Assume that the amount of solid food that an animal consumes daily depends on its weight. suppose that an animal weighing 10.0kgs requires 1.50kgs of food each day, and that an animal weighing 40.0kgs requires 4.50kgs of food each day. assume that the amount a of food required daily and the weight w of the animal are related linearly. a. find a as a function of w. b. how much food will be required each day by an animal that weighs 53.0kgs?

a. A= W/10 +0.5

b. 5.8 kgs will be required each day by an animal that weighs 53.0 kgs.

Step-by-step explanation:

a. Given that,

An animal weighing 10.0 kgs requires 1.50 kgs of food each day and animal weighing 40.0 kgs requires 4.5 kgs of food.

So, we have two ordered pairs (10.0, 1.5) and (40.0, 4.5)

As food required daily of A amount and the weight W of the animal are linearly related. So the equation of line would be as follow:

Step 1: To find the slope m of the line.

m= (4.5-1.5)/(40.0-10.0)=3/30

m= 1/10

Step 2: To find the equation of the line.

Using point-slop form, we get:

A-1.5= 1/10(W-10.0)

A= W/10 – 1 + 1.5

A= W/10 +0.5

b. Putting W= 53.0 kgs, we get:

A= 53.0/10 + 0.5

A= 5.3 +0.5

A= 5.8

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the red lines denote the population of the bacteria after 35 hours. that would be 2,409.24. the y-intercept is the blue line which is 800 because that is the starting point. the equation used to make the graph is 800(1.032)^h, where h is the number of hours.

step-by-step explanation: