New answers tagged algebra-precalculus
0
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Graph Transformation
$y = f(x+c)$ translates the graph by $-c$ in the x direction.
$y = f(x) + c$ translates the graph by $c$ in the y direction.
$y = f(cx)$ stretches the graph by a factor of $\frac{1}{c}
$ in the x ...
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0
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Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction
This answer uses the same sort of reasoning as given in the answer by zyx. I show that we can get a bit more out of the recursion relation satisfied by the function $S(n)$ and their generalizations ...
- 131
1
vote
Accepted
Evaluate ${\cos ^{ - 1}}\sqrt {1 - {x^2}} = $ for $x<0$
Because $x=\sin \theta <0$ and $\cos \theta = \sqrt{1-x^2} >0$, angle $\theta$ is in the fourth quadrant. Thus, $\cos^{-1} |\cos \theta|=-\theta$
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Why am I getting wrong solution to the system $2x+6y-3z=10$, $5x+2y-1z=12$?
Your solution strategy is not wrong, but you go astray before reaching your target.
Given the linear equation system $(1)\:\&\:(2),\,$ you are building the equation
$$-3x+4y-2z \:=\: -2 \tag{3) = (...
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$9\{x\}=2x+[x]$, how many values of $x$?
Let $x=n+q, n \in I, q\in [0,1)$
Then the eq becomes
$9q=2n+2q+n$ so $0<q=3n/7 <1$ are the possible values of n are 0,1,2.
The corresponding of q are $0, 3/7, 6/7$. Finally, $x=0, 10/7,20/7$ ...
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5
votes
Accepted
$9\{x\}=2x+[x]$, how many values of $x$?
For any $x\in\Bbb{R}$,we know
$x=\{x\}+\lfloor x \rfloor$$ \ \text{ and }
\ 0\le \{x\}<1$
$$9\{x\}=2x+\lfloor x \rfloor$$
$\Rightarrow 9\{x\}=2\{x\}+ 2\lfloor x \rfloor + \lfloor x \rfloor$
$\...
- 10k
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Why am I getting wrong solution to the system $2x+6y-3z=10$, $5x+2y-1z=12$?
\begin{align}
2x+6y-3z&=10 \tag 1 \\
5x+2y-1z&=12 \tag 2
\end{align}
Matrix form: $AX=b$
$A=\begin{pmatrix} 2 & 6&-3\\5&2&-1\end{pmatrix}$
$X=\begin{pmatrix}x\\y\\z\end{pmatrix}...
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Solve the equation $6(3^{2x})=2^{(x+1)}$
is it possible to put the two back in the exponent of the three and make the three nine? What I mean:$$2x \ln 3 = x \ln 9$$
Yes, this step is correct.
For the right hand side, would $x \ln 2 + \ln 2 ...
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Solve the equation $6(3^{2x})=2^{(x+1)}$
Another way to approach it for the sake of curiosity:
\begin{align*}
6\cdot3^{2x} = 2^{x+1} & \Longleftrightarrow 3\cdot 9^{x} = 2^{x}\\\\
& \Longleftrightarrow \left(\frac{9}{2}\right)^{x} = \...
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Is there a variant of the dot-product operation that returns $\frac{a_1}{b_1} + \frac{a_2}{b_2}$ from vectors $[a_1,a_2]$ and $[b_1, b_2]$?
To answer the asked question: NO. While there are many inverse simplification formulas that exist, none can be resolved to fulfill your requested outcome for all input values.
I agree that the ...
1
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Solve the equation $6(3^{2x})=2^{(x+1)}$
The answer to your first question is yes: $$2x \ln 3 = x \ln 3^2 = x \ln 9.$$ The answer to your second question is no: $$x \ln 2 + \ln 2 \ne x \ln 4,$$ because if this were true, it would ...
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Find the sum of radicals without squaring ?Is that impossible?
In this answer I tried to generalize the answer given above.
Generalization:
Let,
$$\sqrt {a\pm\sqrt b}=\sqrt m\pm\sqrt n,\,\,\,m\ge n$$
and
$$A=\sqrt {a+\sqrt b}+\sqrt {a-\sqrt b}=2\sqrt m$$
Then we ...
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0
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Find the sum of radicals without squaring ?Is that impossible?
Write,
$$\sqrt {3\pm\sqrt 5}=\sqrt a\pm\sqrt b,\, a\ge b$$
and we obtain
$$A=\sqrt {3+\sqrt 5}+\sqrt {3-\sqrt 5}=2\sqrt a$$
Then, we want to find ratinal $a,b$ such that:
$$\sqrt {3\pm\sqrt 5}=\sqrt a\...
