Amixture contains forty ounces of glycol and water, and it is ten percent glycol. the mixture is to be strengthened to twenty-five percent by adding glycol. how much glycol is in the original mixture?

8 ounces.

Step-by-step explanation:

Let amount of glycol in original mixture = x ounce

Given mixture is of 40 ounces and it contains 10% glycol.

So amount of glycol in that mixture = 10% of 40 = 0.10(40)

Now we are adding x ounces of glycol which is 100% pure.

So amount of glycol added = 100% of x = 1.00(x)

Now the volume of new solution = 40+x ounces

So amount of glycol in new mixture = 25% of (40+x) = 0.25(40+x)

Since there is no loss in amount of glycol then total input = total output

0.10(40)+1.00(x)=0.25(40+x)

Now solve the above equation

4+x=10+0.25x

x-0.25x=10-4

0.75x=6

x=8

Hence final answer is 8 ounces.

4 oz

Step-by-step explanation

The reason why its 4 oz is becuase the question is looking for the amount of glycol is in the original mixture, not a strenghtend mixture.

You would figure this out by doing 10 percent of 40 percent to get 4. The reason you do this is because in the text, it states that there is ten percent glycol in the forty ounces or 10 percent of 40.

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3