All Answers 358129 Answers

$$\sqrt180=13.4164078649987382$$

It's -1.3513513513513511

Like this.

e like the constant or e as a variable?

and are you sure that there are no places where it's supposed to be x^y instead of xy?

there are no natural logarithms for negative numbers.

the only way it will work is if you use an imaginary number.

Example:

$${{log}}_{{\sqrt{-{\mathtt{1}}}}}{\left(-{\mathtt{1}}\right)}$$ which is 2, since $${{\sqrt{-{\mathtt{1}}}}}^{\,{\mathtt{2}}} = -{\mathtt{1}}$$

Yes. It is the same when you ask,

"is 1+1 really 3?"

if you need more help, just ask.

$${\mathtt{1\,987\,654\,321}}{\mathtt{\,\times\,}}{\mathtt{1\,234\,567\,891}} = {\mathtt{2\,453\,894\,203\,114\,007\,011}}$$

no, it's actually 19.

$${\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}} = {\mathtt{19}}$$

4^x+3=10^2-x

$$\\\small{\text{
$ 4^{x+3}=10^{2-x} \quad 4^x*4^3=10^2*10^{-x} \quad (4*10)^x=\frac{5^2}{4^2} \quad 40^x = 1.25^2 \; | \;\ln $
}}
\\
\small{\text{
$ x\ln{(40)}= 2*\ln{(1.25)} \quad x = 2*\frac{\ln{(1.25)}}{\ln{(40)}} \quad x=0.12098175291$
}}$$

4^x+3=10^2-x

$$4^x+3=10^2-x \qquad
4^x=100-3-x \qquad 4^x=97-x$$

$$x \approx 3.275179625$$

I assume this refers to a sphere, in which case volume = (4/3)*pi*radius³

so 113 = (4/3)*pi*radius³

Multiply both sides by 3/(4*pi)

3*113/(4*pi) = radius³

Take the cube root of both sides

radius = (339/(4*pi))^1/3  m

$${\mathtt{radius}} = {\left({\frac{{\mathtt{339}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} \Rightarrow {\mathtt{radius}} = {\mathtt{2.999\: \!139\: \!117\: \!950\: \!116\: \!3}}$$

radius ≈ 3m

Hi,I need your help.I have to do one mathematical task:Volume of proper octogonal prism is 8 cm^3,and height of the prism is sqrt(2)+1.Find surface of prism

$$\small{\text{
Volume $V=8\ cm^3$ and height $h=(1+\sqrt{2})\ cm$. Surface $=2\frac{V}{h}+4\sqrt{2hV(\sqrt{2}-1)}=6.62741699797+16=22.6274169980\ cm^2$
}}$$

Yes indeed. I see that it is only from android, but you can download it from the google play store, which I think that is not safe...

I da no'n ornlig bøgser hva'? I sgu da hav' den i RØV

Just replace x, wherever it occurs in f(x), by whatever value you are interested in and evaluate.

For example if x = 2, then f(2) = 2^3/3 - 2*2^2 + 4*2 = 8/3 - 2*4 +8 = 8/3

No. It is not. It is to know math, for people who don't know. Now, I have a question, anonymous. Instead of the x, do I put 3? Even after the equal sign that where it says,

g(x)=$$\mathbf{(x=3)}$$^3-5?

Let c be old cost and p be old profit, then

s = p + c  where s is selling price.

We are told that p = 3.2c, so s = 3.2c + c or s = 4.2c, so that c = s/4.2

Let nc be the new cost.  We are told that nc = 1.25c, which means that nc = 1.25*s/4.2

The selling price is the same, but now:  s = np + nc, where np is new profit.

Thus s = np + 1.25s/4.2

Subtract 1.25s/4.2 from both sides

s - 1/25s/4.2 = np

s(1 - 1.25/4.2) = np

np/s = 1 - 1.25/4.2 ≈ 0.702

So the new profit as a percentage of the selling price is approximately 70%

Wrong. 

$${\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{4}}$$

And don't make mean comments that hurt people feelings!

Wise words

du kan ikke engang stave bøsse

Äru dansk ellä?

if 2 devides p^2 how to proove that 2 devides p

$$\boxed{ \small{\text{
Even number times even number = even number. }}
\small{\text{Example:}} \\ \quad 8 * 8 = 64}

\boxed{ \small{\text{
Odd number times odd number = odd number. }}
\small{\text{Example:}} \\ \quad 7 * 7 = 49}
\\
\\
\small{\text{
$p^2=p\times p$. if $p^2$ is even then p is even and devides 2 }}$$

$$\begin{array}{rll}
f(x)&=&9(\frac{\sqrt{3}x^4-8}{5}+10)\\\\
y&=&9(\frac{\sqrt{3}x^4-8}{5}+10)\\\\
\mbox{The inverse function will be}\\\\
x&=&9(\frac{\sqrt{3}y^4-8}{5}+10)\\\\
\frac{x}{9}&=&\frac{\sqrt{3}y^4-8}{5}+10\\\\
\frac{x}{9}-10&=&\frac{\sqrt{3}y^4-8}{5}\\\\
5\left(\frac{x}{9}-10\right)&=&\sqrt{3}y^4-8\\\\
\frac{5x}{9}-50&=&\sqrt{3}y^4-8\\\\
\frac{5x}{9}-42&=&\sqrt{3}y^4\\\\
\frac{5x-42*9}{9}&=&\sqrt{3}y^4\\\\
\frac{5x-378}{9}&=&\sqrt{3}y^4\\\\
\frac{5x-378}{9\sqrt3}&=&y^4\\\\
\frac{(5x-378)\sqrt3}{27}&=&y^4\\\\
y^4&=&\frac{(5x-378)\sqrt3}{27}\\\\
y&=&\pm\sqrt[4]{\frac{(5x-378)\sqrt3}{27}}\\\\
f^{-1}(x)&=&\pm\sqrt[4]{\frac{(5x-378)\sqrt3}{27}}\\\\


\end{array}$$

That's what I get anyway.

-1,1,-1,1 how does this series work

$$(-1)^n \quad n\in$$

Try watching this video