+2441 We meet again, Rick!
The shaded region, I would presume anyway, also happens to be the area of a rectangle. Let's take a guess and say that the left figure has more area. To confirm (or disprove) this, one must find the areas of the rectangles.
| \(\text{left }\boxed{\quad}=a(b+2)\) | |
| \(\text{right}\boxed{\quad}=b(a+2)\) | Let's assume that the left rectangle has more area and see what happens. |
| \(a(b+2) < b(a+2)\) | Now, solve. This will allow you to figure out the relationship between a and b. Distribute. |
| \(ab+2a < ab+2b\) | The ab's cancel out. |
| \(2a<2b\) | Divide by 2 on both sides. |
| \(a < b\) | |
What is this result telling you or me? It tells you that, in order for the leftward rectangle to have more area, a must be less than b. Of course, this cannot be; the original problem states that \(a>b\) . We have reached a contradiction. Therefore, the rightward rectangle has more area!
+2441
In expressions (1) and (2), there is only one difference: the signage. Since there is at least one boy, the minimum number that b/2 can equal is 1/2 or 0.5. There is no maximum.
Do you think that subtracting or adding at least 0.5 to the number of girls in the classroom will result in the larger value? Certainly adding it would make it the larger value!
We still have to consider the possibility where they could be equal, though. This prompts the following equation:
| \(g-\frac{b}{2}=g+\frac{b}{2}\) | For the time being, we will assume that these values are equal. Doing this will allow us to discover how many boys or girls will result in an equilibrium. Notice how the g's cancel out. |
| \(-\frac{b}{2}=\frac{b}{2}\) | Add b/2 to both sides. |
| \(b=0\) | According to the answer I just got, there would have to be 0 boys in order for \(g-\frac{b}{2}\) to be equal to \(g+\frac{b}{2}\) . This is impossible because the problem states that there is at least one boy. Therefore, (4) is not possible. |
This means that the original idea stands: \(g+\frac{b}{2}\) always results in the larger value, in this case.
+2441 Hi, Rick, I will find the first two terms, and I want to see you find the next two. See if you can do it!
There are a few aspects of the given explicit formula.
n = the term number
an = the value of the nth term.
The problem asks for the first 4 terms. I will start with the first term. Since n represents the term number and since we want the first term, n=1:
| \(n=1;\\ a_n=6\cdot(2)^{n-1}\) | Replace all instances of n with a 1 and solve for the first term! |
| \(a_1=6\cdot(2)^{1-1}\) | We can now simplify the exponent. |
| \(\)\(a_1=6\cdot2^0\) | Via the exponent properties, \(x^0=1, x\neq 0\). |
| \(a_1=6\) | a1 , the value of the 1st term, is 6. That's what the notation means. |
In order to find the 2nd term, set n to 2 and solve for a2:
| \(n=2;\\ a_n=6\cdot(2)^{n-1}\) | The process is no different than what occurred above. The only difference is the value of n. |
| \(a_2=6\cdot(2)^{2-1}\) | Simplify the expression in the exponent. |
| \(a_2=6\cdot 2^1\) | |
| \(a_2=12\) | |
Now, I am leaving the rest up to you. See what you can do.
+2441 This is a recursive formula. The first term, \(a_1\), is already given; it is -3. The recursive formula is saying that the following term is equal to the previous term minus 6.
The first term is easy to find.
| \(a_1=-3\) | This is the first term. It is given to us! |
Let's find the second term! In order to find the second term, set n equal to 2 in the formula!
| \(n=2;\\ a_2=a_{2-1}-6\) | a2-1 simplifies to a1 . |
| \(a_2=a_1-6\) | Of course, we already know what the value of a1 is. a1=-3. |
| \(a_2=-3-6\) | |
| \(a_2=-9\) | |
See if you can generate the next two terms in the sequence. Well, it looks like Gavin already answered the question I planned to pose to you.
