All Answers 358513 Answers

Thank you CPhill

(-1)^(1/8) is not a real number, it is imagniary. You are effectively trying to sqaure root a negative number (multiple times due to the 8) which leads to imaginary numbers. Despite the 3/8, it is still not possible. 

A polygon is a 2 dimensional shape and therefore does not have a volume  

9+3/4x=7/8x-10

It is easiest if you get rid of all the fractions  first so mult both sides by 8

8*(9+3/4x)=8(7/8x-10)

72+6x=7x-80

Can you take it from there?

"oh the dinosaur got jokes now lol ok. I guess you had got that off your big bookshelf?"

NO Zegroes - I got it out of my big brain.   LOL 

No

$$\left({\mathtt{18}}{\mathtt{\,\times\,}}{\mathtt{6}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{18}}}{{\mathtt{4}}}}\right) = {\frac{{\mathtt{225}}}{{\mathtt{2}}}} = {\mathtt{112.5}}$$

We can do a little more with this anon

$$\\\frac{5e^{4x}}{10e^x}\\\\
=\frac{e^{4x}}{2e^x}\\\\
=\frac{e^{4x-x}}{2}\\\\
=\frac{e^{3x}}{2}\\\\$$

Another excellent answer - thank you Saseflower   

4.9t^2+19.8t-20=0? How do i find t ?

$$\small{\text{
Use the formula:
$
\boxed{
ax^2+bx+c=0 \qquad x_{1,2}=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}
}
$
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm\sqrt{19.8^2-4*4.9*(-20)} }{2*4.9}
$
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm\sqrt{ 392.04+392 } }{9.8}
$
}}
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm\sqrt{ 784.04 } }{9.8}
$
}}
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm 28.001 }{9.8}
$
}}
}}\\\\
\small{\text{
$
t_{1}=\dfrac{-19.8 + 28.001 }{9.8} = 0.8368
$
}}
}}\\\\
\small{\text{
$
t_{2}=\dfrac{-19.8 - 28.001 }{9.8} = -4.8777
$
}}$$

Why would I help you when you have a history of being rude in return?

gar12 you are being rude and I do not think that it is the first time.

BEHAVE YOURSELF !

100%-15%=85%

85% of $30 = 85/100*30  dollars

YES Tiddle they would be.  Some countries use a comma instead of a point when they write decimals.  

19.99/0.8    just put it into the calc on  the home page.

I'd use the quadratic formula  

I think it is 

$$\\2e^{i\pi/6}\\\\
=2(cos\frac{\pi}{6}+isin\frac{\pi}{6})\\\\
=2(\frac{\sqrt3}{2}+i*\frac{1}{2})\\\\
=\sqrt3+i$$  

-----------------

You do not need this pic to answer the question but I think it may help you to understand

2x+3y=1 and -3x+y=15 how do you get points (-4,3)

$$\begin{array}{rrcl|lrcl}
(1) & 2x+3y &=& 1 \quad & \quad (2)& -3x+y &=& 15 \quad | \quad * 3 \\
& & & \quad & \quad (2)& -9x+3y &=& 45 \\
\hline
\end{array}$\\\\\\$
\small{\text{
$
\begin{array}{lrcl}
(1)-(2): & 2x-(-9x) +3y -3y &=& 1-45 \\
& 11x &=& -44 \\
& \textcolor[rgb]{1,0,0}{x} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{-4} \\\\
\hline \\
(1) & 2(-4) + 3y &=& 1 \\
& -8 + 3y &=& 1 \\
& 3y &=& 9 \quad | \quad : 3\\
& \textcolor[rgb]{1,0,0}{ y} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{3}
\end{array}
$
}}$$

Canberra

For the second one, we have

tan(3x)/x  = (sin3x)/x * 1/cos3x = (3sin3x/3x) * 1/(cos 3x) 

Let 3x = Θ

And  using the fact that sinΘ/Θ = 1, and cosΘ = 1   as Θ→ 0..... we have

(3sinΘ/Θ) * 1/cosΘ = (3)(sinΘ/Θ) * 1/cosΘ =

[3(1)] / [1/1]  = 3/1  = 3  

(without using l'hopital's rule) how to do you prove that:

lim x->0 (tan3x)/x =3 ? 

$$\small{\text{
$
\boxed{ \lim\limits_{x\to0}
\left(
\dfrac{ \tan{(3x)} } { x } \right)
\qquad \tan{ (3x) } &= (3x)+\dfrac13 (3x)^3+\dfrac{2}{15}(3x)^5+\dfrac{17}{315}(3x)^7+\dotsb
} % boxed
$
}}$\\\\\\$
\small{\text{
$
\lim\limits_{x\to0}\left(
\dfrac{ \tan{(3x)} } { x } \right)
=
\lim\limits_{x\to0}\left(
\dfrac{
(3x)+\dfrac13 (3x)^3+\dfrac{2}{15}(3x)^5+\dfrac{17}{315}(3x)^7+\dotsb
} { x } \right)
=
\lim\limits_{x\to0}\left(
\dfrac{
x \left( 3+\dfrac13 3^3x^2+\dfrac{2}{15}3^5x^4+\dfrac{17}{315}3^7x^6+\dotsb \right)
} { x } \right)
$
}}$\\\\\\$
\small{\text{
$
=
\lim\limits_{x\to0}\left(
3+\dfrac13 3^3x^2+\dfrac{2}{15}3^5x^4+\dfrac{17}{315}3^7x^6+\dotsb \right)= \textcolor[rgb]{1,0,0}{3}
$
}}$$

See if this helps.

[2nd] asin

arc sine is the same as inverse sine

(without using l'hopital's rule) how to do you prove that:

lim x->0 (sinx)/x=1 ? 

$$\small{\text{
$
\boxed{ \lim\limits_{x\to0}
\left(
\dfrac{ \sin{(x)} } { x } \right)
\qquad \sin (x) = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!} = \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}\mp\dotsb
} % boxed
$
}}$\\\\\\$
\small{\text{
$
\lim\limits_{x\to0}\left(
\dfrac{ \sin{(x)} } { x } \right)
=
\lim\limits_{x\to0}\left(
\dfrac{
\dfrac{x}{1!}-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}\mp\dotsb
} { x } \right)
=
\lim\limits_{x\to0}\left(
\dfrac{
x \left( \dfrac{1}{1!}-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}\mp\dotsb \right)
} { x } \right)
$
}}$\\\\\\$
\small{\text{
$
=
\lim\limits_{x\to0}\left(
\dfrac{1}{1!}-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}\mp\dotsb \right)= \dfrac{1}{1!} = \textcolor[rgb]{1,0,0}{1}
$
}}$$

Thanks for checking that answer Rosala :)

Hey Happy, your explanation is great