All Answers 358098 Answers

i think it is 6 :)

I guess 32) might be:

$$3\sqrt[3]{5y^3}\times 2\sqrt[3]{50y^4}$$

Multiply the terms outside the surds together, and multiply the terms inside the surds together:

$$6\times \sqrt[3]{250y^7}$$ 

Now you want to express the terms inside the surd as something-cubed multiplied by whatever is left over and can't be expressed as a nice cube!

$$250=2\times 5^3\\
y^7=y\times (y^2)^3$$ 

So:

$$6\sqrt[3]{250y^7}=6\sqrt[3]{2\times 5^3\times y\times (y^2)^3}=6\times5\times y^2\times \sqrt[3]{2y}$$

or just:

$$30y^2\sqrt[3]{2y}$$

But you must use brackets more to ensure your meaning is clear and unambiguous Nataszaa!

OK.....use it to do what???

Ths "identity" for the cosine is :  adjacent / hypoteneuse

Do you mean

$$\frac{x^2}{x+2}+4x ?$$     If you do this is how you should have presented it.    x^2/(x+2) +4x

To make double sure we understand you coule REALLY stress you meaning this way.   [x^2/(x+2)] +4x

$$\frac{x^2}{x+2}+4x \\\\
= \frac{x^2}{x+2}+\frac{4x(x+2)}{x+2} \\\\
= \frac{x^2+4x(x+2)}{x+2} \\\\
= \frac{x^2+4x^2+8x}{x+2} \\\\
= \frac{5x^2+8x}{x+2} \\\\
= \frac{5x^2+8x}{x+2} \\\\$$

Tues 22/7/14

♬                                        ♬ ♬ MELODY ♬ ♬

1) Let's end the day with Wonders.  Says it all, dosen't it.  Thanks Rosala.

@@ End of Day Wrap :  Mon 21/7/14   Sydney, Australia   Time 10:30 pm    ♬       

Hi Everyone,

Great answers have been provided by CPhill, NinjaDevo, Rosala, Zegroes, Kitty<3, Alan, Heureka and DragonSlayer554.  A  big thank you to each of you.  

Zegroes said I needed a COOL avatar but Godzilla was already taken.   

So I became the next best thing, a dinosaur!

It has not been busy but here are a couple of interest posts.

1) Rosala's Wrap - Thanks Rosala, we really enjoy these. I love the Goa waterfall! And the fun pics as well!

30) √8x^5*√3x

$$\sqrt8x^5\sqrt3x = \sqrt{24}x^6 \\\\$$

There is no cubed root in this question. 

$$\sqrt{54}$$

You need to find a squared number that goes into 54.  If there is not one then you cannot simplfy!

Square numbers are 4,9,16,25,36,49, etc (1 is not helpful)

9 goes into 54.

$$\sqrt{54}\\
=\sqrt{9*6}\\
=\sqrt9*\sqrt6\\
=3\sqrt6$$

I am sorry but this question does not make sense.   

I am not sure what you are actually asking.  

It would be better if you helped this person learn how to do the question Zegroes.  

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

there  are 2 ways

15% of 32.50 = 15/100*32.50 = $4.875  

$32.50 - $4.875 = $27.625

So either $27.62 or $27.63

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

OR

100% - 15% = 85%

85% of $32.50 = 85/100*32.50 = $27.625

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

There is not enough information here to give any answer.  Many functions and relations would meet these criteria. 

Do you mean an yearly income of $70,000 for ever?  (A perpetuity?)

Is  this your question?

How much will I need to save every year in order to retire with a perpetuity of $70000 paid at the end of each year into perpetuity?  Assume the money earns 5% per annum .

If this is your question then I will answer it.  Let me know.  

It depends on what other information that you have.

