+2440 We are going to be using this theorem: the Triangle Inequality Theorem. I think a picture can tell a thousand words, and I think this picture does just that. It may help you understand more than the words I can produce on the page!
If the sum of the shorter side must be greater than the final side, then we must consider three cases:
Let's deal with the simplest case. Let's assume that 3p-1 is the longest side. Wait! It cannot be the longest side. Why? Well, 3p-1 will always be less than 3p, so this side length cannot possibly be the shortest side length. The subtraction of one causes me two know this for certain. This leaves us with the next viable possibility: 3p is the longest side length.
| \(3p-1+p^2+1>3p\) | If 3p is the longest side, then I am assuming that the other 2 are shorter sides. We can use the relationship aforementioned via the Triangle Inequality Theorem, and we can solve for one possible range for p. Notice how there is a lot that will cancel out in this inequality. |
| \(p^2>0\) | Take the square root of both sides. |
| \(|p|>0\) | An absolute-value inequality, when a greater-than symbol is involved, results in an or-type solution set. Let's write those out. |
| \(p<0\text{ or }p>0\) | This did not really restrict the options that much. Let's consider the next possibility. |
The next possibility is that p^2+1 is the longest side length.
| \(3p-1+3p>p^2+1\) | This inequality is going to be much harder. Let's just everything to one side of the inequality. |
| \(6p-1>p^2+1\) | |
| \(p^2-6p+2=0\) | I think it is easiest to pretend, for now, that this inequality is actually a regular equation. In this case, it is not factorable, so we will have to use another method to solve this quadratic. |
| \(a=1;b=-6;c=2\\ p_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(2)}}{2(1)}\) | The rest is a matter of simplification. |
| \(p_{1,2}=\frac{6\pm\sqrt{28}}{2}\) | The radical can use some simplfiying here. |
| \(p_{1,2}=\frac{6\pm\sqrt{28}}{2}=\frac{6\pm2\sqrt{7}}{2}\) | A common factor between every term is two. Factor that out. |
| \(p_1=3+\sqrt{7}\\ p_2=3-\sqrt{7}\) | Use test points to find the range that satisfies the original inequality. I found that the following range for p satisfied the original equation. |
| \(3-\sqrt{7} | Let's convert these to decimal approximations so that we can determine the number of integer solutions easier. |
| \(\approx0.35 | |
Therefore, every positive integer in between these two numbers is the answer. The integers are 1,2,3,4, and 5. This is a total of 5 integers for p.
+2440 The first equation involves knowledge on how to solve system of equations. There is a chance that there will be two solutions here because the second equation \(x^2+y^2=28\) is a degree-two polynomial. I think it would be easiest to utilize the substitution method here.
| \(x+y=6\) | I will solve for x, but you could just as easily solve for y. |
| \(x=6-y\) | |
Since I know that x=6-y, I can substitute this information into the second equation \(x^2+y^2=28\).
| \(x^2+y^2=28\) | Substitute x for 6-y and solve the equation. |
| \((6-y)^2+y^2=28\) | Let's expand the left hand side of the equation. |
| \(36-12y+y^2+y^2=28\) | Let's do some rearranging and combining of like terms. |
| \(2y^2-12y+36=28\) | Subtract 28 from both sides. |
| \(2y^2-12y+8=0\) | Every term on the left hand side has a factor of 2, so let's fact that out! |
| \(2(y^2-6y+4)=0\) | Unfortunately, the trinomial inside the parentheses is not factorable, so we will have to use alternate methods (like the quadratic formula) to figure out the roots of this equation. |
| \(a=1;b-6;c=4\\ y_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(4)}}{2(1)}\) | The only thing to do now is simplify. |
| \(y_{1,2}=\frac{6\pm\sqrt{20}}{2}\) | Let's simplify the radical completely. |
| \(y_{1,2}=\frac{6\pm2\sqrt{5}}{2}=3\pm\sqrt{5} \) | |
| \(y_1=3+\sqrt{5}\\ y_2=3-\sqrt{5}\) | We now have to figure out what the corresponding x-values are. Let's use the first equation for that (\(x+y=6\)). |
| \(x_1+y_1=6\) | Substitute the value for y_1 and solve for the remaining unknown. |
| \(x_1+3+\sqrt{5}=6\) | Subtract 3 |
| \(x_1+\sqrt{5}=3\) | Subtract \(\sqrt{5}\) |
| \(x_1=3-\sqrt{5}\) | Now, let's figure out the next x-value. |
| \(x_2+y_2=6\) | |
| \(x_2+3-\sqrt{5}=6\) | Now, do the same process. |
| \(x_2-\sqrt{5}=3\) | |
| \(x_2=3+\sqrt{5}\) | |
Therefore, the two solutions for this system are \((3-\sqrt{5},3+\sqrt{5}),(3+\sqrt{5},3-\sqrt{5})\). This is just part one.
+2440 When I solve for m in this equation \(\sqrt{\frac{2m}{5-m}}=t\), I agree with Mark Scheme's answer of \(m=\frac{5t^2}{t^2+2}\).
| \(\sqrt{\frac{2m}{5-m}}=t\) | Squaring both sides is definitely the first step. This will make it easier to make m the subject of this equation. |
| \(\frac{2m}{5-m}=t^2\) | Let's multiply both sides by the denominator 5-m. |
| \(2m=t^2(5-m)\) | Let's distribute. |
| \(2m=5t^2-t^2m\) | It is generally easiest to place all terms with an "m" in it on the same side of the equation. Let's do that by adding \(t^2m\) |
| \(t^2m+2m=5t^2\) | Factor out the greatest common factor of the left hand side of the equation. This is why you should place all terms with an "m" in it on the same side. |
| \(m(t^2+2)=5t^2\) | Divide by \(t^2+2\) |
| \(m=\frac{5t^2}{t^2+2},t\geq 0\) | I place this restriction on t because the original equation had a square root equation in it, which can only result in a nonnegative answer. |
You did not provide any work on how you arrived at your answer, so I do not know which step went awry. Maybe this work will help guide you, and it may help you spot your error. Good luck!
