All Answers 358091 Answers

Because of the positioning of the 2's and the 1/2 on the outside, it looks to me as if the following is what is wanted:

$${\left[{\mathtt{0.05}}{\mathtt{\,\times\,}}{\left({\mathtt{\,-\,}}{\mathtt{34}}{\mathtt{\,-\,}}{\mathtt{4.1}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}{\left({\mathtt{\,-\,}}{\mathtt{22}}{\mathtt{\,-\,}}{\mathtt{4.1}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.5}}{\mathtt{\,\times\,}}{\left({\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{4.1}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}{\left({\mathtt{14}}{\mathtt{\,-\,}}{\mathtt{4.1}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.05}}{\mathtt{\,\times\,}}{\left({\mathtt{28}}{\mathtt{\,-\,}}{\mathtt{4.1}}\right)}^{{\mathtt{2}}}\right]}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)} = {\mathtt{16.976\: \!159\: \!754\: \!196\: \!471\: \!4}}$$

or, ≈ $16.98

(I copied and pasted rather than rewriting this, then inserted ^ and * as appropriate and changed the long horizontal lines to - signs.)

It doesn't matter, it is just a little annoying.  If I wanted to find it badly enough then i would.

When all else fails....rely on CPhill!

Hm..can't really help you with that.

Hope you get that figured out. :)

Thanks Ninja,

A lot of things could have been meant but what is annoying me is that I did it on the calc and I did it myself.  The same question was being answered both times but they are not the same and I have not been able to spot where I went wrong.  

It looks like you took off the 2's after each set of parentheses from step 2 to step 3, Melody.

Maybe he meant these to be ^2? I'm not sure. Could be possible.

Since 6³ is greater than 201, but 5³ is less than 201 there must be 201 - 5 or 196 integers that are not perfect cubes.

Yet again, look at

Again, look at 

Look at the two I've done here: 

NinjaDEvo i understood your trick at the very first moment but though nice trick i have to say!

DragonSlayer , even i've gone through it so i will help you!it would have been much easier to return your points id u had given the link of that post!

But nevermind, i'll give u a thumbs up on ur this question which will indirectly result in the giving u a point back and deleting the minus point!

Thanks Heureka

Ninja you've done a great job of photo shopping that calc display!

DragonSlayer, the answer wasn't really 5 Ninja has changed it for a joke!  

Hi Melody,

this is a arithmetic succession. Order is 3 ( So i hold 6 as a constant number in Delta 3 so the order is 3 ). I got the Polynom from Internet.

Beginning bei n= 0. If i set x = n+1 so the Polygon can written as:

$$p(x) \\
= (x-1)^3+3(x-1)^2+4(x-1)+2 \\
= x^3-3x^2+3x-1+3(x^2-2x+1)+4x-4+2 \\
= x^3-3x^2+3x-1+3x^2-6x+3+4x-4+2\\
= x^3-3x^2+3x^2+3x-6x+4x -1+3-4+2\\
= x^3+x \\
p(x)=x^3+x \quad | \quad x=1,2,3,... \quad Your\quad solution$$

I always draw pictures for these. 

Now cos is positive (1st and 4th quad)  and tan is negative (2nd and 4th quadrant)

so x has to be in the 4th quadrant.  So Sin x is negative

$$Sin(x)=\frac{-\sqrt{77}}{9}$$

Heureka, you have got the same answer as me but I don not understand what you have done.  

$$\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$&1&2&3&4&5\\\hline
$f(x)$&2&10&30&68&\\\hline
\end{tabular}$$

Okay they are not getting bigger by a set multiplication factor - as the x gets bigger, f(x) gets MUCH bigger.

Try squared, no still not big enough -- Try cubed.  That looks close.  Add x and you've got it.

$$\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$&1&2&3&4&5\\\hline
$f(x)$&2&10&30&68&\\\hline
$x^2$&1&4&9&16&\\\hline
$x^3$&1&8&27&64&\\\hline
$x^3+x$&2&10&30&68&\\\hline
\end{tabular}$$

$$\\f(x)=x^3+x\\\\
f(5)=5^3+5=125+5=130$$

\begin{tabular}{|c|c|c|c|c|c|}

\hline
$x$&1&2&3&4&5\\\hline
$f(x)$&2&10&30&68&\\\hline

$$\begin{tabular}{|c|c|c|c|c|c|}

\hline
$x$&1&2&3&4&5\\\hline
$f(x)$&2&10&30&68&\\\hline
\end{tabular}$$

        if     1  =  2

                2 = 10

                3 = 30

                4 = 68

        then 5 =   ?

