(c)=\sum_{m=0}^c\binom{2c}{2m}$, so it reduces to the famous result that even-sized subsets of a given nonempty set of finite size are exactly as numerous as the odd-sized subsets.
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