How many ml of a 20% acid mixture and a 80% acid mixture should be mixed to get 120ml of a 35% acid mixture?

Volume of Mixture A =90 ml

Volume of Mixture B =30 ml

Step-by-step explanation:

Let say, Mixture A + Mixture B = Mixture C

Volume of Mixture A is x

Volume of Mixture B is y

So, Volume of Mixture C is x+y = 120 ml

Now, Acid contain in Mixture A is 20% =0.2x

Acid contain in Mixture B is 80% =0.8y

Also, Acid contain in Mixture C is 35% =(0.35)(x+y) = 0.35×120=42

Now, we know that,

Acid contain of Mixture A + Acid contain of Mixture B=Acid contain of Mixture C

∴ 0.2x+0.8y=42

∴ 2x+8y=420

We get two linear equations

2x+8y=420 and x+y = 120

Solving above equation...

∴ x=120-y

Replacing x value in 2x+8y=420

∴ 2(120-y)+8y=420

∴ 240-2y+8y=420

∴ 6y=180

∴ y=30

Replacing y value in any equation

∴ x=120-y=120-30=90

∴ x=90

Thus,

Volume of Mixture A is x=90 ml

Volume of Mixture B is y=30 ml

Option A.

Step-by-step explanation:

We need to use this identity:

We know that the ramp forms an 8 degrees angle with the ground and the distance between the entry point of the ramp and the building is 7 feet (Observe the figure attached). Then:

To find how many feet from the ground  is the end of the ramp to the nearest hundredth (represented with "x"), we need to substitute values into   and solve for "x".

Then: