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NotSoSmart  Mar 5, 2018

Best Answer 

 #1
avatar+1958 
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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
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1+0 Answers

 #1
avatar+1958 
+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

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