radio

I mean with a unmarked staight edge and compass, how could I contruct an accurate version or not.

Well numbers go on infinately and since primes also theoretically go on infinately, so yes, but the number of primes found by using p+2 will exponetially decrease.

You can use the take away symbol.

$$-{\mathtt{3}}$$

When you do calculations, its best to use brackets around negative numbers:

$$\left(-{\mathtt{3}}\right)$$

Random =/= Funny

If a camera was reduced by 20%, then the equation would be:

$${\mathtt{X}}{\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{0.2}}\right) = {\mathtt{60}}$$

We simply rearrange to find X.

$${\mathtt{x}} = {\frac{{\mathtt{60}}}{{\mathtt{0.8}}}} \Rightarrow {\mathtt{x}} = {\mathtt{75}}$$

So the original price was £75

This is a very simple division problem. Simply divide 1.5x10^-10 by 2.5x10^-11 which is 6 meaning the Mercury atom is 6 times larger than a hydrogen atom. 

I don't know 

To find the midpoint between 2 points, we can do $${\frac{\left({{\mathtt{x}}}^{{\mathtt{1}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{x}}}^{{\mathtt{2}}}\right)}{{\mathtt{2}}}}$$ , $${\frac{\left({{\mathtt{y}}}^{{\mathtt{1}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{y}}}^{{\mathtt{2}}}\right)}{{\mathtt{2}}}}$$ . 

So by inputting the coordinates you gave,  $${\frac{\left({\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)}{{\mathtt{2}}}} = {\mathtt{1}}$$ , $${\frac{\left({\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}}\right)}{{\mathtt{2}}}} = {\mathtt{6}}$$ .

Then we can simply solve for x and y.

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}={\mathtt{2}}\\ {\mathtt{x}}=-{\mathtt{1}}\end{array}\right)} \Rightarrow {\mathtt{x}} = -{\mathtt{1}}$$

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}}={\mathtt{12}}\\ {\mathtt{y}}={\mathtt{14}}\end{array}\right)} \Rightarrow {\mathtt{y}} = {\mathtt{14}}$$

So C= (-1,14)

D**n it so close yet so far

Tau is the number of radians it takes to go around a circle.

An Armstrong number, aka a Narcissistic Number is a number where the digits in the number raised to the power of the number of digits in the number equals the original number. For example:

153

Since it is a 3 digit number, we raise 1, 5 and 3 to the power of 3.

$${{\mathtt{1}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{5}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{3}}}^{{\mathtt{3}}} = {\mathtt{153}}$$

1634

Since this is a 4 digit number, we raise 1, 6, 3 and 4 to the power 4.

$${{\mathtt{1}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{6}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{3}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{4}}}^{{\mathtt{4}}} = {\mathtt{1\,634}}$$

$${\mathtt{y}} = \left({\mathtt{4.956}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{0.586}}{\mathtt{\,\times\,}}{\mathtt{x}}\right){\mathtt{\,-\,}}\left({\mathtt{3.214}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{3}}}\right)$$

This is impossible to solve since we need aat least 2 equations to work out a value for both x and y via a simultaneous equation.

The probability of throwing at least one 4 on a pair of dice is 1 in 6, so the probability of sucess is 1/6. The probability of faliure is 1-1/6 which is 5/6. As it is after 7 rolls, there are 7 trials in the experiment.

Putting these values into an equation for binomial distribution, nCr*P^(r)*Q^(n-r), where P is prob of sucess, q is prob of failure, r is number of sucesses and n is the number of trials, we get...

$${\left({\frac{{\mathtt{7}}{!}}{{\mathtt{1}}{!}{\mathtt{\,\times\,}}({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}^{\left({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{1}}\right)} = {\mathtt{0.390\: \!714\: \!306\: \!127\: \!114\: \!8}}$$

as the probability of getting a 4 in one roll.

This is very likely wrong.

Anyone better at maths should check this first, since I'm very very unconfident in my binomial skills.

Since both are conveniently  $${{\mathtt{x10}}}^{-{\mathtt{11}}}$$ , we can just divide 2.5 by 1.5

$${\frac{{\mathtt{2.5}}}{{\mathtt{1.5}}}} = {\frac{{\mathtt{5}}}{{\mathtt{3}}}} = {\mathtt{1.666\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

So a hydrogen atom is 1.6 times larger than a mercury atom.

To first find the volume of the sphere, you need to first find the radius. To find this, you take the circumference of 29.5 inches and divide by pi to get the diameter.

$${\mathtt{C}} = {\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{D}}$$

$${\frac{{\mathtt{29.5}}}{{\mathtt{3.14}}}} = {\frac{{\mathtt{1\,475}}}{{\mathtt{157}}}} = {\mathtt{9.394\: \!904\: \!458\: \!598\: \!726\: \!1}}$$

Then divide the diameter by 2 to get the radius.

$${\frac{\left({\frac{{\mathtt{29.5}}}{{\mathtt{3.14}}}}\right)}{{\mathtt{2}}}} = {\frac{{\mathtt{1\,475}}}{{\mathtt{314}}}} = {\mathtt{4.697\: \!452\: \!229\: \!299\: \!363\: \!1}}$$

So we get a radius of about 4.70.

Then we use the formula for the volume of a sphere which is 

$${\mathtt{V}} = {\frac{{\mathtt{4}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{r}}}^{{\mathtt{3}}}$$

Substituting in R for 4.69........, we get

$${\frac{{\mathtt{4}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{3.14}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1\,475}}}{{\mathtt{314}}}}\right)}^{{\mathtt{3}}} = {\mathtt{433.965\: \!796\: \!448\: \!807\: \!93}}$$

Rounded to the nearest inch, we get 434 inches cubed.

Sorry that's wrong. Think about it more carefully.

100% is equal to 1 in decimals, so 165 would be 0.16.

You could try downloading the page for offline reading.

11

The only number that factorises into 1 is 1 because 1x1=1

1 + 10 = 11, so 10 less than 11 rooted is 1