Achemist plans to mix 10 liters of a 40% acid solution with some 70% to obtain a 50% solution. how much of the 70% solution should be used?

A chemist plans to mix 10 liters of a 40% acid solution with some 70% to obtain a 50% solution

40% = 0.4

70% = 0.7

50% = 0.5

A chemist plans to mix 10 liters of a 40%. So 10 times of 0.4 = 4

Let x be the some amount  of solution of 70%

70% of x = 0.7x

chemist obtain a 50% of solution

Amount of mixture is 10 + x

So 10+x of 50% = (10+x)*0.5

Now we frame an equation

10 * 0.4  + x * 0.7 = (10+x) * 0.5

4 + 0.7x = 5 + 0.5x

Subtract 4 and 0.5x on both sides

0.2 x= 1

Divide by 0.2

So x= 5

5 liters of 70% of solution is used.

the two may have that, even a breakdown is probably offered

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?