+2446 Let's solve this equation for q
| \(0=80q-0.05q\) | Combine like terms on the right hand side of the equation |
| \(0=79.95q\) | Divide by \(79.95 \) on both sides of the equation |
| \(0=q\) | We're done! |
+2446 I think I can help you evaluate this expression:
\([-2-((-4)^2-2)]*(-4-(-3))\)
| \([-2-((-4)^2-2)]*(-4-(-3))\) | This is the original expression. First, evaluate parentheses from left to right. Of course, you have to order of operations within a set of parentheses. Do \((-4)^2=-4*-4=16\) |
| \([-2-(16-2)]*(-4-(-3))\) | Do \(16-2\) because those are the innermost parentheseses. |
| \([-2-14]*(-4-(-3))\) | Do \(-2-14\) because you evaluate parentheses from left to right! |
| \(-16*(-4-(-3))\) | Now, evaluate \(-4-(-3)=-4+3=-1\). |
| \(-16*-1\) | Evaluate -16*-1. |
| \(16\) | This is your answer |
+2446 Here are the points that are given, \(A(-1,n)\hspace{3mm},B(3,5)\hspace{3mm},C(-4,0)\hspace{3mm},D(0,2)\). Our first task is to fid the value for n wherein \(\overline{AB}\hspace{1mm}||\hspace{1mm}\overline{CD}\). Let's think about what we must do first.
How can we verify if two segments are parallel to each other? We can compare the slopes! If the slopes are the same, then the segments are parallel. If the slopes are different, then the segment, if extended infinitely, will intersect at some point. Let's find the slope of \(\overline{CD}\):
| \(m=\frac{y_2-y_1}{x_2-x_1}\) | Of couse, this is the equation for the slope of a segment. You need to know 2 coordinates of a segment in order to find the slope. Luckily, we already have 2 given |
| \(m\hspace{1mm}\text{of}\hspace{1mm}\overline{CD}=\frac{0-2}{-4-0}\) | Put the fraction in simplest terms now. |
| \(m\hspace{1mm}\text{of}\hspace{1mm}\overline{CD}=\frac{-2}{-4}\) | A negative number in the numerator and denominator is the same as dividing two positive numbers. |
| \(m\hspace{1mm}\text{of}\hspace{1mm}\overline{CD}=\frac{2}{4}\) | 2/4 can be simplified still |
| \(m\hspace{1mm}\text{of}\hspace{1mm}\overline{CD}=\frac{1}{2}\) | |
Let's find the slope of \(\overline{AB}\):
| \(m=\frac{y_2-y_1}{x_2-x_1}\) | I am reminding you of the slope formula again! |
| \(m\hspace{1mm}\text{of}\hspace{1mm}\overline{AB}=\frac{n-5}{-1-3}\) | We can simplify the denominator, but the numerator will need something else |
| \(m\hspace{1mm}\text{of}\hspace{1mm}\overline{AB}=\frac{n-5}{-4}\) | This is the slope of \(\overline{AB}\). |
What did I say above? I said that parallel lines have the same slope! That means that the slope of \(\overline{CD}\) equals the slope of \(\overline{AB}\):
| \(\frac{-4}{1}*\frac{1}{2}=\frac{n-5}{-4}*\frac{-4}{1}\) | Multiply by -4 on both sides to get rid of all the pesky fractions! |
| \(-2=n-5\) | Add 5 to both sides |
| \(3=n\) | Therefore, the coordinates of point A is (-1,3) |
+2446 Our goal is to isolate Q. Let's solve for Q:
| \(0=20-0.05q\) | Subtract 20 on both sides of the equation |
| \(-20=-0.05q\) | Convert 0.05 into a fraction to ease calculation |
| \(-20=-\frac{1}{20}q\) | Multiply by -20 on both sides to get rid of that pesky fraction |
| \(400=q\) | You're done! |
+2446 Take a look at the works of beauty possible inside of the Demos Graphing Calculator. Of course, I do not take credit for any of these:
Keep searching! There are good ones that we haven't seen yet!
+2446 I didn't know that a formula existed either, so I derived it myself. Here's how I derived it. I will warn that hairy algebra is coming. I thought that I would skip that part to answer the question, but you, Cphill, seem interested!
A parabola can be defined as an infinite set of points that is equidistant from a point (the focus) and a line (the directrix). Here is a picture. I think it is hard to visualize and understand without it:
I couldn't find a diagram where the directrix equation is y=k, but that's OK. I'll just use what it in the picture:
My goal now is to find an equation for the focus where its coordinates lie on (a,b) and the directrix where it lies on y=c. Let's do it to derive that equation I used in my answer.
