All Answers 358407 Answers

x^2 = 16x -28

x^2 - 16x +28 =0

(x -14) (x -2) =0

x -14 =0 or x-2 =0

x =14 or x=2

there are two answers for x

they are 14 and 2

sqrt(8-sqrt(55))+sqrt(8+sqrt(55))

$$\sqrt{8-\sqrt{55}}+\sqrt{8+\sqrt{55}} \quad | \quad \sqrt{ x^2 } \\ \\
= \sqrt{ \left( \sqrt{8-\sqrt{55}}+\sqrt{8+\sqrt{55}} \right)^2 } \\ \\
= \sqrt{ (8-\sqrt{55}) +(8+ \sqrt{55}) +2*\left( \sqrt{8-\sqrt{55}} \right)* \left(\sqrt{8+\sqrt{55}} \right)} \\ \\
= \sqrt{ 16+2*\left( \sqrt{8-\sqrt{55}} \right) * \left(\sqrt{8+\sqrt{55}} \right) } \\ \\
= \sqrt{ 16+2* \sqrt{8^2- 55 } }\\ \\
= \sqrt{ 16+2* \sqrt{64-55} } \\ \\
= \sqrt{ 16+2* \sqrt{9} } \\ \\
= \sqrt{ 16+2* 3 }\\ \\
= \sqrt{ 22 }\\ \\
= 4.69041576$$

Very crafty, Melody....!!!

$$\sqrt{8-\sqrt{55}}+\sqrt{8+\sqrt{55}}$$

This is a tough one!

First I am going to consider      $$8-\sqrt{55}$$

I want to express this as a perfect square.

$$\\8-\sqrt{55}\\\\
=\frac{5}{2}-\sqrt{55}+\frac{11}{2}\\\\
=\frac{25}{10}-\frac{10\sqrt{55}}{10}+\frac{55}{10}\\\\
=\frac{25-10\sqrt{55}+55}{10}\\\\
=\frac{5^2-10\sqrt{55}+(\sqrt{55})^2}{10}\\\\
=\frac{(5-\sqrt{55})^2}{10}\\\\
But\;\; (5-\sqrt{55})^2=(\sqrt{55}-5)^2\;\;$and I want the positive one, so $\\\\
=\frac{(\sqrt{55}-5)^2}{10}\\\\$$

$$\\Hence\\\\
\sqrt{8-\sqrt{55}}\\\\
=\sqrt{\frac{(\sqrt{55}-5)^2}{10}}\\\\
=\frac{\sqrt{55}-5}{\sqrt{10}}}\\\\
=\frac{\sqrt{10*55}-5\sqrt{10}}{10}\\\\
=\frac{\sqrt{2*5*5*11}-5\sqrt{10}}{10}\\\\
=\frac{5\sqrt{22}-5\sqrt{10}}{10}\\\\
=\frac{\sqrt{22}-\sqrt{10}}{2}\\\\$$

NOW, BY THE SAME LOGIC,

$$\\Hence\\\\
\sqrt{8}+\sqrt{55}=\frac{\sqrt{22}+\sqrt{10}}{2}\\\\
SO\\\\
\sqrt{8-\sqrt{55}}+\sqrt{8+\sqrt{55}}\\\\
=\frac{\sqrt{22}-\sqrt{10}}{2}+\frac{\sqrt{22}+\sqrt{10}}{2}\\\\
=\frac{2\sqrt{22}}{2}\\\\
=\sqrt{22}\\\\$$

Simple interest

I = PRT = 5600*0.03*2=$336

FV = 5600+336=$5936 

--------------

compound int - Unit trust

FV=5600(1+0.045)^2

FV=5600*(1.045)^2

$${\mathtt{5\,600}}{\mathtt{\,\times\,}}{\left({\mathtt{1.045}}\right)}^{{\mathtt{2}}} = {\mathtt{6\,115.34}}$$  dollars

----------------------

Difference = 6115.34-6025.60

$${\mathtt{6\,115.34}}{\mathtt{\,-\,}}{\mathtt{5\,936}} = {\frac{{\mathtt{8\,967}}}{{\mathtt{50}}}} = {\mathtt{179.34}}$$

so yes the difference is almost $180 

@@ End of Day Wrap    Sun 25/1/15     Sydney, Australia      Time 10:35 pm

Hello everyone,

Our great answerers today were Sasini, CPhill, saseflower, Alan and Heureka.  A big thank you to each of you.

