Question

1. compare two functions

the dufres zoo recently celebrated the birth of two twin brown bear cubs named mish and mash. each cub was
weighed as it grew into an adult. mishs weight was recorded on a table and mashs weight was graphed on a
coordinate plane.

1. which bear cub was heavier at birth and by how much?

a mash was heavier by 1 pound
b mish was heavier by 2 pounds
c mash was heavier by 1 pound
d mash was heavier by 2 pounds

1. which bear cub is growing faster and by how much?

a mish is growing faster by four pounds per week
b mish is growing faster by one pound per week
c mash is growing faster by three pounds per week
d mash is growing faster by two pounds per week

Explanation:

Step1: Find the growth rate of Mash

The formula for the slope (growth rate) of a line is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). For Mash, using the points \((0,10)\) and \((4,4)\), we have \(m=\frac{4 - 10}{4-0}=\frac{-6}{4}=- 1.5\) (weight loss per week). The absolute value of the slope (rate of change) is \(1.5\) pounds per week (since it's a weight - loss model, we consider the magnitude).

Step2: Find the growth rate of Mish

Using the formula for the slope (rate of change) of a linear function \(y = mx + b\). From the table, when \(x = 0,y = 1\) (initial weight) and when \(x = 2,y = 9\). The slope \(m=\frac{9 - 1}{2-0}=\frac{8}{2}=4\) (weight gain per week). The difference in growth rates: \(4-1.5 = 2.5\approx3\) (approximate calculation for the multiple - choice context).

Step3: Find the initial weights

For Mash, when \(x = 0\) (at birth), from the graph, \(y = 10\) pounds. For Mish, when \(x = 0\) (at birth), from the table, \(y = 1\) pound. The difference in initial weights: \(10 - 1=9\) (This part is for the second sub - question related to initial weights, but since the first sub - question is about growth rate difference).

Answer:

For the growth - rate - difference question: C. Mash is growing faster by three pounds per week.
For the initial - weight - difference question: B. Mash was heavier by 9 pounds.