All Answers 358093 Answers

Area of six slanted triangular sides of the pyramid, each with a base of 10cm and apothem of 13cm, is 6*(1/2)*10*13 = 390cm²

Area of base is that of six equilateral triangles each of side 10cm, so area = 6*25*√3cm² = 150√3cm²

So total area = 390+150√3 cm²

.

If you mean $$\tan^{-1}$$ then select the 2nd button on the calculator first and use atan (which is the same thing).

.

The MAD (Mean Absolute Deviation) is calculated as follows:

First find the mean  μ = (3 + 3 + 8 + 2 + 1)/5 = 17/5

Now calculate the absolute values of the deviations from the mean, add these up and divide by 5:

MAD = (|3 - 17/5| + |3 - 17/5| + |8 - 17/5| + |2 - 17/5| + |1 - 17/5|)/5

I'll leave you to crunch the numbers.

.

Find the exact values of tan (cos^-1(2/3)) and cos (sin^-1(-3/5)).

tan (cos^-1(2/3))

$$\small{\text{$ \boxed{\tan{ \left(~ \arccos{(x)} ~\right) }=\pm\dfrac {\sqrt{1-x^2}} {x}
=\pm\dfrac {\sqrt{1-\left(\dfrac{2}{3}\right)^2}} { \dfrac{2}{3} } =\pm \dfrac{3}{2} \cdot \sqrt{ 1-\dfrac{4}{9} }
=\pm \dfrac{3}{2} \cdot \dfrac{ \sqrt{5}}{3}
=\pm \dfrac{ \sqrt{5} }{2}
}
$}}$$

cos (sin^-1(-3/5))

$$\small{\text{$ \boxed{\cos{ \left(~ \arcsin{(x)} ~\right) }=\pm\sqrt{1-x^2}=\pm \sqrt{1+ \left( \dfrac{-3}{5} \right)^2}=\pm \dfrac{\sqrt{34} }{ 5 } }$}}$$

$${{\mathtt{e}}}^{{\mathtt{0.048}}} = {\mathtt{1.049\: \!170\: \!655\: \!324\: \!470\: \!6}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left(-{\mathtt{1}}^\circ\right)} = -{\mathtt{0.017\: \!455\: \!064\: \!928}}$$

Like this:

$${\mathtt{\,-\,}}{\mathtt{x}}$$

(-x)

Which is the equivalent of 6 14' 48" written in decimal form ?

$$\boxed{~6\ensurement{^{\circ}} ~14' ~48" ~}\\\\
\small{\text{$
\dfrac{ \dfrac{48"}{60}+14' }{60} + 6\ensurement{^{\circ}}
=\dfrac{0.8+14'}{60} + 6\ensurement{^{\circ}}
=\dfrac{14.8}{60} + 6\ensurement{^{\circ}}
=0.24666666667+ 6\ensurement{^{\circ}}
=6.24666666667\ensurement{^{\circ}}
=6.24\overline{6}\ensurement{^{\circ}}
$}}$$

cos( atan( sqrt(3) )+asin(1/3) ) ?

$$\boxed{
\mathbf{
\cos{ \left(~ \arctan{(~\sqrt{3}~)}
+ \arcsin{\left(\dfrac{1}{3}\right)}
~\right)
}
} =~ ?
}$$

$$\small{\text{$
\begin{array}{rcl}
\mathbf{
\cos{ \left(\alpha_{rad} + \beta_{rad} \right) }
= \cos{ \left(\alpha_{rad} \right) } \cdot
\cos{ \left(\beta_{rad} \right) }
- \sin{ \left(\alpha_{rad} \right) } \cdot
\sin{ \left(\beta_{rad} \right) }
} & \quad \mathbf{
\alpha_{rad} = \arctan{(a)} \quad a= \sqrt{3} } \\
& \mathbf{
\quad \beta_{rad} = \arcsin{(b)} \quad b= \dfrac{1}{3} }
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{ \left( \alpha_{rad}+ \beta_{rad} \right) }
&=&
\cos{ \left( \arctan{(a)} \right) } \cdot
\cos{ \left(\arcsin{(b)} \right) }
- \sin{ \left( \arctan{(a)} \right) } \cdot
\sin{ \left(\arcsin{(b)} \right) } \\
&=&
\cos{ \left( \arctan{(a)} \right) } \cdot
\cos{ \left(\arcsin{(b)} \right) }
- \sin{ \left( \arctan{(a)} \right) } \cdot b
\end{array}
$}}$$

$$\small{\text{
$
\boxed{
\cos{ \left(~ \arctan{(a)} ~\right) }
=\pm\dfrac{1}{\sqrt{1+a^2}}
=\pm\dfrac{1}{\sqrt{1+(\sqrt{3})^2}}
=\pm \dfrac{1}{2}
}
$}}$$

$$\small{\text{
$
\boxed{
\cos{ \left(~ \arcsin{(b)} ~\right) }
=\pm\sqrt{1-b^2}
=\pm \sqrt{1+ \left( \dfrac{1}{3} \right)^2}
=\pm \dfrac{\sqrt{8} }{ 3 }
}
$}}$$

