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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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  • 149
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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 ...
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1 vote
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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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  • 8,552
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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
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$9\{x\}=2x+[x]$, 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$$ $\Rightarrow 9\{x\}=2\{x\}+ 2\lfloor x \rfloor + \lfloor x \rfloor$ $\...
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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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1 vote

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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2 votes

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 vote

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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1 vote

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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  • 1,287
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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\...
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  • 1,287
4 votes

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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1 vote

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 ...
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  • 1,287
15 votes

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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  • 3,046
1 vote

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
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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 votes

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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1 vote

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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  • 2,006
1 vote

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
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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
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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
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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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  • 56
3 votes

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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0 votes

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 \...
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  • 1,303
2 votes

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 vote

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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  • 1,054
-1 votes

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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  • 174
1 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,\...
1 vote

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 votes

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 ...
1 vote

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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  • 360
0 votes

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 ...
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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 votes

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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  • 6,182
1 vote

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
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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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  • 8,348
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
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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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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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  • 6,182
1 vote

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 vote

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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  • 159
2 votes

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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  • 3,842
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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  • 8,348
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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