+2440 I generally find it to be easier when the equations are put in slope-intercept form (or y=mx+b). In order to convert the original equation to this form, solve for y. The slope-intercept form tells one important information about a certain equation.
| \(7x-2y=-24\) | To solve for y, it is probably easiest to subtract 7x from both sides. |
| \(-2y=-7x-24\) | Finally, divide by -2. |
| \(y=\textcolor{red}{\frac{7}{2}}x+\textcolor{blue}{12}\\ \textcolor{red}{m}=\textcolor{red}{\frac{7}{2}}\\ \textcolor{blue}{b}=\textcolor{blue}{12} \) | |
One special property about perpendicular lines is that their slopes are opposite reciprocals from each other. Since we know the slope of the given line, we can figure out the slope of the perpendicular one.
\(\frac{7}{2}\Rightarrow{-\frac{7}{2}}\Rightarrow-\frac{2}{7}\)
-2/7 is the slope of the perpendicular line. However, we know another requirement about this line: It passes through the coordinate (14,-2). Let's take this information into account. We know that the equation, thus far, is \(y=-\frac{2}{7}x+b\).
| \(y=-\frac{2}{7}x+b\) | b is the only variable remaining in this equation. We know that, based on the previous info, that when x=14, y=-2. This was preset by the given coordinate. Let's plug those values in and solve for the missing variable: b. |
| \(-2=-\frac{2}{7}*14+b\) | 14 and 7 are common factors, so this will divide evenly. |
| \(-2=-4+b\) | Add 4 to both sides to isolate b. |
| \(2=b\) | Now, let's write the final equation in slope-intercept form (because that is the form that the answer choices are given in). |
\(y=-\frac{2}{7}x+2\), which is B in the answer choices provided.
+2440 Well, I guess I committed a cardinal sin: I did not factor completely. I thought there was no way that either the numerator or denominator could have a common factor. How wrong I was!
| \(-3x^7-3x^6-3x^5+3x^4+3x^2+3\) | First, factor out a common factor of -3 from every term. |
| \(-3(x^7+x^6+x^5-x^4-x^2-1)\) | Well, I can use the rational root theorem to determine that 1 is a root of the seven-degree polynomial, so \(x-1\) must also be a factor. This results in the following result. After using synthetic division, I got the following! |
| \(-3\left(x-1\right)\left(x^6+2x^5+3x^4+2x^3+2x^2+x+1\right)\) | Let's worry about the denominator. |
| \(x^3(x^2+x-2)\) | I was able to use factoring here, too, by finding the products of -2 that equal 1: 2 and -1. Let's complete the factoring, then! |
| \(x^3(x-1)(x+2)\) | |
| \(\frac{-3\left(x-1\right)\left(x^6+2x^5+3x^4+2x^3+2x^2+x+1\right)}{x^3(x-1)(x+2)}\) | Look at that! \(x-1\) is a common factor. That's crazy! This leaves us with the following! |
| \(g(x)=\frac{-3\left(x^6+2x^5+3x^4+2x^3+2x^2+x+1\right)}{x^3(x+2)}, x(x+2)\neq0\text{ and }x\neq 1\) | I guess this is better! |
+2440 Asymptotes are always expressed as equations because they are lines, so I would assume that the horizontal asymptote lies at y=0.
Of course, horizontal asymptotes have many rules associated with them. In this case, we are concerned with the following rule: If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptotes exists at the x-axis (also known as y=0).
If we take \(f(x) = \frac{3(x^4+x^3+x^2+1)}{x^2+x-2}\) and multiply it by \(\frac{1}{x^3}\), then we would have finished this problem. Therefore, let's find g(x) that does this.
| \(\frac{3(x^4+x^3+x^2+1)}{x^2+x-2}+g(x)=\frac{3(x^4+x^3+x^2+1)}{x^3(x^2+x-2)}\) | Let's do some subtraction here. |
| \(g(x)=\frac{3(x^4+x^3+x^2+1)}{x^3(x^2+x-2)}-\frac{3(x^4+x^3+x^2+1)}{(x^2+x-2)}\) | Let's convert the rightmost fraction into one with a common denominator so that further simplification is possible. |
| \(g(x)=\frac{3(x^4+x^3+x^2+1)}{x^3(x^2+x-2)}-\frac{3x^3(x^4+x^3+x^2+1)}{x^3(x^2+x-2)}\) | Let's do some distributing. |
| \(g(x)=\frac{3x^4+3x^3+3x^2+3}{x^3(x^2+x-2)}-\frac{(3x^7+3x^6+3x^5+3x^3)}{x^3(x^2+x-2)}\) | Now, subtract! |
| \(g(x)=\frac{-3x^7-3x^6-3x^5+3x^4+3x^2+3}{x^3(x^2+x-2)}\\ \) | Not a lot of simplification happens, though, but let's do it anyway! |
You should see that, when adding g(x) to f(x), this results in a horizontal asymptote of y=0.
