CentsLord

though, piecewise defined functions are very good for rating so grind on

your solution for $(x, y, z)$ is $(-3, 1, 0).$ if you would like steps, please tell. otherwise, hope this helped. 

grinding some piecewise-defined functions today, aren't we?

we have $2x + 7 = -5$, so $x = -6.$

we also have $-x^2 - x + 1 = -5$, and solving for $x$ gives us solutions of $-3$ and $2.$ As $-3$ is not greater or equal than $-2$, we only consider the $2.$

So, the values of $x$ are $-6$ and $2.$ the sum of these is $\boxed{-4}.$ nice

so, we have \(7(4f + 3) - 8f + 6 = 8f + (7 * 5) + 6f \)

distributing and simplifying parentheses gives us \(28f + 21 - 8f + 6 = 8f + 35 + 6f\). combining like terms, we have \(20f + 27 = 14f + 35\).

this can then be rearranged into \(6f = 8\), so the solution for f is \(f = \frac{4}{3}\). and that's pretty good

I'll assume you mean \((8x^6y^5)(5x^5y^{12})\). this can be expressed as \(8 \cdot 5 \cdot x^6 \cdot x^5 \cdot y^4 \cdot y^{12}\). 

\(8\) times \(5\) is \(40\)

using a law of exponents which states that \(a^n \cdot a^m = a^{n + m}\), we can write \(x^6 \cdot x^5 = x^{11}\) and \(y^4 \cdot y^{12} = y^{16}\).

so, we have \(40 \cdot x^{11} \cdot y^{16} = 40x^{11}y^{16}\) 

and there you have it

\(a \cdot 5 = 5a\)

noooooo

i actually thought tho

well done man

basically, sir cphill said that f(n) for any positive value of n would just be 1, as the difference between the sum of the elements of the nth row, and the sum of all elements from the previous rows, is just one. you can read his post at the link the guest above me gave. :)

wow man. respect+. upvoted

let's start this party

here's my answer