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# Help 19 ​

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Help 19

Mar 5, 2018

#1
+2298
+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. $$68d=34$$ $$d=\frac{34}{68}=\frac{1}{2}=0.5$$
Mar 5, 2018

#1
+2298
+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