+2441 a) In order to factor using this method, let's try and identify the GCF first. 9 is the greatest common factor between 36 and 81. a^4 is the greatest common factor between the a's, and b^10 is the factor for the b's. Let's factor it out!
| \(36a^4b^{10}-81a^{16}b^{20}\) | Factor out the GCF, \(9a^4b^{10}\), like I described earlier. |
| \(9a^4b^{10}\left(4-9a^{12}b^{10}\right)\) | Don't stop here, though! Notice that the resulting binomial is a difference of squares. |
| \(9a^4b^{10}\left(2+3a^6b^5\right)\left(2-3a^6b^5\right)\) | |
b) The beginning binomial is a difference of squares to begin with, so it is possible to start with this first!
| \(36a^4b^{10}-81a^{16}b^{20}\) | Let's do this approach this time! |
| \(\left(6a^2b^5+9a^8b^{10}\right)\left(6a^2b^5-9a^8b^{10}\right)\) | Don't stop yet! Both binomials have their own GCF's! |
| \(3a^2b^5\left(2+3a^6b^5\right)*3a^2b^5\left(2-3a^6b^5\right)\) | Combine the multiplication. |
| \(9a^4b^{10}\left(2+3a^6b^5\right)\left(2-3a^6b^5\right)\) | |
Well, these are the two techniques.
+2441 8) This one will require the formula that yields the volume of a cylinder. \(V_{\text{cylinder}}=\pi r^2h\). We can manipulate this formula so that we can find any missing information such as the height, in this case.
| \(V_{\text{cylinder}}=\pi r^2h\) | We know what the volume is, and we know the height, so finding the radius is simply a matter of isolating the variable. | ||
| \(27143=15\pi r^2\) | Divide by 15 pi first. | ||
| \(\frac{27143}{15\pi}=r^2\) | Take the square root of both sides. | ||
| \(|r|=\sqrt{\frac{27143}{15\pi}}\) | The absolute value splits the answer into two possibilities. | ||
| In the context of geometry, negative side lengths are nonsensical, so let's just reject the answer now. | ||
| \(r=\sqrt{\frac{27143}{15\pi}}\approx 24\text{m}\) | The radius is a one-dimensional part of a cylinder, so the units should be in one dimension, too. | ||
9) If the height of the un-consumed soup was 8 centimeters tall and 3-centimeters-worth of soup is consumed, then 5 centimeters of soup remains. We already know the radius of this soup can (that I assume is cylinder-shaped despite not being explicitly stated), so we can determine the volume.
| \(V_{\text{cylinder}}=\pi r^2h\) | Plug in the known values. |
| \(V_{\text{cylinder}}=\pi*12^2*5\) | Now, combine like terms. |
| \(V_{\text{cylinder}}=720\pi\approx 2262\text{cm}^3\) | Volume is always expressed as a cubic unit. |
10) If the town park enlarges its area by a factor of 5, then both dimensions of the park are affected by this scale factor. For example, if we assume, for the sake of understanding, that the park is perfectly rectangular with dimensions 3yd by 100yd, then both dimensions (the length and the width) would be affected by this scale factor. This means that, on area, the scale factor actually affects the area by its square, or 25 in this case.
\(300\text{yd}^2*25=7500\text{yd}^2\)
.+2441 The problem is easier than I first thought.
By the given information, we know that there are three right-angled triangles in the diagram. We know that \(m\angle AEB=m\angle BEC=m\angle CED= 60^{\circ}\). We can use this information to determine that every right triangle is also a 30-60-90 triangle. We also know that \(AE=24\).
A 30-60-90 triangle is a special kind of right triangle where the ratio of the side lengths are \(1:\sqrt{3}:2\). \(\overline{AE}\) is the longest side length because it is the hypotenuse of the largest right triangle in the diagram. \(\overline{BE}\) is the shortest side length of \(\triangle ABE\) because this side is opposite the smallest angle. We mentioned earlier what the ratio of the side lengths are, so we can determine the length of \(\overline{BE}\) without doing anything too computationally demanding.
| \(\frac{BE}{AE}=\frac{1}{2}\) | We already know the ratio of the side lengths of a 30-60-90 triangle, so we can apply this relationship and create a proportion. We already know what AE equals, so let's fill that in. |
| \(\frac{BE}{24}=\frac{1}{2}\) | In order to solve a proportion, simply cross multiply. |
| \(2BE=24\) | Divide by 2 on both sides to determine the unknown length of the side. |
| \(BE=12\) | |
Of course, the ultimate goal is to figure out the length of \(\overline{CE}\). If you look at \(\triangle BCE\), carefully, you will notice that we are in an identical situation to when we solved for \(BE\). Notice that \(\overline{BE}\) is the hypotenuse of this triangle, and \(\overline{CE}\) is the shortest side length since it is opposite the 30º angle. We can use the same \(1:\sqrt{3}:2\) relationship of the side lengths to find the missing length.
| \(\frac{CE}{BE}=\frac{1}{2}\) | Just like before, we know what the value of BE is, so let's plug it in! |
| \(\frac{CE}{12}=\frac{1}{2}\) | Just like before, cross multiplying is the way to go! |
| \(2CE=12\) | Divide by 2 on both sides to solve this problem. |
| \(CE=6\) | |
+9466 1.
\(y=\frac{20}{1+9e^{3x}} \\~\\ y(1 + 9e^{3x})=20 \\~\\ 1 + 9e^{3x}=\frac{20}{y} \\~\\ 9e^{3x}=\frac{20}{y}-1 \\~\\ 9e^{3x}=\frac{20-y}{y} \\~\\ e^{3x}=\frac{20-y}{9y} \\~\\ 3x=\ln(\frac{20-y}{9y}) \\~\\ x=\frac{\ln(\frac{20-y}{9y})}{3}\)
There is an asymptote when 9y = 0 which is when y = 0
There is an asymptote when \(\frac{20-y}{9y}\) = 0 which is when y = 20
Here's a graph: https://www.desmos.com/calculator/9czjghjp9s
