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 factoring and tell whether the parabola opens  up or down. Also Identify the minimum and how does the axis of symmetry  relate to the x-intercepts? 

What do you need help with here? Graphing? Factoring? Finding the roots? You haven't made it clear.

Wait! There is another solution, as well! Heureka was so close to the second solution, too. I think the user forgot about taking the square root always results in a positive and negative answer. Anyway, I like Heureka's method, but here is my method of solving.

The original equation is \(\sqrt[3]{x^2-4x+4}=16\). Now let's solve for x:

Normally, with equations inside radicals, you would have to check both solutions, but here it isn't necessary. This is because after cubing both sides, you end up with a quadratic, and an answer from a quadratic is always right--assuming no arithmetic error was made when solving. Therefore, your solution set is:

\(x_1=-62\)

\(x_2=66\)

What is that question?

I thought the same, actually. The correct interpretation is \(|-|-11||\), but I feel as if this is not what the guest meant. 

Let's say...

log₈16  =  n         In exponential form, this is...

8ⁿ  =  16              And  8 = 2³  and  16 = 2⁴  .

(2³)ⁿ  =  2⁴

2³ⁿ  =   2⁴

3n  =  4

n  =  4/3

log₈16  =  4/3

You can check this on the home page by typing in  log(16,8)  .

For some reason...it rounds  4/3  off with a  5  ???

This innovative game is most popular card game now a days.

Matrix Problem: A \(\mathbf{\in \mu}\) (2x2).

Find the possible conditions and possible cases in the real numbers for which A² = 0

In linear algebra, a nilpotent matrix is a square matrix A such that

\({\displaystyle A^{k}=0\,}\)

for some positive integer k.

If A is a nilpotent matrix then det(A) = 0

\(\begin{array}{|rcll|} \hline det(A) &=& \begin{vmatrix} a & b \\ c&d \end{vmatrix} = ad-bc = 0 \\\\ \Rightarrow \qquad \mathbf{bc} & \mathbf{=} & \mathbf{ad} \\ \hline \end{array}\)

\(A^2 = 0 \\ \begin{array}{|rcll|} \hline A^2 &=& \begin{pmatrix} a & b \\ c&d \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c&d \end{pmatrix} \\\\ &=& \begin{pmatrix} a^2+bc & b\cdot (a+d) \\ c\cdot (a+d) & d^2+bc \end{pmatrix} \quad & | \quad \mathbf{bc=ad} \\\\ &=& \begin{pmatrix} a^2+ad & b\cdot (a+d) \\ c\cdot (a+d) & d^2+ad \end{pmatrix} \\\\ &=& \begin{pmatrix} a\cdot (a+d) & b\cdot (a+d) \\ c\cdot (a+d) & d\cdot(a+d) \end{pmatrix} \\\\ A^2 &=& \underbrace{(a+d)}_{=0}\cdot \underbrace{\begin{pmatrix} a & b \\ c & d \end{pmatrix}}_{A\ne 0} = 0\\\\ a+d &=& 0 \\ \mathbf{a} &\mathbf{=}& \mathbf{-d} \\ \hline \end{array}\)

if \(A\ne 0 \) and \(A^2 = 0\), then \( a = -d\) and \(det(A) = 0\).

\(\text{Example } 1:\\ \begin{array}{|rcll|} \hline A &=& \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} \\ det(A) &=& \begin{vmatrix} 0 & 1 \\ 0&0 \end{vmatrix} = 0\cdot 0 - 0 \cdot 1 = 0 \\ a &=& -d \\ 0 &=& 0\ \checkmark \\\\ A^2 &=& \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} = 0\ \checkmark \\ \hline \end{array}\)

\(\text{Example } 2:\\ \begin{array}{|rcll|} \hline A &=& \begin{pmatrix} 4 & 8 \\ -2 & -4 \end{pmatrix} \\ det(A) &=& \begin{vmatrix} 4 & 8 \\ -2 & -4 \end{vmatrix} = 4\cdot(-4) - (-2) \cdot 8 = -16+16 = 0 \\ a &=& -d \\ 4 &=& -(-4) \\ 4 &=&4\ \checkmark \\\\ A^2 &=& \begin{pmatrix} 4 & 8 \\ -2 & -4 \end{pmatrix} \cdot \begin{pmatrix} 4 & 8 \\ -2 & -4 \end{pmatrix} = 0\ \checkmark \\ \hline \end{array}\)

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Fixed :D

Trully! This is a terrfic card game of rummy player.

Input: 2+2=?

Intepretation: \(2+2=?\)

Answer: \(2+2=5\) Because the Big Brother said so

(Just kidding, it is \(4\))

Input: 12/m=2/4+1x10^33!= x i know the answer!

Intepretation: \(\frac{12}{m}=\frac{2}{4}+1\cdot 10^{33}\)

Simplify: \(\frac{12}{m}=\frac{1}{2}+10^{33}\)

\(m=\frac{8}{666666666666666666666666666666667}\)

\(m≈1.2\cdot10^{-32}\)

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If you are using Microsoft Word, try clicking the  " ¶ "  button at the top in the  "Paragraph"  category....here:

Hi  Jeffes,

Thanks for answering this and other questions, it is appreciated.

