Dr. Khan is combining a 20% acetic acid solution with a 60% acetic acid solution. She wants to obtain 10 liters of a 40% acetic acid solution, how many liters of the 20% acetic acid solution does Dr. Khan need to create the 40% acetic acid solution?

Solution:

Let the amount of 20% acetic acid, he added to get 10 liters of 40% solution = 'a' liters

Volume of 60% solution used will be = (10 - a) liters

Equation will be,

Amount of solution (1) × 20% + Amount of solution (2) × 60% = Amount of final solution × 40%

a*20/100+(10-a)*60/100=10*40/100

20a + 60(10 - a) = 400

20a + 600 - 60a = 400  

600 - 40a = 400

40a = 600 - 400            

a=200/40

a = 5 liters

Thus, Dr Khan  need to create 5 liters of acetic acid to get 40% acetic acid solution.

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Dr Khan will add 5 liters of 20% acetic acid solution.

Step-by-step explanation:

Dr Khan is combining two solutions of acetic acid with the given concentrations of 20% and 60% respectively.

Let the amount of 20% acetic acid, he added to get 10 liters of 40% solution = 'a' liters

Volume of 60% solution used will be = (10 - a) liters

Now the equation will be,

Amount of solution (1) × 20% + Amount of solution (2) × 60% = Amount of final solution × 40%

20a + 60(10 - a) = 400

20a + 600 - 60a = 400

600 - 40a = 400

40a = 600 - 400

a =

a = 5 liters

Therefore, Dr Khan will add 5 liters of acetic acid to get 40% acetic acid solution.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}