How much pure alcohol should be added to a 360 ml of a 30% alcohol solution to raise the concentration to 40%? lol

  60 mL

Step-by-step explanation:

Let x represent the amount of pure alcohol to be added (in mL). The total amount of alcohol in the mix is ...

  30% × 360 mL + 100% × x mL = 40% × (x +360) mL

  108 + x = 0.4x + 144 . . . . simplify

  0.6x = 36 . . . . . . . . . . . . . subtract 0.4x +108

  36/0.6 = x = 60 . . . . . . . . divide by the coefficient of x

60 mL of pure alcohol must be added to 360 mL of 30% to raise the concentration to 40%.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

a_n = a_1 + (n - 1)d

the first term is 6.

the common difference d = ?

when you find d, replace n with 7 and d with the common difference to find your answer.