$$\begin{array}{ccccc}
n+1& p(n)& \Delta_1 & \Delta_2 & \Delta_3
\\
\\
1&2&
\\
&&8\\
2&10&&12
\\
&&20&&\textcolor[rgb]{1,0,0}{6}\\
3&30&&18&\downarrow
\\
&&38&&(\textcolor[rgb]{1,0,0}{6})\\
4&68&&(24)
\\
&&(62)\\
5&(130)&
\end{array}$$

 5 =   130

Polynom

$$p(n) = n^3+3*n^2+4*n+2 \quad | \quad n = 0,1,2,3, ...$$

Yes as Alan has shown you,

If the domain is confined to  $$0\le x\le4$$

then the range is   $$0\le y \le 16$$

[0.05(-$34 – $4.1)2 + 0.20(-$22 – $4.1)2 + 0.50($12 – $4.1)2 + 0.20($14 – $4.1)2 + 0.05($28 – $4.1)2]½

=[0.05(-38.10)2 + 0.20(-26.10)2 + 0.50(7.90)2 + 0.20(9.90)2 + 0.05(23.90)2]*0.5

=0.05*-38.10 + 0.20*-26.10 + 0.50*7.90 + 0.20*9.90 + 0.05*23.90

=0.05(-38.10+7.9+23.9)+0.20(-26.10+9.9)

=0.05(-38.10+31.8)+0.20(-16.2)

=0.05(-8.10+1+0.8)+-3.24

=0.05(-7.10+0.8)-3.24

=0.05(-6.3)-3.24

=-0.315-3.24

= -3.555     I suppose i have made an error but I cannot find it.   

$$\left({\mathtt{0.05}}{\mathtt{\,\times\,}}{\mathtt{\,-\,}}\left({\mathtt{38.1}}{\mathtt{\,\times\,}}{\mathtt{2}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}{\mathtt{\,-\,}}\left({\mathtt{26.1}}{\mathtt{\,\times\,}}{\mathtt{2}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{0.5}}{\mathtt{\,\times\,}}{\mathtt{7.9}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}{\mathtt{9.9}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.05}}{\mathtt{\,\times\,}}{\mathtt{23.9}}{\mathtt{\,\times\,}}{\mathtt{2}}\right){\mathtt{\,\times\,}}{\mathtt{0.5}} = {\mathtt{0}}$$

$$\left({\mathtt{0.05}}{\mathtt{\,\times\,}}\left(-{\mathtt{38.1}}\right){\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}\left(-{\mathtt{26.1}}\right){\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.5}}{\mathtt{\,\times\,}}\left({\mathtt{7.9}}\right){\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}\left({\mathtt{9.9}}\right){\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.05}}{\mathtt{\,\times\,}}\left({\mathtt{23.9}}\right){\mathtt{\,\times\,}}{\mathtt{2}}\right){\mathtt{\,\times\,}}{\mathtt{0.5}} = {\mathtt{0}}$$

Cos (a+b) sin a=-3/5 sin b=-5/13 a & b in Q3  ?

$$\boxed{\cos(a+b)=\;?} \quad \sin{(a)} =-{3\over5} \qquad \sin {(b)}=-{5\over13}$$

$$\cos{(a+b)}=\cos{(a)}*\cos(b)-sin{(a)}*\sin{(b)}$$

cos(a)=?   and   cos(b)=?

$$\textstyle{
\cos{(a)}=\sqrt{1-\sin^2{(a)}}=
\sqrt{1-({3\over5})^2}= {\sqrt{5^2-3^2}\over5}={\sqrt{16}\over5}={\pm4\over5}=\pm{4\over5}
}$$

$$\textstyle{
\cos{(b)}=\sqrt{1-\sin^2{(b)}}=
\sqrt{1-({5\over13})^2}= {\sqrt{13^2-5^2}\over13}={\sqrt{144}\over13}={\pm12\over13}=\pm{12\over13}
}$$

$$\cos{(a+b)}=\pm
({4\over5})
\times
({12\over13})
-
(-{3\over5})
\times
(-{5\over13})$$

$$\cos{(a+b)}=\pm
({4\over5})
\times
({12\over13})
-
({3\over5})
\times
({5\over13})$$

$$\cos{(a+b)}=({\pm(4*12)\over5*13})
-
({3*5\over5*13})$$

$$\cos{(a+b)}={\pm(4*12)-3*5\over5*13}$$

$$\cos{(a+b)}={\pm48-15\over65}$$

$$\text{1.) }\cos{(a+b)}={48-15\over65}={33\over65}=0.50769230769$$

$$\text{2.) }\cos{(a+b)}={-48-15\over65}=-{63\over65}=-0.96923076923$$