I just wanted to point out that you are using QED incorrectly.

QED is used at the end of proofs, it means.. 'thus it has been demonstrated'.

Here you are simplifying an expression, you are not proving anything. So it is not appropriate to add QED at the end :)

x|x|=2x+1 

The lines are absolute value

I'd do it in 2 parts.

If x is positive then

\(x^2=2x+1\\ x^2-2x-1=0\\ x=\frac{2\pm\sqrt{4--4}}{2}\\ x=\frac{2\pm\sqrt{8}}{2}\\ x=\frac{2\pm2\sqrt{2}}{2}\\ x=1\pm\sqrt{2}\\ \text{Since x is positive the only answer is }\\ x=1+\sqrt2\)

If x is negative then

\(-x^2=2x+1\\ -x^2-2x-1=0\\ x=\frac{2\pm\sqrt{4-4}}{2}\\ x=1\\ \text{but since x is negative there is no solution }\\ \)

\(\text{So the only solution is }\quad x=1+\sqrt{2}\)

\(If \;\;A=\frac{1}{4} \;\;and \;\;B=-\frac{1}{2}\)

what is the absolute value of the difference between $18A$ and $18B$? Express your answer as a decimal to the nearest tenth.

18|1/4--1/2| = 18*3/4 = (9*3)/2 = 13.5

Solve for the positive value of x such that  \sqrt[3]{x^2 - 4x + 4} = 16

(\(\sqrt[3]{x^2 - 4x + 4}\) ) .

Formula:

\(\begin{array}{|rcll|} \hline && \sqrt[n]{a^m} \\ &=& (a^m)^\frac{1}{n} \\ &=& a^{m\cdot \frac{1}{n}} \\ &=& a^{\frac{m}{n}} \\ &=& a^{\frac{1}{n}\cdot m } \\ &=& (a^{\frac{1}{n}})^m \\ &=& (\sqrt[n]{a})^m \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \sqrt[3]{x^2 - 4x + 4} &=& 16 \quad & | \quad x^2 - 4x + 4 = (x-2)^2 \\ \sqrt[3]{(x-2)^2} &=& 16 \quad & | \quad \sqrt[n]{a^m} = (\sqrt[n]{a})^m \\ \left(~ \sqrt[3]{x-2} ~\right)^2 &=& 16 \quad & | \quad \sqrt{} \\ \sqrt[3]{x-2} &=& 4 \quad & | \quad cube \\ x-2 &=& 4^3 \\ x-2 &=& 64 \\ x &=& 64+2 \\ \mathbf{x} & \mathbf{=} & \mathbf{66} \\ \hline \end{array}\)

x|x|=2x+1 

The lines are absolute value

\(\begin{array}{|rcll|} \hline x|x| &=& 2x+1 \quad & | \quad \text{square both sides} \\ x^2\cdot |x|^2 &=& (2x+1)^2 \quad & | \quad |x|^2 = x^2 \\ x^2\cdot x^2 &=& (2x+1)^2 \\ x^4 &=& 4x^2+4x+1 \\ \mathbf{x^4 - 4x^2+4x+1} & \mathbf{=} & \mathbf{0} \\ \hline \end{array} \)

solutions of \(x^4 - 4x^2+4x+1 = 0\)

What, percent, is the gray circle's area out of the larger circle's area?

\(\frac{\text{gray circle area}}{\text{larger circle area}}\,=\,\frac{\frac{\pi OD^2}{4}}{\pi OD^2}\,=\,\frac{\pi OD^2}{4}\,*\,\frac{1}{\pi OD^2}\,=\,\frac14\,=\,\frac{25}{100}\,=\,25\%\)

Rhombus  ABCD has perimeter 148, and one of its diagonals has length 24.

How long is the other diagonal?

Let e one of its diagonal = 24

Let f the other diagonal = ?

Let a be the side of the rhombus. All sides are equal.

\(\mathbf{a= \ ? }\\ \begin{array}{|rcll|} \hline 4a &=& \text{perimeter} \\ a &=& \frac{ \text{perimeter} } {4} \\ a &=& \frac{ 148 } {4} \\ \mathbf{a} & \mathbf{=} & \mathbf{37} \\ \hline \end{array}\)

\(\mathbf{f= \ ? }\\ \begin{array}{|rcll|} \hline \mathbf{4a^2} & \mathbf{=} & \mathbf{e^2+f^2 } \\\\ 4\cdot 37^2 &=& 24^2+f^2 \\ 4\cdot 1369 &=& 576 + f^2 \\ 5476 &=& 576 + f^2 \quad & | \quad -576 \\ 5476-576 &=& f^2 \\ 4900 &=& f^2 \\ 70 &=& f \\\\ \mathbf{f} & \mathbf{=} & \mathbf{70} \\ \hline \end{array}\)

The other diagonal is 70