tertre

We just find for 10P4. Thus, the answer is \(\frac{10!}{6!}=10*9*8*7=\boxed{5040}.\)

Help on solving, right?

The negative sign changes 6 to -6 and -x to x. Thus, 9x-6+x=10x-6. 

We divide: \(\frac{2400}{24}=100.\)

Close...

We can obtain the graph of \(y=g(x)\) by taking the graph of \(y=f(x)\) and stretching it horizontally by a factor of 2, then shifting it down by 4 units. Therefore, \(g(x) = \boxed{f \left( \frac{x}{2} \right) - 4}.\)

Isn't this correct?

It should be \(81+256=337.\) First, change the fourth root into \(\frac{1}{4}.\)

Now, without words:

\(x^{\frac{1}{4}}=\frac{12}{7-x^{\frac{1}{4}}}\)

\(x^{\frac{1}{4}}\left(7-x^{\frac{1}{4}}\right)=\frac{12}{7-x^{\frac{1}{4}}}\left(7-x^{\frac{1}{4}}\right)\)

\(x^{\frac{1}{4}}\left(7-x^{\frac{1}{4}}\right)=12\)

Expand, \(7x^{\frac{1}{4}}-\sqrt{x}=12\)

\(7x^{\frac{1}{4}}-\left(x^{\frac{1}{4}}\right)^2=12\)

Plug variables instead of \(x^\frac{1}{4}\)

to get \(x=81,\:x=256\)

Thus, te answer is \(81+256=\boxed{337}.\)

I'll post a solution:

We just apply the sum of cubes formula, so, we get \(\left(\sqrt{2}\right)^3+3\left(\sqrt{2}\right)^2\sqrt{3}+3\sqrt{2}\left(\sqrt{3}\right)^2+\left(\sqrt{3}\right)^3\) . 

We just simplify this and add like terms, to reach, \(11\sqrt{2}+9\sqrt{3}.\)

From here, we know that \(a=11, b=9, c=0\)  , thus  \(11+9+0=\boxed{20}.\)

Welcome! I just learned this a few days ago under Pascal's triangle.

I can clearly read the second one. Thanks!

Hint: Apply sum of cubes formula.

There is actually a trick for this! \( {n \choose k} \)+\( {n \choose k+1} \)=\( {n+1 \choose k+1} \), so in this case, we have \( {8 \choose 3} + {8 \choose 4} = {9 \choose 4} \) or \( {9 \choose 5} \). Thus, \(\boxed{n=5}.\)

The first choice and fourth choice. 

We could find the least common multiplier, turn things around and get \(x\left(x+3\right)+2=5\left(x+2\right)\) .

Then, solving, we get \(x=4,\:x=-2\) and \(\boxed{x=4}\)  as    \(x=2\) , creates a zero in the denominator.