+2440 MrPatel, this particular problem is not as labor-intensive as you probably think. As the inequalities become more complicated, though, this process will probably not be as simple. I would recommend treating the inequality \(y>2x^2\) as if it were an equation for now. This should help you get started. I will make a table of values with 5 x-values. This should not be too difficult.
| Input (x-coordinate) | Output (y-coordinate) | Coordinate | |
| \(-2\) | \(8\) | \((-2,8)\) | |
| \(-1\) | \(2\) | \((-1,2)\) | |
| \(0\) | \(0\) | \((0,0)\) | |
| \(1\) | \(2\) | \((1,2)\) | |
| \(2\) | \(8\) | \((2,8)\) | |
Plot these coordinates on a coordinate plane. Then, connect them with a dotted curve. Do this because the inequality in a less-than type. The original inequality indicates that we want to include all outputs that are greater than the output given by 2x^2. For example, a valid output for the given equality for the input 0 is 1 or 2 or \(\frac{5}{12}\) or \(\sqrt{30}\) or \(10.123\). Notice how these are all greater than 0. This means that we want to shade all values above the given outputs.
Does this make sense to you?
+2440 This is a finite summation of an arithmetic series because each subsequent term is simply subtracting two. We can figure out what the first term of the sequence is. Of course, you could just add up all the terms from 1 to 11, but it is probably better to use or derive a formula. In order to find the sum of an arithmetic sequence, you must find three values:
We know that the first term of a summation is when the variable, n, equals its first possible value, which is one, in this case.
| \(a_1=98-2*1=96\) | |
an can be found by using a formula.
| \(a_n=a_1+d(n-1)\) | Of course, we already know what a1 equals. We know the common difference is -2. The coefficient of the variable indicates this information. |
| \(a_{11}=96-2(11-1)\) | Time to simplify! |
| \(a_{11}=96-20\) | |
| \(a_{11}=76\) | |
What about the number of terms? Well, this can be determined using the term numbers indicated in summation formula.
\(n=11-1+1=11\)
Let's put all this information together and use the formula to find the sum.
\(S_n=n\left(\frac{a_1+a_n}{2}\right)\)
| \(S_n=n\left(\frac{a_1+a_n}{2}\right)\) | Plug in the values and simplify. |
| \(S_{11}=11(\frac{96+76}{2})\) | |
| \(S_{11}=946\) | |
+2440 11A recursive rule allows one to generate future terms of a sequence if one knows one or more of the previous terms. We know that the explicit rule can be written in the following form:
\(a_n=a_1+d(n-1)\)
There are two variables that we must identify in order to finish this formula. Those are:
We can easily identify both of these with some observation. It is given that a1 =21 since that information was given in the recursive formula. In the recursive formula, one must add 9 to obtain the next term in the sequence. This would be the common difference. Let's fill that in and simplify completely.
| \(a_n=a_1+d(n-1)\) | Substitute in the known values and simplify. |
| \(a_n=21+9(n-1)\) | Distribute the 9 into the binomial. |
| \(a_n=21+9n-9\) | Combine like terms. |
| \(a_n=9n+12\) | This answer corresponds to the first answer choice. |
+2440 I hope this helps you! This is just algebraic manipulation.
| \(x^2-9y^2=0\) | |
| \(x^2=9y^2\) | |
| \(\frac{x^2}{y^2}=9\) | |
| \(\left(\frac{x}{y}\right)^2=9\) | |
| \(\frac{x}{y}=\frac{3}{1}\\ x:y=3:1\) | Be careful! Do not take the negative root. "x" and "y" must both be positive. This is given information. |
+2440 Let w = width of shoebox
Let l = length of the shoebox
Since the length is 8 centimeteres more than twice the width, we can conclude that l = 2w+8.
The area of a rectangle is given, so we can solve for the now one-variable equation.
| \(w(2w+8)=72\) | Expand completely. |
| \(2w^2+8w=72\) | |
| \(2w^2+8w-72=0\) | Factor out the GCF of every term, which is 2 in this case. |
| \(2(w^2+4w-36)=0\) | Unfortunately, this trinomial is not factorable. Let's use the quadratic formula instead. |
| \(a=1,b=4,c=-36;\\ x_{1,2} = {-b \pm \sqrt{b^2-4ac} \over 2a}\) | Now, this is a matter of simplification. |
| \(x_{1,2} = {-4 \pm \sqrt{4^2-4(1)(-36)} \over 2(1)}\) | |
| \(x_{1,2} = {-4 \pm \sqrt{160} \over 2}\) | It is possible to simplify the radical. |
| \(x_{1,2} = {\frac{-4\pm4\sqrt{10}}{2}}\) | Simiplify the fraction as every term has a common factor of 2. |
| \(x_{1,2} = -2\pm2\sqrt{10}\) | Reject the negative answer since a width can never be negative in the context of geometry. |
| \(x=-2+2\sqrt{10}\approx 4.32 \text{cm}\) |
When I read the question more closely, I realized that all this solving was completely superfluous. You can still use the equation from the beginning, though! It represents the situation perfectly.
