Question

(a) A dog owner records the weight of her dog. She finds that from the age of 20 weeks to the age of 48 weeks, the dog’s weight can be modelled by the equation w = 0.92t−0.15 (20 ≤ t ≤ 48), where w is the weight of the dog in kilograms, and t is the age in weeks.

(i) Find the weight of the dog at age 26 weeks according to this model.
(ii) Explain the inequality (20 ≤ t ≤ 48) that follows the equation.
(iii) Using algebra, calculate the age at which the dog’s weight is 40kg
(iv) Write down the gradient of the straight line represented by the equation w = 0.92t−0.15. What does this measure in the practical situation being modelled? (v) Either using Graphplotter or by hand, sketch the graph of w = 0.92t−0.15, putting w on the vertical axis and covering the time interval 20 ≤ t ≤ 48.
(vi) Explain why the model cannot be extended to model accurately the dog’s weight at birth

Answer 1

(i)23.77kg

(ii)(ii) the inequality 20≤t≤48 means the weight of the dogs determined is for dogs that have the age of 20 weeks or less to the age of 48 weeks not more.

(iii)44 weeks

(iv)0.92

(v)Attached

(vi)The value of weight will be negative,which is unrealistic.

Explanation:

(i)Given that the weight is modeled by the equation;

w = 0.92t−0.15 (20 ≤ t ≤ 48)

then substitute the value of t in the equation given that t=26 weeks

w=0.92*26-0.15=23.77kg

(ii) the inequality 20≤t≤48 means the weight of the dogs determined is for dogs that have the age of 20 weeks or less to the age of 48 weeks not more.

(iii)Apply the equation for weight

w=0.92t-0.15

40=0.92t-0.15

40+0.15=0.92t

40.15=0.92t

40.15/0.92=t

43.6⇒⇒44 weeks

(iv) Here , write the equation in slope intercept form of y=mx+c where m is the slope and c is the y-intercept

w=0.92t-0.15

y=mx+c

m=0.92

(v)sketch attached

(vi)

At birth, t=0 hence to find weight

w=0.92t-0.15

w=0.92*0-0.15

w=0-0.15

w=-0.15

The model cannot be extended to model accurately the dog's weight at birth because you will get a negative value for weight which is not realistic.