#2**+1 **

Hey guest,

Let's try and solve for the question mark. Note that I have replaced the question mark with a more standard symbol, *x*. Below is a representation of the equation in LaTeX form:

\(1\frac{3}{5}(1999*\frac{1}{24}+x)-3\frac{5}{11}*3\frac{2}{3}=127\)

\(1\frac{3}{5}(1999*\frac{1}{24}+x)-3\frac{5}{11}*3\frac{2}{3}=127\) | Mixed numbers may be nice to look at, but they are fairly clumsy when algebraic computation is involved. I will convert all mixed numbers into improper fractions. |

\(\frac{8}{5}(\frac{1999}{24}+x)-\frac{38}{11}*\frac{11}{3}=127\) | Let's do some simplification and distribution. |

\(\frac{8}{5}*\frac{1999}{24}=\frac{1}{5}*\frac{1999}{3}=\textcolor{red}{\frac{1999}{15}}\\ \frac{38}{11}*\frac{11}{3}=38*\frac{1}{3}=\textcolor{blue}{\frac{38}{3}}\) | |

\(\textcolor{red}{\frac{1999}{15}}+\frac{8}{5}x-\textcolor{blue}{\frac{38}{3}}=127\) | In general, I recommend for situations like this to multiply both sides by the LCD. This will eliminate all instances of fractions. The LCD, in this case, is 15. |

\(1999+24x-190=1905\) | Let's do some more simplification on the left-hand side of the equation. |

\(24x+1809=1905\) | Subtract 1809 on both sides. |

\(24x=96\) | Divide by 24 on both sides to isolate x completely. |

\(x=4\) | |

TheXSquaredFactor
Aug 17, 2018

#8**0 **

I think that the easiest way to prove that \((3n)^2+(4n)^2=(5n)^2\) has infinitely many solutions is to solve for n:

\((3n)^2+(4n)^2=(5n)^2\) | Just solve for n. |

\(9n^2+16n^2=25n^2\) | Combine like terms on the left hand side of the equation. |

\(25n^2=25n^2\) | Notice how both sides of the equation are equal. |

\(0=0\) | \(0=0\) is a true statement, so every value for n results in a true statement. Since infinitely many solutions exist, more than 2005 solutions exist, and we have solved the problem. |

TheXSquaredFactor
Jul 26, 2018

#1**+1 **

I already answered this question moments ago.

Visit https://web2.0calc.com/questions/math_67590 to view the response.

TheXSquaredFactor
Jul 24, 2018

#1**+1 **

If \(f(x)=\sqrt{x-3}\), then, in order to evaluate \(f(12)\), replace all instances of x with 12.

\(f(x)=\sqrt{x-3}\) | This is the original definition of the function. |

\(f(12)=\sqrt{12-3}\) | Simplify. |

\(f(12)=\sqrt{9}\) | |

\(f(12)=3\) | |

TheXSquaredFactor
Jul 24, 2018

#5**+1 **

Hello Guest,

You have been proactive in attempting to simplify the numerator, and you are continuing to make significant headway. However, there is a slight algebraic error lying somewhere in your simplification, but I cannot pinpoint where because of the lack of work provided. Computations like these require extreme perserverance and accuracy. Any small imperfection will result in an answer that is horribly askew.

\(3(2x^2+9x-5)+(4x^2-1)(x-5)\) | For now, I am worrying about the simplification of the numerator. Let's distribute the 3 into the trinomial. |

\(6x^2+27x-15+(4x^2-1)(x-5)\) | It is time to expand the product of two binomials. |

\(\textcolor{red}{6x^2}\textcolor{blue}{+27x}\textcolor{green}{-15}+4x^3\textcolor{red}{-20x^2}\textcolor{blue}{-x}\textcolor{green}{+5}\) | Combine all like terms. I have utilized color coding to show which terms combine. |

\(\)\(4x^3-14x^2+26x-10\) | |

After this algebra, we are left with \(\frac{4x^3-14x^2+26x-10}{(x+5)(x-5)(2x-1)}\). Notice the discrepancies and try and locate the mistake.

From here, you must factor the numerator and determine whether or not any of the factors of \(4x^3-14x^2+26x-10\) match those of the denominator. Factoring by grouping does not appear to be a valid option here, so you will most likely have to resort to methods such as the rational root theorem and the Descartes' Rules of Signs. To start, though, notice how each term is even, so one can factor out a 2 from every term.

Good luck!

TheXSquaredFactor
Jul 23, 2018