#1**+1 **

a)

If we suppose that *c* varies inversely with *d*, then this is what that statement is telling you:

- If d increases, then c decreases.
- If d decreases, then c increases.

The most basic example of this occurence is \(y=\frac{1}{x}\). As you think of larger values of x, the input, then the output, y, will decrease.

There is only one difference from the previous function to this particular problem: We do not know the rate in which the inverse relation occurs. For example, if the numerator of the previous function was a 2, then the rate of the inverse function would double. Let's name the rate of the inverse relation "k." We, therefore, make the following equation.

\(c=\frac{k}{d}\)

b)

We can use the previous equation to figure out the rest. It involves basic substitution.

\(c=\frac{k}{d}\) | Substitute in the known values for c and d. |

\(17=\frac{k}{2}\) | Solve for k by multiplying by 2 on both sides. |

\(k=34\) | |

Now, let's find d when c=68:

\(c=\frac{k}{d}\) | We now know the values of c and k are this time. |

\(68=\frac{34}{d}\) | Now it is a matter of simplifying. |

\(68d=34\) | |

\(d=\frac{34}{68}=\frac{1}{2}=0.5\) | |

TheXSquaredFactor
Mar 5, 2018

#1**+1 **

Best Answer

a)

If we suppose that *c* varies inversely with *d*, then this is what that statement is telling you:

- If d increases, then c decreases.
- If d decreases, then c increases.

The most basic example of this occurence is \(y=\frac{1}{x}\). As you think of larger values of x, the input, then the output, y, will decrease.

There is only one difference from the previous function to this particular problem: We do not know the rate in which the inverse relation occurs. For example, if the numerator of the previous function was a 2, then the rate of the inverse function would double. Let's name the rate of the inverse relation "k." We, therefore, make the following equation.

\(c=\frac{k}{d}\)

b)

We can use the previous equation to figure out the rest. It involves basic substitution.

\(c=\frac{k}{d}\) | Substitute in the known values for c and d. |

\(17=\frac{k}{2}\) | Solve for k by multiplying by 2 on both sides. |

\(k=34\) | |

Now, let's find d when c=68:

\(c=\frac{k}{d}\) | We now know the values of c and k are this time. |

\(68=\frac{34}{d}\) | Now it is a matter of simplifying. |

\(68d=34\) | |

\(d=\frac{34}{68}=\frac{1}{2}=0.5\) | |

TheXSquaredFactor
Mar 5, 2018