22) omar is running a chemical reaction that can be modeled by a quadratic function. when he begins the…

22) omar is running a chemical reaction that can be modeled by a quadratic function. when he begins the reaction there are 20.3 grams of sodium chloride present. at 2 minutes there are 47.6 grams remaining. at 5 minutes there are 59.8 grams, at 8 minutes there are 36.1 grams, and at 9 minutes there are 19.7 grams remaining.\n\n|time (min)|0|2|5|8|9|\n|--|--|--|--|--|--|\n|sodium chloride present (g)|20.3|47.6|59.8|36.1|19.7|\n\nuse a graphing calculator to find a quadratic model that represents the table of values.\n\nquadratic model: y = -1.98x²+17.83x + 20.15\nuse your model to determine the time all the sodium chloride will be used up in the reaction.

22) omar is running a chemical reaction that can be modeled by a quadratic function. when he begins the reaction there are 20.3 grams of sodium chloride present. at 2 minutes there are 47.6 grams remaining. at 5 minutes there are 59.8 grams, at 8 minutes there are 36.1 grams, and at 9 minutes there are 19.7 grams remaining.\n\n|time (min)|0|2|5|8|9|\n|--|--|--|--|--|--|\n|sodium chloride present (g)|20.3|47.6|59.8|36.1|19.7|\n\nuse a graphing calculator to find a quadratic model that represents the table of values.\n\nquadratic model: y = -1.98x²+17.83x + 20.15\nuse your model to determine the time all the sodium chloride will be used up in the reaction.

Answer

Explanation:

Step1: Set up the quadratic - function equation

The general form of a quadratic function is $y = ax^{2}+bx + c$, where $y$ is the amount of sodium - chloride present and $x$ is the time. We are given the quadratic model $y=-1.98x^{2}+17.83x + 20.15$. When all the sodium chloride is used up, $y = 0$. So we need to solve the quadratic equation $-1.98x^{2}+17.83x + 20.15=0$.

Step2: Use the quadratic formula

The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=-1.98$, $b = 17.83$, and $c = 20.15$. First, calculate the discriminant $\Delta=b^{2}-4ac=(17.83)^{2}-4\times(-1.98)\times20.15$. [ \begin{align*} \Delta&=317.8089+159.576\ &=477.3849 \end{align*} ] Then, $x=\frac{-17.83\pm\sqrt{477.3849}}{2\times(-1.98)}=\frac{-17.83\pm21.85}{-3.96}$. We have two solutions for $x$: $x_1=\frac{-17.83 + 21.85}{-3.96}=\frac{4.02}{-3.96}\approx - 1.02$ (rejected since time cannot be negative) $x_2=\frac{-17.83-21.85}{-3.96}=\frac{-39.68}{-3.96}\approx10.02$ minutes.

Answer:

Approximately 10.02 minutes