If you have another angle then you can use the angle sum of a triangle (180degrees)

If you have 2 sides then you use inverse trigonometry.

$$\frac{4}{(2-3x)}-\frac{1}{(1+x)}+\frac{3}{(1+x^2)}\\\\
=\frac{4(1+x)(1+x^2)}{(2-3x)(1+x)(1+x^2)}-\frac{1(2-3x)(1+x^2)}{(1+x)(2-3x)(1+x^2)}+\frac{3(1+x)(2-3x)}{(1+x^2)(1+x)(2-3x)}\\\\
=\frac{4(x^3+x^2+x+1)}{(2-3x)(1+x)(1+x^2)}-\frac{1(-3x^3+2x^2-3x+2)}{(1+x)(2-3x)(1+x^2)}+\frac{3(-3x^2-x+2)}{(1+x^2)(1+x)(2-3x)}\\\\
=\frac{(4x^3+4x^2+4x+4)}{(2-3x)(1+x)(1+x^2)}+\frac{(3x^3-2x^2+3x-2)}{(1+x)(2-3x)(1+x^2)}+\frac{(-9x^2-3x+6)}{(1+x^2)(1+x)(2-3x)}\\\\
=\frac{(4x^3+4x^2+4x+4)+(3x^3-2x^2+3x-2)+(-9x^2-3x+6)}{(2-3x)(1+x)(1+x^2)}\\\\
=\frac{7x^3-7x^2+4x+8}{(x^2+1)(-3x+2)(x+1)}\\\\$$

I think that is ok.  It hasn't simplified very nicely - maybe I made an error somewhere 

Point G is located at (3, −1), and point H is located at (−2, 3). Find the x value for the point that is 2 over 3 the distance from point G to point H.

$$\\x=(1-\lambda)*3 + \lambda*(-2) \quad | \quad x_g=3 \quad and \quad x_h = -2 \\\\
\text{if } \lambda=\textcolor[rgb]{1,0,0}{0} \Rightarrow x = (1-\textcolor[rgb]{1,0,0}{0})*3+\textcolor[rgb]{1,0,0}{0}*(-2) = 3 = x_g \\
\text{if } \lambda=\textcolor[rgb]{1,0,0}{1} \Rightarrow x = (1-\textcolor[rgb]{1,0,0}{1})*3+\textcolor[rgb]{1,0,0}{1}*(-2) = -2 = x_h\\\\
\text{if } \lambda=\textcolor[rgb]{1,0,0}{{2\over3 }} \Rightarrow x = (1-\textcolor[rgb]{1,0,0}{{2\over3}} )*3+\textcolor[rgb]{1,0,0}{{2\over3}} *(-2)\\
\Rightarrow x={1\over3}*3 - {4\over3}\\
\Rightarrow x=-{1\over 3}$$

Nice thinking CPhill!

Yeah !for me too!this has been happening with most of the questions!i think!but i can't figure out the actual problem!!!!

$$\begin{array}{rclclcl}
x &=& x^1 &=& x^{(1)} &=& x^1\\
x*x &=& x^1*x^1 &=& x^{(1+1)} &=& x^2\\
x*x*x &=& x^1*x^1*x^1 &=& x^{(1+1+1)} &=& x^3\\
x*x*x*x &=& x^1*x^1*x^1*x^1 &=& x^{(1+1+1+1)} &=& x^4\\
x*x*x*x*x &=& x^1*x^1*x^1*x^1*x^1 &=& x^{(1+1+1+1+1)} &=& x^5
\end{array}$$

This is an empirical formula for the wind-chill temperature (in °F) where the ambient temperature is T°F and the wind-speed is V miles/hr.  It's normally written as:

$$$$WCT=35.74+0.6215\times T-35.75\times V^{0.16}+0.4275\times T\times V^{0.16}$$

(Notice that 0.16 = 4/25)

More info needed.  How many people?  How many are coupled?

There is more than one answer, depending on where the parentheses should go in your question. 

It could be:  (8 * 10)^67 

Or: 8 * (10^67)

33.78^(1/19)  = 1.2035268145828129

Thanks CPhil.

I had to have my dad help me with this one, I was getting confused on the whole different bases deal.

So I give a tip of the hat to my dad. :)