Interest posts:

Mon 26/1/15

$$\\\frac{\sqrt{6}\times 2}{ \sqrt {3}}\\\\
=2\times\frac{\sqrt{6}}{ \sqrt {3}}\\\\
=2\times\sqrt{2}\\\\
=2\;\sqrt{2}\\\\$$

Standard form is what the English call scientific notation.

$${\frac{{\mathtt{59\,639}}}{{\mathtt{0.087}}}} = {\mathtt{685\,505.747\: \!126\: \!436\: \!781\: \!609\: \!2}}$$

$$6.85505747.... \times 10^5$$

$$1.230\times 10^3$$

You could write this as 12300 but then you would not know if it had 3 or 4 or 5 significant figures/

what is -3xto the power of 2 -x to the power of 2 + 11x to the power of 2

$$\\-3x^2-x^2+11x^2\\
=-3x^2-1x^2+11x^2\\
=(-3-1+11)x^2\\
=7x^2$$

What graph?     

I suppose for any graph you go across to x=0.0001  then go up or down to find the y value that is on your graph.    

do you mean

$$\\log_3 x + 11 log_3 x^3 = 14\\\\
log_3 x + 33 log_3 x = 14\\\\
34 log_3 x = 14\\\\
log_3 x = \frac{14}{34}\\\\
\frac{logx}{log3} = \frac{14}{34}\\\\
logx = \frac{14}{34}\times log3\\\\
10^{logx} =10^{\left( \frac{14}{34}\times log3\right)}\\\\
x =10^{\left( \frac{7}{17}\times log3\right)}\\\\$$

$${{\mathtt{10}}}^{\left({\frac{{\mathtt{7}}}{{\mathtt{17}}}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right)\right)} = {\mathtt{1.572\: \!033\: \!125\: \!399\: \!619\: \!1}}$$

$${\mathtt{28}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{67\,313}}}} = {\mathtt{7\,264.529\: \!716\: \!368\: \!431\: \!237\: \!3}}$$

I entered it in the input line like this     28*sqrt(67313)

You cannot put  *^    together - it has no meaning    

Thanks Alan  

178=17800%

I think that her marginal tax rate is 40% 

It is what she has to pay just on the top $1000 of her income.

It's possible this means there are "x" and "y" values given for years (?) 1995 and 1996, such that the following holds:

.

Your answer is good anon, thank you.  

Definitely zero 

.

If tan(α/2) =  3 then α/2 = tan^-1(3) and α = 2tan^-1(3).  Most calculators will give the 1st quadrant value for tan^-1(3), so just finding sin(2tan^-1(3)) will give the correct result:

$${\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{3}}\right)} = {\mathtt{143.130\: \!102\: \!354\: \!156}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{3}}\right)}\right)} = {\frac{{\mathtt{3}}}{{\mathtt{5}}}} = {\mathtt{0.6}}$$

.

a=27     r=1/3

27+9+3+1+(1/3)+(1/9)=364/9

$$\boxed{S_n=\frac{a(1-r^n)}{1-r}}\\\\\\
\frac{364}{9}=\frac{27(1-(1/3)^n)}{1-(1/3)}}\\\\
\frac{364}{9*27}=\frac{1-(1/3)^n}{2/3}}\\\\
\frac{364*2}{9*27*3}=1-(1/3)^n\\\\
\frac{728}{729}-1=-(1/3)^n\\\\
\frac{-1}{729}=-\frac{1}{3^n}\\\\
729=3^n\\\\
log_3(729)=log_3 3^n\\\\
n=log_3(729)\\\\
n=log729/log3\\\\
n=6$$

Thank you Chris for helping me spot my previous error   

You need to put a 3 in front of the NaOH and the NaCl terms to make this balance.

.