Thank you

Thanks

The whole matrix (A) cant be 0. But "a"  of course it can, I guess.

a, b, c, d cant be 0 at the same time . Because Thats the whole matrix. 

Is there something to fix? 

Where you say that couldnt find the minus plus sign. Where is it should placed? 

Thanks, heureka.....I like your method of solving this one...!!!

 60 = 30 sin 60 (3-d) + 30   subtract 30 from each side

30  = 30sin (180 - 60d)       divide both sides by 30

1=  sin (180 - 60d)         the sine will = 1 at 90°

Therefore .....    180 - 60d   = 90     subtract 180 from each side

-60d  = - 90     divide both sides by -60

d  = 9/6  =  3/2

Note that in  the 12 hours after 12 noon, the clock hands pass each other 11 times until they meet again at 12 midnight

So....that means that they pass each other every 12 / 11 ths hours...

So....in terms of minutes ......  12/11  * 60   ≈  65.45 minutes  =  65 minutes and [ .45 *60 ] seconds  ≈

65 minutes and 27 seconds

So....they meet again at about   27 seconds after 1:05 PM

\(\begin{bmatrix} a && b \\ c && d \end{bmatrix}*\begin{bmatrix} a && b \\ c && d \end{bmatrix}=\begin{bmatrix} a^2+b*c &&a*b+b*d \\ c*a+d*c && d^2+c*b \end{bmatrix}= \begin{bmatrix} a^2+bc && b*(a+d) \\ c*(a+d) && d^2+bc \end{bmatrix}=\begin{bmatrix} 0 && 0 \\ 0 && 0 \end{bmatrix}\)

(The multiplication of the matrices is derived from the formula: \((A*B)i,j =\sum_{k=1}^{n}a(i,k)*b(k,j)\)

so we have a set of equations:

1. a²+bc=0

2. b*(a+d)=0

3. c*(a+d)=0

4. d²+bc=0

using the first and the fourth equations, we can find that :(a²+bc)-(d²+bc)=(0)-(0)=0=a²-d²=(a-d)*(a+d). therefore, we can divide our answer to two cases:

first case- a=d

that means b*(a+d)=0=b*(2a) and c*(a+d)=0=c*(2a). now we have a new set of equations:

1. a²+bc=0

2. b*(2a)=0

3. c*(2a)=0

we can make it simpler:

1. a²+bc=0

2. b*a=0

3. c*a=0

now we can divide that case to two different cases:

a=0:

now we have 1 equation:

1. a²+bc=0=0²+bc=bc=0. that means b or c must be equal to 0 (that or is a mathematical or: it means if both of them are 0 it works too)

that means the matrices that maintain the equations are of the following form:

\(\begin{bmatrix} 0 && b \\ c && 0 \end{bmatrix}\)(where c*b=0)

a is not 0:

that means we can divide the second and the third equations by a and get the set of equations:
 

1. a²+bc=0

2. b=0

3. c=0

that means a²+bc=a²+0*0=a². but that means a=0, and that doesnt make any sense because we know that a CANNOT be 0. So there is no solution

second case- d=-a

now we can derive a simpler set of equations:

1. a²+bc=0

2. b*(a-a)=0=b*0=0

3. c*(a-a)=0=c*0=0

now we have one equation- a²+bc=0:

a²+bc=0/ -a²= 

bc=-a²

that gives us the next set of matrices: \(\begin{bmatrix} (-1)^n\sqrt{-bc} && b \\ c && -(-1)^n\sqrt{-bc} \end{bmatrix} \)

(where n is a natural number, couldnt find the plus minus sign). but as you can see, every matrice from the first set of matrices is also a member of the second set of matrices, therefore we can simply say that every matrice that is a member of the second set is a solution (keep in mind that every matrice from that set is a solution)

but we dont want A=0 as a solution, so we simply have to add that b*c is not 0.

solution: \(\begin{bmatrix} (-1)^n\sqrt{-bc} && b \\ c && -(-1)^n\sqrt{-bc} \end{bmatrix} \)

(where b*c is not 0)

Yes, a positive number multiplied by a negative number always results in a negative number. For example:

1) \(3*-2=-6\)

2) \(5*-9=-45\)

3) \(6*-6=-36\)

4) \(8*-1=-8\)

This list continues indefinitely, but here are a few examples demonstrating this phenomenon.

