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Just enter 3.14^39 in the calculator that is on the home page of this site and hit enter and you will get your answer!

If you are talking about base 10 then it is just 10 to the power of

log x=7

x=10^7

answer is 58.1

1097509546666.6666666666666667

√x :)

P(a least 1 is defective)

=1-P(none are defective)

$$\\=1-\left(\frac{11}{15}\times \frac{10}{14} \times \frac{9}{13}\right)$$

thanks Chris :)

No Mellie,

Chris answered this one correctly mine is not correct.    

(Unless perhaps you were given a pic to help eliminate confusion :/  )

Your resorces often seem to give you incorrect answers :)

Or at least they often give you the correct answer to a slightly different question from the one that they actually asked.     

 75.36:230.79

7536:23079

$${\frac{{\mathtt{7\,536}}}{{\mathtt{23\,079}}}} = {\frac{{\mathtt{16}}}{{\mathtt{49}}}} = {\mathtt{0.326\: \!530\: \!612\: \!244\: \!898}}$$

So the answer is     16:49

Beats me, is the price of each book proportional to their weight ??

I suppose if you were going to use them as fuel for a fire this would be reasonable :/

Actually they are exercise books so I guess that is not really so unreasonable.  

I still need more info though.       

sin x^0=7/9,find x

x^0=1    

(sinx)^0=1

so the x is irrelvant

OH I get you - that is not a power of 0 that is a degree sign - silly me.  It would have been better just to write degrees :))

$$\\sin x^0=7/9\\
asin(7/9)=x\\$$

arc sin is the same as inverse sin

If you need to find an angle then you need to use inverse (or arc) trig    :)

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{7}}}{{\mathtt{9}}}}\right)} = {\mathtt{51.057\: \!558\: \!731\: \!019^{\circ}}}$$

Actually, Melody was correct, but still thank you so much for helping me!!

$$The total area of the dartboard is $4\times4=16$ square inches. The area of the shaded region is the area of the $3\times3$ region after subtracting the area of the $2\times2$ region, or $9-4=5$ square inches. The shaded region is $5$ square inches out of the possible $16$ square inches the dart could hit, so the probability of the dart hitting the shaded region is $\boxed{\frac{5}{16}}$.$$

$${\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{2.4}}{\mathtt{\,\times\,}}{{\mathtt{6}}}^{{\mathtt{2}}} = {\frac{{\mathtt{581}}}{{\mathtt{10}}}} = {\mathtt{58.1}}$$

$$\\10 logx=-62\\
logx=-6.2\\
10^{logx}=10^{-6.2}\\
x=10^{-6.2}\\$$

$${{\mathtt{10}}}^{\left(-{\mathtt{6.2}}\right)} = {\mathtt{0.000\: \!000\: \!630\: \!957\: \!344\: \!5}}$$

On the web2 calc you do not need to do all that - it will do it for you.  Tricky isn't it.   

$${\mathtt{10}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{x}}\right) = -{\mathtt{62}} \Rightarrow {\mathtt{x}} = {{\mathtt{e}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{31}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{10}}\right)}}{{\mathtt{5}}}}\right)} \Rightarrow {\mathtt{x}} = {\mathtt{0.000\: \!000\: \!630\: \!957\: \!344\: \!5}}$$

What is the rest of the question?

$$slope \;(gradient)=\frac{7-0}{-4--15}=\frac{7}{-4+15}=\frac{7}{11}$$

This net might help you visualise the cylinder

Hi Atoombii        

I think that you are saying that there are 2 peices of paper.

Each one is 8.5 cm and 11cm   and they are each wrapped to make cylinders.

One is 8.5cm high and a circumference of 11cm

and the other is 11 cm high wirh a circumference of 8.5cm

Well, since they both use the same sized peice of paper the surface are of the curved surfaces must be identical.

So if you are refering to the SA of  the open cylinders then they are identical.

If you put ends on your cylinder then the one with the biggest circumference will have the biggest 'end' 

So  the short squat one will have the biggest surface area.     

this is not a quadratic equation.

Perhaps you want to rationalize the denominator?

$$\\\frac{4}{\sqrt3-5}\\\\
=\frac{4}{\sqrt3-5}\times \frac{\sqrt3+5}{\sqrt3+5}\\\\
=\frac{4(\sqrt3+5)}{3-25}\\\\
=\frac{4(\sqrt3+5)}{-22}\\\\
=\frac{-2(\sqrt3+5)}{11}\\\\
=\frac{-2\sqrt3-10}{11}\\\\$$

So it seems you have rationalized the denominator correctly :)

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