+2440 The general formula for volume for any pyramid or cone is \(V=\frac{1}{3}Bh\)
Let V = volume of the cone or pyramid
Let B = area of the base
Let h = the perpendicular height of the cone or pyramid
Since all these figures are cones, we know that the area of the base equals \(V=\frac{1}{3}\pi r^2h\). Ok, let's start solving for the individual volumes.
| Cone 1 | Cone 2 | Cone 3 |
| \(V_1=\frac{1}{3}\pi r^2h;\\ r=11,h=9 \) | \(V_2=\frac{1}{3}\pi r^2h;\\ r=6,h=14 \) | \(V_3=\frac{1}{3}\pi r^2h;\\ r=14, h=8\) |
| \(V_1=\frac{1}{3}\pi*11^2*9\) | \(V_2=\frac{1}{3}\pi*6^2*14\) | \(V_3=\frac{1}{3}\pi*14^2*8\) |
| \(V_1=3\pi*121\) | \(V_2=12\pi*14\) | \(V_3=\frac{1}{3}\pi*1568\) |
| \(V_1=363\pi\text{cm}^3\) | \(V_2=168\pi\text{cm}^3\) | \(V_3=\frac{1568\pi}{3}\text{cm}^3\approx 500\text{cm}^3\) |
I estimated for cone 3 because 3 does not divide evenly with 1568. This approximation is good enough for these purposes anyway because we only care about their order of greatness. The otder, therefore, from least to greatest is the following:
Cone 2, Cone 1, Cone 3 (option 2)
+2440 I think I will complete a and c. I think you can manage the rest on your own. You can use my work as a model.
a)
| \(10^{x^2}=320\) | Take the logarithm base 10 of both sides, which is exactly that the directions say! |
| \(\log_{10}\left(10^{x^2}\right)=\log_{10}(320)\) | Because of the properties of logarithms, the exponent can be extracted, and it will become the coefficient of the logarithm. |
| \(x^2\log_{10}(10)=\log_{10}(320)\) | The \(\log_{10}(10)\) simiplifies to 1 because the base and the argument are the same value. |
| \(x^2=\log_{10}(320)\) | Take the square root of both sides. |
| \(|x|=\sqrt{\log_{10}(320)}\) | Of course, the absolute value creates two solutions: the positive and negative one. |
| \(x=\pm\sqrt{\log_{10}(320)}\) | |
c)
| \(5^{2x}=200\) | Take the logarithm base 10 of both sides again. |
| \(\log_{10}(5^{2x})=\log_{10}(200)\) | Yet again, the exponent becomes the coefficient. This is a basic property of logarithms. |
| \(2x\log_{10}(5)=\log_{10}(200)\) | Divide by \(2\log_{10}(5)\) to isolate x. |
| \(x=\frac{\log_{10}(200)}{2\log_{10}(5)}\) | |
+2440 1) There is one characteristic about the given graph that will allow one to determine the equation of the original function: the vertical asymptotes. Vertical asymptotes are vertical lines that a function approaches but never reaches or crosses. I like to call this type of asymptote an absolute asymptote since the function will never cross this. Horizontal asymptotes are not absolute; there are times where the function crosses a horizontal asymptote. In fact, there is a horizontal asymptote located at \(y=0\) here, but that is beyond the scope of this particular problem. There are three vertical asymptotes on the graph. They are located at \(x=-2,x=1,\text{ and }x=2\). The function never reaches this point because of a division-by-zero error.
| \(x=-2\) | \(x=1\) | \(x=2\) |
| \(x+2=0\) | \(x-1=0\) | \(x-2=0\) |
Let's multiply these factors together. We know that their product equals 0 since the right-hand side of all the equations equal zero.
\((x+2)(x-2)(x-1)=0\)
Let's expand. I decided to rearrange to make the difference of squares more obvious. Let's expand.
| \((x+2)(x-2)(x-1)=0\) | Let's do the first two binomials first. Since they will result in a difference of squares, we can take the shortcut. |
| \((x^2-4)(x-1)=0\) | Now, let's expand completely. |
| \(x^3\textcolor{red}{-\hspace{1mm}1}x^2\textcolor{blue}{-\hspace{1mm}4}x\textcolor{green}{+\hspace{1mm}4}=0\\ x^3\textcolor{red}{+A}x^2\textcolor{blue}{+B}x\textcolor{green}{+C}\\ A=-1,B=-4,C=4\) | This is the denominator of our function! Notice how this function perfects aligns with the question. The only thing left to do is find their sum. |
| \(\hspace{3mm}A+B+C\\ -1-4+4\\ -1\) |
