A scientist needs 10 liters of a 20% acid solution for an experiment, but she has only a 5% solution and a 40% solution. To the nearest tenth of a liter, about how many liters of the 5% and the 40% solutions should she mix to get the solution she needs? Choose the equation to match the situation

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In this question, you want to make 10 liters 20% concentration solution by mixing 5% and 40% concentration. There should be multiple answers to this question. Assuming that the total volume of 5% and 40% solution equal to 10L(no solution wasted), then you can determine the volume of solution needed. The formula should be:

if x would be the of volume 40% solution, then the volume of 5% would be:

10 L= 40% solution volume+ 5% solution volume

10 L= x + 5% solution volume

5% solution volume= 10l-x

Then the volume calculation would be:

volume20%*concentration20%= volume40%*concentration40% + volume5%*concentration5%

10 liter * 20%= x liter*40% + (10- x) liter*5%

10l * 20%= x liter * 35% + 10 liter*5%

x liter *35 = 10l * 20 - 10l *5= 10l*15

x liter = 10l *15/35= 4.28 L= 4.3L

The volume of 40% solution= 4.3L

The volume of 5% solution= 10L- 4.3L= 5.7L

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!