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Solution A is an 80% acid solution. Solution B is a 30% acid solution.

(a) How many mL of Solution A must be added to 500 mL of Solution B in order to produce a 70% acid solution?

(b) How many mL of Solution A and how many mL of Solution B must be combined to form a 100 mL solution that is 50% acid?

(c) Is there a combination of Solution A and Solution B that is 90% acid?

Part a) 2,000 mL of solution A

Part b) 40 mL of solution A and 60 mL of solution B

Part c) Is not possible

Step-by-step explanation:

Part a) How many mL of Solution A must be added to 500 mL of Solution B in order to produce a 70% acid solution?

Remember that

Let

x ---->mL of solution A n 80% acid solution

y ---->mL of solution B n 30% acid solution

we know that

Remember that

----> equation B

substitute equation B in equation A

solve for x

Part b) How many mL of Solution A and how many mL of Solution B must be combined to form a 100 mL solution that is 50% acid?

Remember that

Let

x ---->mL of solution A n 80% acid solution

y ---->mL of solution B n 30% acid solution

we know that

----> equation A

-----> equation B

Solve the system by graphing

Remember that the solution is the intersection point both graphs

using a graphing tool

The solution is (40,60)

see the attached figure

therefore

40 mL of solution A and 60 mL of solution B

Part c) Is there a combination of Solution A and Solution B that is 90% acid?

Is not possible , because 90% is greater than 80% of solution A and greater than 30% of solution B

The percentage of the final concentration must be less than 80% and more than 30%

the correct answer is 45x

9×5=45 then add the x

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.