- 1,287
4
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Find the sum of radicals without squaring ?Is that impossible?
A polynomial approach.
Note that $a_1,a_2=\sqrt{3\pm\sqrt 5}$ are two of the roots of $x^4-6x^2+4=0.$ The other roots are $-a_1,-a_2.$
So $$x^4-6x^2+4=(x-a_1)(x-a_2)(x+a_1)(x+a_2).$$
Now, $a_1a_2=\...
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1
vote
Accepted
$\displaystyle \frac{h_a}{l_a} + \frac{h_b}{l_b} + \frac{h_c}{l_c} \leq \frac{\sqrt{3(p^2 + r^2 - 8Rr)}}{2R} + \sqrt{3\frac{2R - r}{2R}}$
Notation
To make formulas more concise, I'll use some notation.
Angles of the triangle are $\alpha$, $\beta$, $\gamma$.
Sides of the triangle are $\bar{a}$, $\bar{b}$, $\bar{c}$. (Mnemonic: line over ...
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Find the sum of radicals without squaring ?Is that impossible?
Use Vieta's formulas:
Let, $$x_1=\sqrt {3-\sqrt 5},\,\,\,x_2=\sqrt {3+\sqrt 5}$$
$$x^2-px+2=0\\p=\frac{x^2+2}{x}$$
and we have,
$$p=\frac{5+\sqrt 5}{\sqrt {3+\sqrt 5}}=\frac{5-\sqrt 5}{\sqrt {3-\sqrt ...
- 1,287
15
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Find the sum of radicals without squaring ?Is that impossible?
Trick:
$$
\sqrt {3-\sqrt{5}}=\frac{\sqrt {6-2\sqrt{5}}}{\sqrt 2}=\frac{\sqrt {(\sqrt 5)^2-2\sqrt{5}+1}}{\sqrt 2}=\frac{\sqrt {(\sqrt 5-1)^2}}{\sqrt 2}
=\frac{\sqrt 5-1}{\sqrt 2}.
$$
Similarly we see $\...
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Find the rational number of a, b, c, solving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}+ \sqrt[3]{b}+\sqrt[3]{c}$
A systematic approach is to apply the denesting formula
$$\sqrt[3]{\sqrt[3]{A}-B} = \sqrt[3]{x_1}+ \sqrt[3]{x_2 }+ \sqrt[3]{x_3 } $$
where $x_1$, $x_2$ and $x_3$ are the roots of the cubic equation
$$...
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2
votes
Accepted
Why is $\arctan \left(\frac{\cos(\frac{\alpha}{2}t)-\cos(\frac{\alpha}{2})}{\sin(\frac{\alpha}{2}t)-t*\sin(\frac{\alpha}{2})}\right)$ a line?
Assuming that additional $t$ in the end of your denominator is a typo, let
$$f(t) = \arctan\left( \frac{\cos(at)-\cos(a)}{\sin(at)-\sin(a)} \right)\tag 1$$
where $a=\alpha/2$.
The addition theorems ...
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2
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When can we discuss the 'truth value' of a statement?
We don't assign a value to $x$ to ensure that $x+x = 2x$ is true because like you said, the statement is true for all $x$ (assuming we are working with basic algebra we learn in grade school).
It's ...
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When can we discuss the 'truth value' of a statement?
Suppose we have a formula $\phi(x)$ with some “unassigned variable” $x$. Formally, this $x$ is called a “free variable”. For example, your formula $\phi(x) : \equiv x + x= 2x$. Note I am using “$: \...
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analytical proof for a property of parabolas
The parabola has cartesian equation $y^2=4ax$. We have
$P=(at^2,2at)$, the focus is $F=(a,0)$, note the point $K=(ka,0)$.
Line $PQ: \dfrac{y}{x-a}=\dfrac{2t}{t^2-1}$ and
Line $PR: \dfrac{y}{x-ka}=\...
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1
vote
Accepted
Determine for what value(s) of m that the equation $ |2x-1| = mx+1$ has exactly one solution only.
CASE $\bf1$:$\ \ x>1/2$
$2x-1=mx+1$
$x=2/(2-m)$
So, $x=2/(2-m)$ AND $x>1/2$
$\Rightarrow 2/(2-m)>1/2$
$\Rightarrow m>-2, m\neq2$
$\Rightarrow m\in(-2,\infty)\sim2$
CASE $\bf2$: $\ \ x\...
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1
vote
Accepted
If a quadratic equation in $x\in R$ is $\lt 0$ then is $D\lt 0$ true?
The question as you quote it seems a little ill-posed to me. It should say something like: "Find all values $a$ such that $-3< \frac{x^{2}+ax-2}{x^{2}-x+1}<2$ for all real $x$."