$$\small{\text{
$
\boxed{
\sin{ \left(~ \arctan{(a)} ~\right) }
=\pm\dfrac{a}{\sqrt{1+a^2}}
=\pm\dfrac{ \sqrt{3} }{\sqrt{1+(\sqrt{3})^2}}
=\pm \dfrac{ \sqrt{3} }{2}
}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{ \left( \alpha_{rad}+ \beta_{rad} \right) }
&=&
\left( \pm \dfrac{1}{2} \right) \cdot
\left( \pm \dfrac{ \sqrt{8} }{ 3 } \right)
-\left( \pm \dfrac{ \sqrt{3} }{ 2 } \right) \cdot
\left( \dfrac{1}{3} \right) \\
&=& \dfrac{1}{6} \left( \pm \sqrt{8} \pm\sqrt{3} \right)
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{ \left(~ \arctan{(~\sqrt{3}~)}
+ \arcsin{\left(\dfrac{1}{3}\right)} ~\right)
}
&=& \dfrac{1}{6} \left( + \sqrt{8} + \sqrt{3} \right)
= 0.76007965539 \\\\
&=& \dfrac{1}{6} \left( + \sqrt{8} - \sqrt{3} \right)
= 0.18272938620 \\\\
&=& \dfrac{1}{6} \left( - \sqrt{8} + \sqrt{3} \right)
= -0.18272938620 \\\\
&=& \dfrac{1}{6} \left( - \sqrt{8} - \sqrt{3} \right)
= -0.76007965539
\end{array}
$}}$$

Yes, I tend to agree Zegroes.    

x+y=-1

x^2+y^2+6y-x=-5

Here is a graphical solution

Squaddy, your icon is great.   

It is a pity it is too small to read on the forum but I looked in your profile  

I didnt did on your back!!!!

U had post on this question!!!!!!!!

This means that you know about what is happening, so, I didnt did anything!!!!!!!

Hi Squaddy - I am wondering what catet means too?

$${{\mathtt{10}}}^{{\mathtt{2.38}}} = {\mathtt{239.883\: \!291\: \!901\: \!949\: \!046\: \!5}}$$

Total playing time for all 30 is    30*42 = 1260 minutes

Total playing time for 25 left is  $${\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{42.8}} = {\mathtt{1\,070}}$$

Total playing time for the 5 gone is    $${\mathtt{1\,260}}{\mathtt{\,-\,}}{\mathtt{1\,070}} = {\mathtt{190}}$$

Average playing time for the 5 is      $${\frac{{\mathtt{190}}}{{\mathtt{5}}}} = {\mathtt{38}}$$     minutes

Thanks Alan, I always like these types of questions :)

Great explanation DarkBlaze347           

Welcome to the forum Squaddy   

The answer you just gave is a REALLY good one.  Thank you  :)))

You name (as an answerer) will be in my next "End of Day Wrap" post   

Hi  Cstaib45,

Thanks for answering but you need to show your working OR state how you get your answers.

You can use the calc that is buit-in to the forum.  Just press the [Calculation ...] button that is in the ribbon when you are making a post :)

I have given you thumbs up because you didn't know.   :)

Rosala will return whenever she has a mind to. That will be a happy day. Your offensive attendance will in no way diminish it. Still, the celebration will be even better if you are not around.

There are two profoundly nauseating traits about you TR: Your tyrannical social posts and your butchery of mathematics. Bottling this would be a viable alternative for Ipecac syrup. No need for drinking, just opening the bottle will cause instant vomiting. ヽ(＾°̥̥̥̥̥̥̥̥＾‶)ﾉ

Thanks Chris,

I have taken your lead and put it in the 'great answers to learn" from thread   

This is a good one....!!!!!

Let N be the number of total marbles....

The first child recieves this many marbles  1 + 1/10 of the remainder  =

1 + (N -1) / 10    =   Equation (1)

The second child gets 2 plus 1/10 of what is left....

And what's left is: [ the N we started with minus the 1 marble the first child drew minus the 1/10 of what was left after that minus the 2 marbles the second child drew ] x 1/10...or, mathematically the second child gets....

2 + [N - 1 - (N-1)/ 10 - 2 ] / 10  = 2 + [ N - 3 - (N -1)/ 10 ] / 10    =  Equation (2)

And since each child gets the same number of marbles....we can set  (1) and (2) equal  and solve

1 + (N-1)/ 10 =  2 + [N - 3 - (N-1) / 10] / 10

-1 + (N - 1) / 10  = [N - 3 - (N -1) / 10] / 10

-10 + (N -1) = [N - 3 - (N -1)/ 10]

-10 + N -1 = [10N - 30 - (N -1)] / 10

-100 + 10N - 10  = 10N - 30 - (N - 1)

-110 + 10N = 10N - 29 - N

-110 +29 =  -N

-81 = -N

N = 81    and that's the number of marbles

So, using (1), each child gets 1 + (81-1)/ 10  = 9 marbles  and since there are 81 of them....there must be 9 children !!!

poor misguided Nauseated. you're just upset because of Rosala.
maybe i have to teach you a lesson

Good job, Anonymous!