All Answers 358516 Answers

Infinity is a concept NOT  a number.

$${\frac{{\mathtt{934\,500}}}{{\left({\mathtt{1.07}}\right)}^{{\mathtt{42.3}}}}} = {\mathtt{53\,412.820\: \!234\: \!701\: \!481\: \!951\: \!6}}$$

just a little info here.

In many countries a comma is used instead of a decimal point.  

Thank you anon2 for answering and question asker anon for giving feed back on the answer.

Lets take a look.

$$sin36=\frac{7}{x}$$

multiply both sides by x

$$\\x\times sin36=\not{x}\times \frac{7}{\not{x}}\\\\
x\times sin36=\not{x}\times \frac{7}{\not{x}}\\\\
x\times sin36=7\\\\$$

divide both sides by sin36    

$$\\x\times sin36=7\\\\
\frac{x\times sin36}{sin36}=\frac{7}{sin36}\\\\
x=\frac{7}{sin36}\\\\$$

Now when you put this into the web2 calc you have to put the 36 in brackets I think

7/sin(36)=

$${\frac{{\mathtt{7}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{36}}^\circ\right)}}} = {\mathtt{11.909\: \!111\: \!316\: \!938\: \!145\: \!1}}$$

so 11.9 degrees is the correct answer.

What do you think the answer should be?  Maybe we can clear up the misunderstanding?

Assuming this is 

$$6x^{1.7}+4=14$$

subtract 4 from both sides

$$6x^{1.7}=10$$

divide by 6

$$x^{1.7}=\frac{10}{6}$$

Take logs of both sides and use the fact that log(a^b) = b*log(a) and log(a/b) = log(a) - log(b)

$$1.7\log{x}=\log{10}-\log{6}$$

divide by 1.7

$$\log{x}=\frac{\log{10}-\log{6}}{1.7}$$

Use the fact that log₁₀a = b means a = 10^b

$$x=10^{\frac{\log{10}-\log{6}}{1.7}}$$

and since log₁₀10 = 1

$$x=10^{\frac{1-\log{6}}{1.7}}$$

You can complete the calculation.

.

$${\frac{\left({\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{16}}}}\right)}{\left({\mathtt{5}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\right)}}$$

$${\frac{\left({\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}{\left({\mathtt{5}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\right)}}$$

$${\frac{{\mathtt{7}}}{\left({\mathtt{5}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\right)}}$$

$$\left({\frac{{\mathtt{7}}}{\left({\mathtt{5}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\right)}}{\mathtt{\,\times\,}}{\frac{\left({\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{2}}}}\right)}{\left({\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{2}}}}\right)}}\right)$$

$$\\\frac{35+7\sqrt{2}}{25-2}\\\\
\frac{35+7\sqrt{2}}{23}\\\\$$

The probability that you wouldn't win at all is (2/3)⁵, so the probability you would win at least once is 1 - (2/3)⁵

$${\mathtt{p}} = {\mathtt{1}}{\mathtt{\,-\,}}{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}^{{\mathtt{5}}} \Rightarrow {\mathtt{p}} = {\mathtt{0.868\: \!312\: \!757\: \!201\: \!646\: \!1}}$$

(1/3)⁵ gives the probability that you win in all five competitions!

.

The total area under a normal curve is 1. 

So you need to know, or work out, the  relevant z scores for your question.

if you colour the bit of the curve that makes your question correct you can look up the z table table and work out the percent probability from that.

I haven't explained well, I know this.  If you give an exact question it might be easier.

On the web 2 calc you have to do it the long way

5 degrees 7 minutes and 8 seconds would be entered as  (5+7/60+8/3600)

Look at the topics in the sticky notes on the bottom right of your page.

Perhaps you are looking for the principal?

simple intrest what is the intrest for I=54 R=6% and t= 9months

I=Prt

54=P*0.06*9/12

$${\frac{{\mathtt{0.06}}{\mathtt{\,\times\,}}{\mathtt{9}}}{{\mathtt{12}}}} = {\frac{{\mathtt{9}}}{{\mathtt{200}}}} = {\mathtt{0.045}}$$

54=P*0.045

54/0.045 = P

$${\frac{{\mathtt{54}}}{{\mathtt{0.045}}}} = {\mathtt{1\,200}}$$

the principal would have been $1200

I think

PE = m*g*h

10 = 10*9.8*h

h = 1/9.8 

$${\frac{{\mathtt{1}}}{{\mathtt{9.8}}}} = {\frac{{\mathtt{5}}}{{\mathtt{49}}}} = {\mathtt{0.102\: \!040\: \!816\: \!326\: \!530\: \!6}}$$

I think that is 0.102 metres.   

(on Earth, at sea level)

Look in the sticky topics on the lower right side of your page.

One is on fractions and the calc

$${\frac{{\mathtt{2\,977.824\: \!44}}}{{\mathtt{5\,298.62}}}} = {\frac{{\mathtt{281}}}{{\mathtt{500}}}} = {\mathtt{0.562}}$$

Use a statistical Standard Distribution Chart.

I believe that you are missing part of the data. What was the amount of money on which the interest was charged?

$${\frac{\left({\mathtt{7}}\right)}{\left({\mathtt{5}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\right)}} = {\mathtt{1.952\: \!151\: \!953\: \!765\: \!724\: \!6}}$$

$${\frac{{\mathtt{15}}}{\left({\frac{{\mathtt{5}}}{{\mathtt{7}}}}\right)}} = {\mathtt{21}}$$

$$4\frac{5}{8}$$

is not a whole number. If you mean, round to the nearest whole number, then it is 5.

LOG 5^ 1/5=X

$$\small{\text{ $ x=\log( 5^{\frac{1}{5}}) $ }}\\
\small{\text{ $ x=\frac{1}{5}\cdot \log( 5) $ }}\\
\small{\text{ $ x=0.2\cdot \log( 5) $ }}\\
\small{\text{ $ x=0.2\cdot 0.69897000434 \quad | \quad \log( 5)= 0.69897000434 $ }}\\
\small{\text{ $ x=0.13979400087 $ }}\\$$

$$\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{5\,823}}{\mathtt{\,\times\,}}{\mathtt{2}}\right){\mathtt{\,-\,}}\left({\mathtt{31\,323}}{\mathtt{\,\times\,}}{\mathtt{55}}\right) = -{\mathtt{1\,699\,473}}$$

Hard to answer this without knowing what you mean. Instead of a comma, should there be another sign of operation?

$${\frac{\left({\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{3}}}\right)}{\left({{\mathtt{2}}}^{{\mathtt{2}}}\right)}} = {\mathtt{32}}$$

53412.8202347014819516$$\mathrm{\ }$$