In the diagram above, I marked one arbitrary point on the parabola with coordinates (x,y). Of course, as I mentioned before, this point must be equidistant from the focus and from the directrix because of the definition of a parabola. What does that mean? It means that the distance from (x,y), our arbitrary point, to the directrix and to the focus is the same. Let's calculate the distance from the focus to the directrix first. We'll use the distance formula:
| \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) | Of course, this is the distance formula. Let's plug in the points of our arbitrary point and the directrix. |
| \(d=\sqrt{(x-x)^2+(y-c)^2}\) | The arbitrary point meets the directrix perpendicularly when measuring distance, so the x-coordinates are going to cancel out. |
| \(d=\sqrt{0^2+(y-c)^2}\) | |
| \(d=\sqrt{(y-c)^2}\) | Yes, I could simplify further, but I am going to leave this as is for now. |
Now, let's calculate the distance from the arbitrary point, (x,y) to the focus (a,b). Let's use the distance formula for this, too:
| \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) | Just like above, this is the distance formula. Let's plug in the values. |
| \(d=\sqrt{(x-a)^2+(y-b)^2}\) | This cannot be simplified any further |
Remember what I mentioned earlier? These distances must be equal for a parabola to be a parabola; it defines a parabola. Therefore, \(\sqrt{(y-c)^2}=\sqrt{(x-a)^2+(y-b)^2}\). Technically, I could stop here and say that this is the formula, but we can simplify this. Also, I wanted it to be in the equation of a parabola, which means I would have to solve for y. Let's do this:
| \(\sqrt{(y-c)^2}=\sqrt{(x-a)^2+(y-b)^2}\) | This is what was deduced above. Let's square both sides to get rid of the radical |
| \((y-c)^2=(x-a)^2+(y-b)^2\) | My goal is to solve for y. Let's expand the terms that contain a y in them. Of course, I'll use the rule that says that \((a-b)^2=a^2-2ab+b^2\). |
| \(y^2-2yc+c^2=(x-a)^2+y^2-2yb+b^2\) | Look at that! The y^2 terms cancel out! Subtract y^2 on both sides of the equation. |
| \(-2yc+c^2=(x-a)^2-2yb+b^2\) | Let's get the c^2 term on the right of the equation. Subtract c^2 on both sides |
| \(-2yc=(x-a)^2-2yb+b^2-c^2\) | Add -2yb to both sides so that the terms containing a y are on the left and all the terms that do not are on the right. |
| \(2yb-2yc=(x-a)^2+b^2-c^2\) | Factor out a y in both terms on the left hand side. |
| \(y(2b-2c)=(x-a)^2+b^2-c^2\) | Divide by (2b-2c) on both sides |
| \(y=\frac{(x-a)^2+b^2-c^2}{2b-2c}\) | I'll split off the \(b^2-c^2 \) from the fraction so I can simplify it separately. |
| \(y=\frac{(x-a)^2}{2b-2c}+\frac{b^2-c^2}{2b-2c}\) | The GCF of 2b-2c is 2. Factor it out. |
| \(y=\frac{(x-a)^2}{2(b-c)}+\frac{b^2-c^2}{2(b-c)}\) | I'll split up b^2-c^2 by using the rule that \(a^2-b^2=(a+b)(a-b)\). |
| \(y=\frac{(x-a)^2}{2(b-c)}+\frac{(b+c)(b-c)}{2(b-c)}\) | In the term \(\frac{(b+c)(b-c)}{2(b-c)}\). The (b-c) will cancel out. |
| \(y=\frac{(x-a)^2}{2(b-c)}+\frac{b+c}{2}\) | This equation should look familiar. Yes, it is the equation I used to calculate the equation of the parabola. Just replace the c's with k's. |
Well, now we have an equation that will always find the equation of a parabola when only the directrix and focus are given. As you can see, there was much more to this problem than what I originally showed.
+2446 There is a a formula for the equation for a parabola when only the focus and directrix are given. It is the following:
Let a= x-coordinate of the focus
Let b= y-coordinate of the focus
Let k= equation of the directrix
\(y=\frac{(x-a)^2}{2(b-k)}+\frac{1}{2}(b+k)\)
Plug the values in to get the equation of a parabola:
a=4
b=6
k=2
| \(y=\frac{(x-a)^2}{2(b-k)}+\frac{1}{2}(b+k)\) | Plug in the appropriate a,b, and k values into the equation. |
| \(y=\frac{(x-4)^2}{2(6-2)}+\frac{1}{2}(6+2)\) | \(2(6-2)=8\hspace{3mm}\text {and} \hspace{3mm}\frac{1}{2}(6+2)=4\) |
| \(y=\frac{(x-4)^2}{8}+4\) | Square the term x-4 by using the rule stating that \((a-b)^2=a^2-2ab+b^2\). |
| \(y=\frac{x^2-8x+16}{8}+4\) | You can eliminate the +4 by converting to a common denominator. Convert 4 to 32/8. |
| \(y=\frac{x^2-8x+16}{8}+\frac{32}{8}\) | \(\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}\), so let's apply it. |
| \(y=\frac{x^2-8x+48}{8}\) | Now, I'll actually split up the fraction so that we can make this look a tad nicer. Remember that \(\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}\) |
| \(y=\frac{x^2}{8}+\frac{-8x+48}{8}\) | Now it may be clear why I split up the terms; this way, I can get rid of the 8 in one part since the GCF of -8x and 48 is 8. |
| \(y=\frac{x^2}{8}-x+6\) | |
| \(y=\frac{1}{8}x^2-x+6\) | Now, you are done. |
