Question

A spring stretches to 22 cm with a 70 g weight attached to the end. With a 105 g weight attached, it stretches to 27 cm. Which equation models the distance y the spring stretches with weight x attached to it?

Answer 1

Simple form of the equation y= ax+b
y= (1/7)x + 12
come from 2 equation :
22= 70 a + b
27= 105 a + b

Answer 2

Answer:

$y=\frac{1}{7}x+12$

Step-by-step explanation:

Let y represent the distance the spring stretches and x be the weight attached to the spring.

We have been given that a spring stretches to 22 cm with a 70 g weight attached to the end. With a 105 g weight attached, it stretches to 27 cm.

First of all, we will find slope of line using the points (70,22) and (105,27) as shown below:

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{27-22}{105-70}$

$m=\frac{5}{35}$    

$m=\frac{1}{7}$    

Now, we will substitute $m=\frac{1}{7}$ and coordinates of point (70,22) in slope-intercept form of equation to find the y-intercept for our equation.

$y=mx+b$

$22=\frac{1}{7}*70+b$

$22=1*10+b$

$22=10+b$

$22-10=10-10+b$

$12=b$

Upon substituting $b=12$ and $m=\frac{1}{7}$ in slope-intercept form of equation we will get our required equation as:

$y=\frac{1}{7}x+12$

Therefore, the equation $y=\frac{1}{7}x+12$ models the distance y the spring stretches with weight x attached to the spring.