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I know, I know       

Thanks for this fun question anon :))

I'm going to put this is my wrap tonight but only little kids are allowed to answer.   :))

$$\left({\mathtt{1}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{125\,754.42}}}{{\mathtt{3\,500}}}}{\mathtt{\,\times\,}}{\mathtt{0.016}}\right)\right) = {\mathtt{0.425\: \!122\: \!651\: \!428\: \!571\: \!4}}$$

Since we have two different answers I might as well do it as well

$$\\x=\sqrt{(\sqrt{15}-\sqrt{3})^2+(\sqrt{15}-\sqrt{3})^2}\\\\
x=\sqrt{2*(15-2\sqrt{45}+3)}\\\\
x=\sqrt{2*(18-6\sqrt{5})}\\\\
x=\sqrt{36-12\sqrt{5}}\\\\$$

$${\sqrt{{\mathtt{36}}{\mathtt{\,-\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}} = {\mathtt{3.027\: \!735\: \!832\: \!268\: \!483\: \!3}}$$

check

$${\sqrt{{\mathtt{2}}{\mathtt{\,\times\,}}{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}}} = {\mathtt{3.027\: \!735\: \!832\: \!268\: \!483\: \!3}}$$

Seems anon is correct on this one.           Sorry Chris.   

Visually.... imagine connecting both ends of the written 5.... you'll get roughly shaped 8 :-)

Sarafina is making monthly payments into an annuity. She wants to have $600 in the fund to buy a new convection range in six months, and the accounts pays 4.8% annual interest. What are her monthly payments to the account?

This is a future value of an ordinary annuity (the first payment is immediate, the first interest is paid at the end of the month.

C=?      FV=600,        i=0.048/12=0.004,      n=6

You can sub the values in and get the answer  :)

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{60}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{30}}}{{\mathtt{60}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{278.5}} = {\mathtt{242.394\: \!061\: \!319\: \!29}}$$

You do not pu the degree sign in you just set your calculator to degrees.

this is done differenly on diff calculators.

On the web2 calc there is a toggle button on tthe bottom left of the calc.

more answers

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Please anon, follow the rules for reposting.

If Alan was kind enough to answer you,   it is polite for you to question his answer on the original thread.

None of us mind our answers being questioned, we like it, it means we really are teaching, and you really are trying to understand. 

REMEMBER:

If you do not understand then PLEASE ask us to explain more. !!!

Is this the new 2+2   or   10+9     ?

I must have seen this question at least a half dozen time in the last few days!

Nauseated and Chris, do you think we have the next permutation for a troll post? 

Well, if the gradient of the line joining one pair is the same as the gradient of the line joining another pair then the points must be collinear.  (that is , on the same line)

Thanks anon 

Please look in the sticky topics on the right.

Thanks Alan,  I am learning/revising too.    

Potential energy = mass* gravity * height

[mass   kg]        [ gravity   m/s^2]           [ height  m]        Energy is measured in Joules

the amount of juice will be

(1-0.1)(0.65+0.4+0.35)

I'll let you, or someone else research the properties.   

You must multiply the original price with the 40%. After that, you add the orginal price with the answer you got after multiplying the original price with 40%.

If this wasn't the answer you were looking for... I don't know what to tell you.

The answer is 2100

$${\frac{{\mathtt{7}}}{{\mathtt{10}}}}{\mathtt{\,\times\,}}{\mathtt{7\,000}} = {\mathtt{4\,900}}$$, 1) This is the first step: to find 7/10ths of 7000

$${\mathtt{7\,000}}{\mathtt{\,-\,}}{\mathtt{4\,900}} = {\mathtt{2\,100}}$$    , 2) Then Subract

If you look at this cotangent graph perhaps you will able to 'see' why    $$cot(\infty)$$    is undefined.

3000 people 

Here are the values....

$${\mathtt{\,-\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{133}}}}{\mathtt{\,-\,}}{\mathtt{5}}\right)}{{\mathtt{18}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{133}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right)}{{\mathtt{18}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.362\: \!920\: \!144\: \!148\: \!377\: \!5}}\\
{\mathtt{x}} = {\mathtt{0.918\: \!475\: \!699\: \!703\: \!933\: \!1}}\\
\end{array} \right\}$$

Since this equation is easy to solve using the quadratic formula (see below), what is your criterion for "approximate"?  Is it just a number of decimal places, or are you expected to guess a few values and home in on the results, or what?

$${\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{58}}}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{7}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{58}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{{\mathtt{7}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.659\: \!396\: \!157\: \!980\: \!558\: \!3}}\\
{\mathtt{x}} = {\mathtt{1.516\: \!539\: \!015\: \!123\: \!415\: \!5}}\\
\end{array} \right\}$$

To two decimal places the values are x = -0.66 and x = 1.52

.

a+b=4ab   b+c=5bc   c+a=3ac

a= b(4a - 1)     a = c(3a - 1)

b = a/(4a -1)  c = a/(3a -1)

b + c = 5bc

a/(4a -1) + a/(3a -1)  = 5a^2/[(4a -1)(3a -1)]

[a(3a -1) + a(4a -1)]/ [(4a -1)(3a -1)]  = 5a^2/[(4a -1)(3a -1)]

3a^2 - a + 4a^2 - a  = 5a^2

2a^2 - 2a = 0

a^2 - a = 0

So

a(a -1) = 0     0 is trivial....so a = 1

So  b = a/[4(a) -1] = 1 / [4(1) -1] = 1/3 

And  c = a/[3(a) -1] = 1 /[3(1)-1] = 1/2

Check

a + b = 1 + 1/3 = 4/3 = 4ab =4(1)(1/3) = 4/3  

b + c = 1/2 + 1/3  = 5/6 =5bc = 5(1/3)(1/2) = 5/6

a + c = 1 + 1/2 = 3/2 = 3ac = 3(1)(1/2) = 3/2

Alternatively, you could plot a graph:

.