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We use the fact that $\sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{ab}$ to get that $\sqrt[3]{2} \cdot \sqrt[3]{14} = \sqrt[3]{2 \cdot 14} = \boxed{\sqrt[3]{28}}$.

Hoping this helped,

asdf334

Thanks! I've edited my answer with the correction.

I think that you might need physics for this, as the answers for each is 90 degress and 12 feet. I am possibly incorrect, but this question's wording does not make sense.

Hoping this helped,

asdf334

You are almost right asdf334, but you slipped up at the end! The probability of three heads is (1/2)³ = 1/8 (i.e. the same as getting no tails).

The formula for the sum of the angles of an n-gon (in this case, a 10-gon), is 180(n - 2). So, the sum of the angles is 1440. Now, we just calculate 1440 - (156 + 178 + 124 + 132 + 138 + 142 + 116 + 178 + 159) = 1440 - 1323 = 117.

Hoping this helped,

asdf334

Since I don't know LaTeX that complicated, I will tell you the probability of rolling 0, 1, 2, and 3 heads.

When you roll 0 heads, that means you can't roll a head each time, or $\left(\frac{1}{2}\right)^3$, which equals $1\over8$.

When you roll 1 head, that means you roll HTT, THT, and TTH. The probably is still the same, but we multiply by 3 for each ordering, for a probability of $3\over8$.

When you roll 2 heads, we can use the same logic to get $3\over8$ again.

When you roll 3 heads, you get a head each time, or $\left(\frac{1}{2}\right)^3={1\over8}$.

Hoping this helped,

asdf334

(2)^(1/3)  * (14)^(1/3)   =    (28)^1/3  

I get the same as Chris

One machine added  11 scoops      the other machine took 38 scoops

Do you know what the correct answer is?  

If you do you should share it with us.  

Thank You Daisy!  

Dang that is still not the correct answer 

I beilive I have heard that  a decagon equals 1440 degree's what you can do lets see you have 9 angles degree mesaurments on here add them then subtract the sum from 1440 after that your diffrence will be your answer..... 

like this 

$156 + 178 + 124 +132 +138+142+116+178 +159 = sum$

$1440 - sum = difference$

$Difference = answer.$

Thanks Alan,

I did think there might be a higher answer than mine.

I guess I should have said so   :)

I also should have thought to use calculus. That never even crossed my mind  

Correct Melody! In fact we have:

$\frac {1}{x} + \frac {1}{y} = \frac {1}{12}\\~\\ \frac{1}{x}<\frac{1}{12}\qquad \qquad \frac{1}{y}<\frac{1}{12} \\ x>12\qquad \qquad y>12 \\ so\\ x+y>24$

That is not correct Nickolas.

Why did you post the same question twice!     How annoying!

1.    2 0[1/((sqrt1+x^3)) dx

$\displaystyle\int_0^2\;\frac{1}{\sqrt1+x^3} dx$

This is what you asked for, I suspect it is not what you really want.  Be careful with your brackets.

Is this what you want?

Integral from 0 to 2 of    [1/(sqrt[1+x^3]) dx

$\displaystyle\int_0^2\;\frac{1}{\sqrt{1+x^3}} \;dx$

It's a mess...

$\displaystyle \int \dfrac{1}{\sqrt{1+x^3}}~dx = \\ \dfrac{2 \sqrt[6]{-1} \sqrt{-\sqrt[6]{-1} \left(x+(-1)^{2/3}\right)} \sqrt{(-1)^{2/3} x^2+\sqrt[3]{-1} x+1} F\left(\sin ^{-1}\left(\frac{\sqrt{-(-1)^{5/6} (x+1)}}{\sqrt[4]{3}}\right)|\sqrt[3]{-1}\right)}{\sqrt[4]{3} \sqrt{x^3+1}}$

$\displaystyle \int_0^2 \dfrac{1}{\sqrt{1+x^3}}~dx =2 \cdot \, _2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-8\right) \approx 1.40218$

yes!!!