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.
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)
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. |
\(x=\frac{\log_{10}(200)}{2\log_{10}(5)}\) | |
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. |
\(\hspace{3mm}A+B+C\\ -1-4+4\\ -1\) |