This question has been answered at least 3 times on this forum. 6/2(1+2) evaluates to 9. Since this question has been asked multiple times on this forum, I have redirected you to 3 different explanations on how to get the answer:

Two altitudes of a triangle have lengths 12 and 14. What is the longest possible length of the third altitude,

if it is a positive integer?

Let h = the length of the third altitude.

Let a, b, and c be the sides corresponding to the altitudes of length 12, 14, and h

Let A = Area of the triangle.

Let 2A = a*12
Let 2A = b*14
Let 2A = c*h

1.

\(\begin{array}{|rcll|} \hline a &=& \frac{2A}{12} \\ b &=& \frac{2A}{14} \\ c &=& \frac{2A}{h} \\ \hline \end{array}\)

2.

A triangle exists with side lengths a, b, and c
if and only if they satisfy the three triangle inequalities:

\(\begin{array}{|lrcll|} \hline (1) & a & < & b + c \\ (2) & b & < & c + a \\ (3) & c & < & a + b \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & \frac{2A}{12} & < & \frac{2A}{14} + \frac{2A}{h} \quad & | \quad :2A \\ & \frac{1}{12} & < & \frac{1}{14} + \frac{1}{h} \\\\ (2) & \frac{2A}{14} & < & \frac{2A}{h} + \frac{2A}{12} \quad & | \quad :2A \\ & \frac{1}{14} & < & \frac{1}{h} + \frac{1}{12} \\\\ (3) & \frac{2A}{h} & < & \frac{2A}{12} + \frac{2A}{14} \quad & | \quad :2A \\ & \frac{1}{h} & < & \frac{1}{12} + \frac{1}{14} \\ \hline \end{array}\)

(1):

\( \begin{array}{|rcll|} \hline \frac{1}{12} & < & \frac{1}{14} + \frac{1}{h} \\ \frac{1}{12}-\frac{1}{14} & < & \frac{1}{h} \\ \frac{12-14}{12\cdot 14} & < & \frac{1}{h} \\ \frac{2}{168} & < & \frac{1}{h} \\ \frac{1}{84} & < & \frac{1}{h} \quad & | \quad \cdot 84h \\ \mathbf{ h } & \mathbf{ < } & \mathbf{ 84 } \\ \hline \end{array} \)

(2):

\(\begin{array}{|rcll|} \hline \frac{1}{14} & < & \frac{1}{h} + \frac{1}{12} \\ \frac{1}{14} - \frac{1}{12} & < & \frac{1}{h} \\ \frac{12-14}{14\cdot 12} & < & \frac{1}{h} \\ -\frac{2}{168} & < & \frac{1}{h} \\ -\frac{1}{84} & < & \frac{1}{h} \quad & | \quad \cdot 84h \\ -h & < & 84 \quad & | \quad \cdot (-1) \qquad switch "<" to ">" \\ \mathbf{ h } & \mathbf{ > } & \mathbf{ -84 } \\ \hline \end{array} \)

(3):

\(\begin{array}{|rcll|} \hline \frac{1}{h} & < & \frac{1}{12} + \frac{1}{14} \\ \frac{1}{h} & < & \frac{14+12}{12\cdot 14} \\ \frac{1}{h} & < & \frac{26}{168} \\ \frac{1}{h} & < & \frac{13}{84} \quad & | \quad \cdot \frac{84}{13}\ h \\ \frac{84}{13} & < & h \\ \mathbf{ 6.46153846154 } & \mathbf{ < } & \mathbf{ h } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 6.46153846154 < h < 84 \\ \hline \end{array}\)

The longest possible length of the third altitude, if it is a positive integer is 83

Assuming each resident have a total body mass of \(M\space(kg)\), they all have a density of \(D\) (\(D\le 1g/cm^3\)) , and the swimming pool being a square prism (i.e. A square with height) with side \(L (meters)\).

