In Purdue's Chemistry department, the chemists have found that in a water based solution containing 4 grams of certain undissolved chemicals, the rate of change of the amount of chemicals dissolved in the solution is proportional to the amount of the undissolved chemicals. Let Q(t) (in grams) be the amount of dissolved chemicals at time t and let k be the positive proportionality constant. The differential equation describing the given situation is:

Step-by-step explanation:

Given that in Purdue's Chemistry department, the chemists have found that in a water based solution containing 4 grams of certain undissolved chemicals, the rate of change of the amount of chemicals dissolved in the solution is proportional to the amount of the undissolved chemicals.

Let Q(t) (in grams) be the amount of dissolved chemicals at time t and let k be the positive proportionality constant.

We have from the given information Q(0) = amount of dissolved chemicals at time 0 = 4 gms

Also rate of change of Q is

is the differential equation with initial condition as Q(0) = 4 gm

By separation of variables we can solve this equatin.

do i do it on a graph?

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!