From a full 50-liter container of a 40% concentration of acid, x liters are removed and replaced with 100% acid. (A) Write the a

Question

1 month ago

6

From a full 50-liter container of a 40% concentration of acid, x liters are removed and replaced with 100% acid. (A) Write the a

mount of acid in the final mixture as a function of x (B) Determine the domain and range of the function (C) Determine if the final mixture is 50% acid PLEASE EXPLAIN...I DON'T UNDERSTAND THIS AT ALL

Mathematics

1 answer:

tester [12.3K] · Answer 1 · 2026-02-24T19:54:25+03:00

Answer:

A) 20+0.6 $x$

B) Domain is [0, 50] (inclusive)

C) 8.33 litres

Step-by-step explanation:

Assuming there is a 40% acid concentration in a 50-litre container.

The quantity of acid in this container amounts to 40% of 50 litres.

Quantity of acid = $\frac{40}{100} \times 50 = 20\ litre$

$x$ litres are extracted.

The acid volume removed equals 40% of $x$ litre.

Remaining acid in the container thus becomes = (20 - 40% of $x$ ) litre

This is then substituted with pure acid.

Final acid content in the container is therefore = (20 - 40% of $x$ + 100% of $x$ ) litre

Quantity of acid in the final mixture:

$20 - \dfrac{40}{100} \times x + \dfrac{100}{100} \times x\\\Rightarrow 20 +\dfrac{100-40}{100}x\\\Rightarrow 20 +\dfrac{60}{100}x$

Answer A) Final mixture acid content = 20+0.6 $x$

Answer B) $x$ cannot exceed 50 litres (the container's initial volume) and must be at least 0 litres.

Thus, the domain is [0, 50] (inclusive)

Answer C)

For the final mixture to contain 50% acid.

The acid volume = 50% of 50 litres = 25 litres

By applying the equation:

$20+0.6x =25\\\Rightarrow 0.6x =5\\\Rightarrow \bold{x =8.33\ litres}$