All Answers 358108 Answers

Very nice Titanium :)

25^x=5^x+7

$$\begin{array}{rcl}
25^x &=& 5^x+7 \\
5^{2x} &=& 5^x+7 \\
5^{2x} - 5^x-7 &=& 0 \qquad |\qquad u = 5^x \\
u^2 -u-7 &=& 0\\
u_{1,2} &=& \frac{1}{2}\cdot [ 1\pm \sqrt{ 1-4\cdot (-7) } ] \\
u_{1,2} &=& \frac{1}{2}\cdot [ 1\pm \sqrt{ 29 } ] \\\\
u_1 &=& \frac{1+\sqrt{29}}{2} \qquad u_2 = \frac{1-\sqrt{29}}{2}\\\\
5^x&=& u \qquad |\qquad \ln() \\
x\ln{(5)} &=& \ln{(u)} \\
x &=& \frac{\ln{(u)}} {\ln{(5)}}\\\\
x_1 &=& \frac{ \ln{ ( \frac{1+\sqrt{29}}{2} ) } } {\ln{(5)}}\\
x_1 &=& 0.72126430677\\\\
x_2 &=& \frac{ \ln{ ( \frac{1-\sqrt{29}}{2} ) } } {\ln{(5)}} \qquad \text{no solution}\\
\end{array}$$

-x+4

x=6

so -6+4=-2

I notice that someone is thumbing down my comments

i will figure out who did, cuz i got my eyes on you

i have many suspicions on a few people, very specific people that seem to be nice when they write down their comments, but are rude and selfish at giving points 

well

the dragon shall be punished and shall not pass

Hi Heureka,

Yes Alan Heureka's 2nd answer is more elegant but Mellie may understand the first one much better 

I like both!  

For once anonymous agrees with me that siblings are worshippers of the devil

reward is 300 gold coins

Siblings are the evil ones.........not most of the time......all the time

What is the bar at the front for?

A quadrant can also just be a quarter of any circle :)

This is what Alan is describing

$${\mathtt{360}}{\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right) = {\mathtt{600}}$$     degrees

Thanks Alan,       I should have thought of that    

Thanks Heureka :)

When multiplying two of the same number (with varying powers), simply add the powers together:

A^x * A^y = A^x+y

Inversely, when dividing, subtract the powers:

A^x / A^y = A^x-y

To find how many times greater something is, you divide the number you are checking with the number you are comparing it to. So in this case, you will want 7^-2 / 7⁶ to see how many times greater 7² is of 7⁶

Using the above rules of multiplying/dividing powers, this equation is:

7^-2 / 7⁶ = 7^{(-2 - 6)} = 7^-8 = 0.0000001734665256

6 - 12 = -6

$${\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{12}} = -{\mathtt{6}}$$

To expand on what anonymous said: the imaginary number is the square root of -1, this is because there is no possible "real" answer to it, but you can use the imaginary number to calculate things (often used in engineering).

$${i} = {\sqrt{-{\mathtt{1}}}}$$

Sometimes you might come across something with a negative surd. To show this in terms of "i", just seperate the square root of -1 from the rest of the surd.

Example:

Let's so I want 7*square-root(-16 * 9):

$${\mathtt{7}}{\mathtt{\,\times\,}}{\sqrt{\left(-{\mathtt{16}}\right){\mathtt{\,\times\,}}{\mathtt{9}}}}$$

$${\mathtt{7}}{\mathtt{\,\times\,}}{\sqrt{-{\mathtt{25}}}}$$

$${\mathtt{7}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{25}}}}{\mathtt{\,\times\,}}{\sqrt{-{\mathtt{1}}}}$$

$${\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{i}$$

$${\mathtt{35}}{i}$$

Hello Anonymous 

Pi is the circumference of any circle, divided by its diameter. Nobody knows its exact value, because no matter how many digits you calculate it to, the number never ends. For most practical uses, you can assume sum it up as 3.14.

integrate ∫(2cos3x + 3sinx) / sin^3x dx

$$\small{\begin{array}{l|lll}
\int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx
\quad & \quad \cos{(3x)}= \cos{(x)}\cos{(2x)} -\sin(x)\sin{(2x)}\\
\quad &\quad \cos{(3x)} = \cos{(x)}[1-2\sin^2{(x)}]-\sin{(x)}\cdot 2\sin{(x)}\cos{(x)}\\
\quad &\quad \cos{(3x)} = \cos{(x)}[1-2\sin^2{(x)}]-2\sin^2{(x)}\cos{(x)}\\
\quad &\quad \cos{(3x)} = \cos{(x)}-2\cos{(x)}\sin^2{(x)}-2\sin^2{(x)}\cos{(x)}\\
\quad &\quad \cos{(3x)} = \cos{(x)}-4\cos{(x)}\sin^2{(x)}\\
=\int \dfrac{2[\cos{(x)}-4\cos{(x)}\sin^2{(x)}] + 3\sin{(x)} }{ \sin^3{(x)} }\ dx & \\
=\int \dfrac{2\cos{(x)}-8\cos{(x)}\sin^2{(x)} + 3\sin{(x)} }{ \sin^3{(x)} }\ dx & \\
= 2 \int \dfrac{ \cos{(x)} } { \sin^3{(x)} }\ dx
-8 \int \dfrac{ \cos{(x)}\sin^2{(x)} } { \sin^3{(x)} }\ dx
+3 \int \dfrac{ \sin{(x)} } { \sin^3{(x)} }\ dx & \\
= 2 \int \dfrac{ 1 } { \sin^2{(x)} }\cdot\cot{(x)}\ dx
-8 \int \dfrac{ \cos{(x)} } { \sin{(x)} }\ dx
+3 \int \dfrac{ 1 } { \sin^2{(x)} }\ dx &\\
\quad & \quad \text{Formula:} \\
\quad & \quad \boxed{ \int \frac{f'(x)}{f(x)}=\ln{(f(x))} ~~ \int \frac{\cos{(x)}}{\sin{(x)}}\ dx = \ln {( \sin{(x)} )} } \\
\quad & \quad \boxed{ (\cot{(x)})' = -\frac{1}{\sin^2{(x)}} ~~\int \frac{1}{\sin^2{(x)}}\ dx = -\cot{(x)}}\\
\quad & \quad \boxed{ \int f'(x) \cdot [f(x)]^1 = \frac{ [f(x)]^2}{2} ~~\int \dfrac{ 1 } { \sin^2{(x)} }\cdot\cot{(x)}\ dx = -\frac{\cot^2{(x)}}{2} }\\
= 2 (-\frac{\cot^2{x}}{2}) - 8 \ln {( \sin{(x)} )} + 3 (-\cot{(x)} )&\\\\
= -\cot^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + c_1&\\
\end{array}}\\\\\\$$

$$\small{\begin{array}{rcl}
\int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx
&=& -\cot^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + c_1 \quad | \quad -\cot^2{(x)}=1-\csc^2{(x)}\\\\
\int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx
&=& 1-\csc^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + c_1 \\\\
\int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx
&=& -\csc^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + (c_1+1) \quad | \quad c = c_1+1 \\\\
\int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx
&=& -\csc^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + c \\\\
\end{array}}$$

I should note that this result applies for x>0.  The sign changes for x<0.

You just plug in the numbers 1 to 10 for x and calculate the corresponding values of y:

.

Melody, 

$$\\
\sin^2+\cos^2=1 \text{ Divide through by }\sin^2 \text{ to get }1+\cot^2=\csc^2$$

So cot² differs from csc² by 1.  This 1 is wrapped up in the constant of integration.  i.e. your constant of integration differs from Wolfram Alpha's by 1.

. 

That is fabulous Brittany  

I prefer your second solution heureka; it's much more elegant!

.

Angles are often thought of as being in quadrants.  Angles between 0° and 90° are in the first quadrant; those between 90° and 180° are in the second quadrant; those between 180° and 270° are in the third quadrant and those between 270° and 360° are in the fourth quadrant.  

You could think of these as being quarters of a circle, taken anticlockwise.

.

the i stands for imaganery number. for examply, 2i= 2 imaganery number.

How many numbers between 1000 and 2000 leave a remainder of 3 when divided by 11 ?

$$\small{
\begin{array}{rcl}
n_1 \cdot 11 + 3 &\ge& 1000 \\
n_1 &\ge& \dfrac{1000-3}{11} = 90.6363636364 \\
n_1 &=& 91 \\\\
n_2 \cdot 11 + 3 &\le& 2000 \\
n_2 &\le& \dfrac{2000-3}{11} = 181.545454545 \\
n_2 &=& 181 \\\\
n&=& n_2-n_1+1\\
n&=& 181-91+1\\
\mathbf{n}&\mathbf{=}& \mathbf{91}\\\\
\hline
\end{array}
}$$