If for ...
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2
votes
Accepted
Doubts over x power y power z
Notice that $a^{b^c}$ with no parenthesis is $a^{(b^c)}$ and it is different from ${(a^b)}^c$ which is $a^{bc}$.
Therefore $3^{3^{1/2}}\neq3^{3^/2}$
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Prove if $\sqrt{x+1}+\sqrt{y+2}+\sqrt{z+3}=\sqrt{y+1}+\sqrt{z+2}+\sqrt{x+3}=\sqrt{z+1}+\sqrt{x+2}+\sqrt{y+3}$, then $x=y=z$.
The solution of tehtmi is wonderful, and I have a similar approach.
For each parameter $t \in \{x, y, z\}$ and each $1 \leq i \leq 3$, let $t_i = \sqrt{t + i}$. For example $x_2 = \sqrt{x + 2}$. So we ...
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How do we show that $(10^{10})!$ is larger than $10^{10^{10}}$?
For integer valued $a$, $a! > a^{10}$ for $a\geq 21$; ($ a= 10^{10}\geq 21$) .
$$a! = a(a-1)(a-2)(a-3) .... (a-19)...1 > a*[(a-1)(a-19)] *[(a-2)(a-18)]*...*[(a-9)(a-10)] > a [(a-1)(a-19)]^9 \...
- 1,303
2
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How to make sense of graphing relationships?
If
3 dollars = 2 euros
then the proportionality constant is
$$
\frac{3}{2}\frac{\text{dollars}}{\text{euro}}
$$
which is just another way to write the number $1$.
Then the relationship is
$$
\text{...
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1
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How to make sense of graphing relationships?
Your equation 3x=2y is wrong, if x is the number of apples, put in x=3 and you know y=2 so you would have $3*3=2*2$ and you see it is wrong.
the relation of number of apples to number of oranges x/y=...
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-1
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How to make sense of graphing relationships?
You are misinterpreting what $3x=2y$ means. If x=(number of apples) and y=(number of oranges), then $3x=2y$ does not means that 3 apples are equivalent to 2 oranges. This tells you that 3 times the ...
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vote
Accepted
Find the largest set $A,B$ such that $f:A\rightarrow B, \, \,f(x)=x^x$ has an inverse
Yes, but some parts semm to be more complicated than necessary.
TL;DR: The largest interval to define an inverse on is $[1/e^{1/e},\infty)$.
Note: The question has been changed after I wrote the ...
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4
votes
Accepted
If $f(x) =ax^3+bx^2+cx+d$ is a cubic equation with roots $\alpha,\beta,\gamma.$ Is there a way to find $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha?$
You know the elementary symmetric polynomials evaluated at the roots (by the Vieta relations):
$$
\begin{aligned}s_1&=\alpha+\beta+\gamma=-b/a,\\
s_2&=\alpha\beta+\beta\gamma+\gamma\alpha=c/a,\...
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If $f(x) =ax^3+bx^2+cx+d$ is a cubic equation with roots $\alpha,\beta,\gamma.$ Is there a way to find $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha?$
Quantity is not-symmetric. This results in rather complex formula:
$$\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha=t_{1}\,t_{3}^2-{{b\,t_{3}^2}\over{3\,a}}+t_{2}^2\,t_{3}-{{2\,b\,
t_{2}\,t_{3}}\over{3\,...
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2
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If $f(x) =ax^3+bx^2+cx+d$ is a cubic equation with roots $\alpha,\beta,\gamma.$ Is there a way to find $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha?$
A function of two or more variables is said to be symmetric function if $f$ remains unaltered by an interchange of any two of it's variables.
If $\alpha, \beta, \gamma$ are roots of a cubic equation ...
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Why does $(2^x -1)/x$ not have a vertical asymptote?
mweiss gave already a nice explanation, but let me give a proof requiring as little calculus as I can.
Let $f:\mathbb{R} \setminus \{0\} \to \mathbb R$ be $f(x) = (2^x-1)/x$.
For function $f$ to have ...
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How to solve a function without Graphing the model
Well, one minute is one-sixtieth of an hour. So five minutes is five-sixtieths or $\frac{1}{12}$ of an hour.
If 9am is $h=0$, then 1:05pm is $h=4\frac{1}{12}=\frac{49}{12}$.
Plug in this $h$ to get ...
- 83
10
votes
Accepted
Solve $(x^3+1)=2\sqrt[3]{2x-1}$ algebraically?
Rearrange as $$\underbrace{\frac{x^3+1}{2}}_{f(x)}=\underbrace{\sqrt[3]{2x-1}}_{g(x)}$$
Since $f(x)$ is a bijective function on $\mathbb R$, it must have an inverse. But note that the inverse of $f(x)$...
