A16 liter radiator is filled with a solution of 40 % antifreeze. how much should you drain from the radiation and replace with pure antifreeze to obtain a 60 % antifreeze solution?

liters is the amount to be drained out and replaced

Solution:

40 % antifreeze solution in 16 liter radiator

Let "x" be the amount drained from radiation and replaced with pure antifreeze

To obtain a 60 % antifreeze solution

The original solution is 16 liter, 40% of which is antifreeze

You want the solution to be 60% antifreeze:

60 % x 16 =

You will remove x liters of the 40% solution and replace it with x liters pure (100%) antifreeze.

Let us solve expression for "x"

Thus liters is the amount to be drained out and replaced

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation: