Question

question #7
six ounces of mixed nuts containing 18% peanuts were mixed with 12 ounces of another kind of mixed nuts which contain 72% peanuts. what percent of the new mixtures will be peanuts? round your answer to the nearest hundredth.
60%
58%
35%
54%
question #8
solve the equation. remember to check for extraneous solutions.
\\(\frac{1}{x^{2}-9}+\frac{x^{2}+x - 2}{x^{2}-9}=\frac{1}{x - 3}\\)
\\(x = 2\\) and \\(x=-2\\)
\\(x = 2\\)
\\(x = 0\\) and \\(x=-1\\)
\\(x=-2\\)

Explanation:

Question #7

Step1: Calculate peanut - amount in first mix

The first mix is 6 ounces and contains 18% peanuts. So the amount of peanuts in the first mix is $6\times0.18 = 1.08$ ounces.

Step2: Calculate peanut - amount in second mix

The second mix is 12 ounces and contains 72% peanuts. So the amount of peanuts in the second mix is $12\times0.72=8.64$ ounces.

Step3: Calculate total peanut - amount

The total amount of peanuts is $1.08 + 8.64=9.72$ ounces.

Step4: Calculate total mixture - amount

The total amount of the new mixture is $6 + 12 = 18$ ounces.

Step5: Calculate percentage of peanuts in new mixture

The percentage of peanuts in the new mixture is $\frac{9.72}{18}\times100\%=54\%$.

Step1: Combine left - hand side fractions

Since the denominators of $\frac{1}{x^{2}-9}$ and $\frac{x^{2}+x - 2}{x^{2}-9}$ are the same, we can combine them: $\frac{1+(x^{2}+x - 2)}{x^{2}-9}=\frac{x^{2}+x - 1}{x^{2}-9}$. And $x^{2}-9=(x + 3)(x - 3)$. So we have $\frac{x^{2}+x - 1}{(x + 3)(x - 3)}=\frac{1}{x - 3}$.

Step2: Cross - multiply

Cross - multiplying gives $(x^{2}+x - 1)(x - 3)=(x + 3)(x - 3)$. Expanding the left - hand side: $x^{3}-3x^{2}+x^{2}-3x - x + 3=x^{3}-2x^{2}-4x + 3$. Expanding the right - hand side gives $x^{2}-9$. So $x^{3}-2x^{2}-4x + 3=x^{2}-9$.

Step3: Rearrange to form a quadratic equation

Bring all terms to one side: $x^{3}-3x^{2}-4x+12 = 0$. Since we are dealing with rational functions and the original equation is a rational equation, we can also work with the non - cubic form after cross - multiplying and simplifying. Starting from $\frac{x^{2}+x - 1}{(x + 3)(x - 3)}=\frac{1}{x - 3}$ (assuming $x
eq3$ and $x
eq - 3$), we can multiply both sides by $(x + 3)(x - 3)$ to get $x^{2}+x - 1=x + 3$.
Subtracting $x$ from both sides gives $x^{2}-4 = 0$.

Step4: Solve the quadratic equation

Factor $x^{2}-4=(x + 2)(x - 2)=0$. So $x=\pm2$.

Step5: Check for extraneous solutions

When $x = 2$, the original equation $\frac{1}{2^{2}-9}+\frac{2^{2}+2 - 2}{2^{2}-9}=\frac{1}{2 - 3}$. $\frac{1}{-5}+\frac{4}{-5}=\frac{1}{-1}$, $-\frac{5}{5}=-1$, the solution is valid.
When $x=-2$, the original equation $\frac{1}{(-2)^{2}-9}+\frac{(-2)^{2}-2 - 2}{(-2)^{2}-9}=\frac{1}{-2 - 3}$. $\frac{1}{-5}+\frac{4 - 2 - 2}{-5}=\frac{1}{-5}$, $-\frac{1}{5}+0=-\frac{1}{5}$, the solution is valid.

Answer:

54%

Question #8