The law of physics tell us that any object with volume displaces water in water, and the displaced water causes the container's water level to rise, in this case, a swimming pool.

The remaining non-overflowing volume of the pool will be \(0.4L^2\space(meter\space cubed)\)

Since \(Volume=Mass/Density\)

Each resident's volume is equal to \(\frac{M}{D}\).

Remember that \(D\le 1g/cm^3\), that means part of their body won't sink under the water when they are in the pool, therefore:

Each residents' displaced water volume: \(\frac{M}{D}\cdot D=M\space(cm^3)\) (Because the density of water = \(1g/cm^3\))

For a total of 3,000 residents, their total volume would be \(3000M\)

In order for the pool to not overflow, the water displaced must be smaller than the remaining volume of the pool:

\(3000M\space(cm^3)\le 0.4L^2\space(m^3)\)

\(0.003M\space(m^3)\le0.4L^2\space(m^3)\)

\(L^2\ge 0.0075 M\)

\(L\ge \frac{1}{20}\sqrt{3} M\)

So the side of the pool have to be longer than \(\frac{1}{20}\sqrt{3}\) of each resident's mass in kilograms.

(Somehow I feel the units have some fatal mistakes, correct me if anything is wrong in my process :P)

And their product is \({144\over5}\)

It could be

\({1+15 \over 1+14}\)

And then abcd = 210...

This problem is probably unsolvable.

(After downloading printscreen and photo-editing plugins later...)
(Wait I think I messed the pictures up...)

I did just type in the numbers. :D.  By the way, not to sound criticizing, you did not graph the answer as per the question.

Given that \(f(5)=5\) and it crosses \(y=0\) at the following points:

\(x=200, x=100,x=0,x=-100,x=-200\)

Since the crossing points of this function is rational, I deduct that the function is factorizable.

Perform reverse-factorization:

The function need to be in this form:

\(g(x)=(x+200)(x+100)(x+0)(x-100)(x-200)\)

For it to have roots at \(x=200, x=100, x=0, x=-100,x=-200\)

There are a total of five zero-points at:

\(1.x=-200 , y=0\)

\(2.x=-100 , y=0\)

\(3.x=0 , y=0\space(Origin)\)

\(4.x=100 , y=0\)

\(5.x=200, y=0\)

There are a total of four critical points at:

Local Maxima:

\(1.x=\sqrt{15000-1000\sqrt{145}} , y=200000000\left(\sqrt{5}+5\sqrt{29}\right)\sqrt{30-2\sqrt{145}}\)

\(2.x=-\sqrt{15000+1000\sqrt{145}} , y=-200000000\left(\sqrt{5}-5\sqrt{29}\right)\sqrt{30+2\sqrt{145}}\)

Local Minima:

\(3.x=\sqrt{15000+1000\sqrt{145}} , y=\left(\sqrt{5}-5\sqrt{29}\right)\sqrt{30+2\sqrt{145}}\)

\(4.x=-\sqrt{15000-1000\sqrt{145}} , y=-200000000\left(\sqrt{5}+5\sqrt{29}\right)\sqrt{30-2\sqrt{145}}\)

Since \(f(5)=5\), Just divide every y-value of maximas and minimas above by a factor of \(g(5)/5=398750625\)

\(f(x)=\frac{1}{398750625}\left(x+200\right)\left(x+100\right)x\left(x-100\right)\left(x-200\right)\)

Q.E.D.

(I bet you just randomly typed the numbers in, didn't you? (Because the \(x\) and \(y\) values are pretty ugly to be honest))

When changing the base of a logarithm, from 

logb(a) = log(a)/log(b)

Is it possible for the bases of the new logarithms to be any number? or only base 10?