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2
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Solve $(x^3+1)=2\sqrt[3]{2x-1}$ algebraically?
$$f(x)=x^9+3x^6+3x^3-16x+9$$
$$f(x)=0\Rightarrow (x-1)(x^2+x-1)(x^6+2x^4+2x^3+4x^2+2x+9)=0$$
where $$x^6+2x^4+2x^3+4x^2+2x+9=x^6+x^4+x^2(x+1)^2+2x^2+(x+1)^2+8>0$$ for $\forall x\in \mathbb{R}$, so ...
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1
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Why does $(2^x -1)/x$ not have a vertical asymptote?
This is a really wonderful question!
For small values of $x$, $2^x$ is approximately equal to $1 + (\ln 2)x$. This is not obvious (unless you know Calculus), but if you graph both functions and zoom ...
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1
vote
Accepted
I'm almost 100% sure that the answer is 60 and yet I got the question wrong.
$AB$ in this case means the length from point $A$ to point $B$. It doesn't mean the area of something like a sub-rectangle of the main rectangle, $A\times B=60$.
To get the length from $A$ to $B$ you ...
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2
votes
Accepted
Algebraic curves--Elementary doubts on their singular points, solutions and inclined asymptotes
"Points on the curve given by $f=0$" is totally correct. "The zero locus of $f$" works too. "The curve cut out by $f$" is also fine. There are also other choices.
Most ...
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1
vote
Accepted
Question on the natural logarithm laws
Yes. Just like the comment to your question, if we know two objects are equal, then the "reverse case" is also true.
Here is a more rigorous proof for fun: "Let $n \in \mathbb{R}$, $a \...
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0
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Is it valid to turn $i^n$ into $((−1)^n)^{1/2}$?
Your solution.1 is correct.
$i=i,~i^2=-1,~i^3=-i,~i^4=1$
$$i^{83}=i^3\cdot i^{80}=i^3=-i$$
Your solution.2 is wrong, because you need to specify the branch. If you take $-1=e^{i\pi}$, then $\sqrt{-1}=...
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1
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Question on the natural logarithm laws
Alternative approach:
Assume that $\log(x) = r \implies e^r = x.$
Then $\displaystyle e^{(3r)} = \left[e^r\right]^3 = x^3.$
Thus, $3r = \log(x^3).$
Thus $3 \times \log(x) = \log(x^3).$
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1
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Question on the natural logarithm laws
Let $x \in \mathbb{R}^+$. Since $3 \cdot log(x) = log(x^3)$, it follows that $log(x^3) = 3 \cdot log(x)$.
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How do we show that $(10^{10})!$ is larger than $10^{10^{10}}$?
Using algebra, you want to find when
$$(a^a)! \geq a^{a^a}$$ To make life easier because of the huge numbers, when is
$$f(a)=\log\Bigg[\frac{(a^a)!}{a^{a^a}}\Bigg] >0$$
By inspection or plotting, ...
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8
votes
How do we show that $(10^{10})!$ is larger than $10^{10^{10}}$?
$10^{10^{10}}$ is the product of $10^{10}$ factors of $10$ while the first $\frac 12 \cdot 10^{10}$ factors of $(10^{10})!$ are larger than $10^2$ so have a product larger than $10^{10^{10}}$.
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5
votes
How do we show that $(10^{10})!$ is larger than $10^{10^{10}}$?
$$\left(10^{10}\right)! = 100!\cdot\left(\prod\limits_{i=101}^{10^{10}}i\right) \gt 100!\cdot\left(\prod\limits_{i=101}^{10^{10}}100\right) = 100!\cdot 100^{10^{10}-100} = 100!\cdot 10^{2\cdot10^{10} -...
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6
votes
Finding the range of $\frac{5 \cos x-2 \sin ^{2} x+4 \sin x-3}{6|\cos x|+1}$
If $x\in[-\pi/2,\pi/2]$ then $|\cos x|=\cos x$ so
\begin{equation}
\frac{5 \cos x-2 \sin ^{2} x+4 \sin x-3}{6|\cos x|+1}
={5\cos x+5/6\over6\cos x+1}+{-2\sin^2 x+ 4\sin x -3\frac56\over6\cos x+1}\\
=5/...
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0
votes
perfect square special formula
Shouldn't the formula be: $$(a - b)^2 = a^2 - 2a\lvert b\rvert + b^2$$
No! Let $a=1, b=-1$ then if
$(a-b)^2=a^2-2a|b|+b^2$
then
$$2^2=(1-(-1))^2=(a-b)^2=1-2\times1\times|-1|+(-1)^2=0.$$
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