Chapter 8.5, Problem 3CYU

(a)

To determine

To write: a direct variation equation

Answer to Problem 3CYU

The equation is: $y=0.25x$

Explanation of Solution

Given:

if the weight of 8-ounce is attached then spring starches by 2 inches.

Calculation:

Step 1: The weight is denoted as x and amount of stretch is y

Now, if there is direct variation then there is some constant variable k such that it satisfies the equation: $y=kx$

So, put the value of $x=8$ and $y=2$ in equation to find the value of k

  $2=8k$

Divide the equation by 8.

  $\begin{array}{l}\frac{2}{8}=\frac{8k}{8}\\ 0.25=k\end{array}$

Step 2: use k to write an equation.

  $y=\mathrm{0.25.}x$

Conclusion:

Therefore, the equation is: $y=0.25x$ .

(b)

To determine

To estimate:the stretch of the spring if it has 20-ounce weight attached.

Answer to Problem 3CYU

The stretch value is $y=5$ inches.

Explanation of Solution

Calculation:

Now, on order to calculate the length of stretched spring when weight is 20-ounce put the value of $x=20$ in equation.

  $\begin{array}{l}y=0.25\left(20\right)\hfill \\ =5inches\hfill \end{array}$

Conclusion:

Therefore, the stretch value is $y=5$ inches.