You can use any number:

Change of Base Formula

The change of base formula for logarithms is:

\(\begin{array}{|rcll|} \hline \log_a{(x)} &=& \dfrac{ \log_b{(x)} } { \log_b{(a)} } \\ \hline \end{array}\)

Example:

\(\begin{array}{|rcll|} \hline \log_4{(16)} &=& \log_4{(4^2)} \\ &=& 2 \\\\ \log_4{(16)} &=& \frac{ \log_2{(16)} }{ \log_2{(4)} } \\ &=& \frac{ \log_2{(2^4)} }{ \log_2{(2^2)} } \\ &=& \frac{ 4\cdot \log_2{(2)} }{ 2\cdot \log_2{(2)} } \\ &=& \frac{ 4 }{ 2 } \\ &=& 2 \quad \checkmark \\ \hline \end{array} \)

Input:log3_(log2_x)=2.

Intepretation:\(\log _3\left(\log _2\left(x\right)\right)=2\)

\(\log _2\left(x\right)=3^2=9\)

\(x=2^9\)

\(=512\)

Q.E.D.

Find all solutions to the equation \sqrt{3x+6}=x+2 (\(\sqrt{3x+6}=x+2\)).
If there are multiple solutions,
order them from least to greatest, separated by comma(s).

\(\begin{array}{|rcll|} \hline \sqrt{3x+6} &=& x+2 \quad & | \quad \text{square both sides} \\ 3x+6 &=& (x+2)^2 \\ 3x+6 &=& x^2+4x+4 \\ x^2 +x -2 &=& 0 \\ (x+2)(x-1) &=& 0 \\\\ x_1 &=& -2 \\ x_2 &=& 1 \\ \hline \end{array}\)

Solution Set: {-2,1}

Proof:
\(x=-2:\qquad \sqrt{3\cdot(-2)+6} = -2+2\quad \checkmark \\ x=1: \qquad \sqrt{3\cdot(1)+6} = 1+2 \quad \checkmark \)

The last one becasue your expression eqaluates to (3-4)+5= -1+5=4 an the last option is 9-5=4 so it works.

                          \(|\frac{12}{x}+3|=2\)

\(\frac{12}{x}+3=2\)                    \(-(\frac{12}{x}+3)=2\)

\(\frac{12}{x}+3-3=2-3\)     \(-\frac{12}{x}-3=2\)

\(\frac{12}{x}+0=2-3\)             \(-\frac{12}{x}-3+3=2+3\)

\(\frac{12}{x}=2-3\)                    \(-\frac{12}{x}+0=2+3\)

\(\frac{12}{x}=-1\)                        \(-\frac{12}{x}=2+3\)

\(\frac{12}{x}\times x=-1\times x\)        \(-\frac{12}{x}=5\)

\(12=-1\times x\)                 \(-\frac{12}{x}\times x=5\times x\)

\(12=-x\)                        \(-12=5\times x\)

\(\frac{12}{-1}=\frac{-x}{-1}\)                        \(\frac{-12}{5}=\frac{5x}{5}\)

\(-12=\frac{-x}{-1}\)                      \(-\frac{12}{5}=\frac{5x}{5}\)

\(-12=x\)                          \(-\frac{12}{5}=x\)

\(x=-12\)                          \(x=-\frac{12}{5}\)

Chiquita

(Source:https://en.wikipedia.org/wiki/Banana_industry)

The equation between a circle's radius and its circumference is as followed:

\(Circumference=2πr\), with r being the radius.

Plug \(6π\) into the equation:

\(2πr=6π\)

\(2r=6\)

\(r=3\)

We obtained the radius of your circle, now use the equation between a circle's radius to its area:

\(Area=πr^2\)

Plug \(r=3\) into the equation

\(Area=π*3^2\)

\(=9π\)

Q.E.D.

Area = Pi x Radius^2

Diameter = Circumference / Pi

                  = 6Pi / Pi

                  = 6

Radius = Diameter / 2

              = 6 / 2

              = 3

Area = Pi x 3^2

         = 9Pi  square units.

That depends on whaat you mean perhaps.

3D movies are not really 3D, but they give people the illusion of 3 dimensions. Your picture does this too.

But

Real 3 dimensional objects are things that you can touch and if they are small you can pick them up. They are not drawings on a page. So no, no picture can ever be 3 dimensional. At best they can be 2D representations of 3D objects.  :)