+2441 I would convert both the multiplicand and multiplier to scientific notation and then multiply.
\(5000\Rightarrow5*10^3\\ 30000\Rightarrow3*10^4\)
Now, multiply them together.
| \(\left(\textcolor{red}{5}*\textcolor{blue}{10^3}\right)\left(\textcolor{red}{3}*\textcolor{blue}{10^4}\right)\) | Multiply these together. The numbers in red and blue can be combined together. The blue numbers can be multiplied because they both have identical bases. |
| \(15*10^{7}\) | Of course, we must remain proper with scientific notation rules. This means that the beginning number must be bounded between 1 and 10. |
| \(1.5 ×10^8\) | I find that the traditional multiplication symbol is used with scientific notation, so I used that particular symbol; however, it is not required. |
+2441 | parallelogram EFGH | Given |
| \(\overline{EF}\cong\overline{HG}\) | Property of a Parallelogram (If a quadrilateral is a parallelogram, then its opposite sides are congruent) |
| \(\overline{EF}\parallel\overline{HG}\) | Definition of a Parallelogram (A quadrilateral that has opposite sides parallel is a parallelogram) |
| \(\textcolor{red}{\angle FEG\cong\angle HGE\\ \angle EFH\cong\angle FHG}\) | Alternate Interior Angles Theorem |
| \(\triangle EKF\cong\triangle GKH\) | ASA Congruence Postulate |
| \(\textcolor{red}{\overline{EK}\cong\overline{KG}\\ \overline{FK}\cong\overline{KH}}\) | CPCTC |
+2441 Yes, it is possible. After "Y1=" command line, input \(x^2/(x\geq0)\).
Let me explain how this works, so you understand, and you don't just use rote memorization. To access the inequality symbols, press, in order, the blue second button followed by the button reading "math."
Upon doing the steps above, your interface should look like the following:
TEST LOGIC
1: =
2: ≠
3: >
4: ≥
5: <
6: ≤
Use your directional arrows to navigate to the individual inequalities symbols.
Another skill you must know how to do is storing values as variables. Use this syntax as a guideline for you.
[number]→[variable]
Let's perform some extremely simple logic calculations with the calculator.
1. Input 10→X on the homescreen command line.
2. Input X≥0 on the homescreen command line.
What did the calculator output? That's right! The calculator outputted a 1. This is the calculator method of saying "true." Now, let's perform another experiment. It is nearly the same process as before.
1. Input -10→X on the homescreen command line.
2. Input X≥0 on the homescreen command line.
The calculator outputted a "0" this time to signify a false statement. An interesting fact about this output is that you can also do arithmetic with this information, too. Syntax is extremely important here, so use parentheses even if you think it is completely unnecessary.
1. Input 9*(X≥0)+4
The calculator outputted 4. How did it do that, you ask? Well, let me explain.
| \(9*\textcolor{red}{(x\geq0)}+4\) | The calculator first evaluates inside of the parentheses so that it adheres to the order of operations. |
| \(\textcolor{red}{9*0}+4\) | Of course, x is set to be equal to -10, so the calculator outputs a "0" because \(-10\ngeq0\). Now, the calculator continues like normal. It, of course, does multiplication first. |
| \(\textcolor{red}{0+4}\) | Now, the calculator adds. |
| \(4\) | |
So, how does this help me understand what occurs for the restricted domain? Well, let's look at it. Let's pick a number that is less than 0. I'll choose -2, in this case.
| \(y_1=x^2/(x\geq0)\) | Since x=-2, substitute that in for every value of x. |
| \(y_1=(-2)^2/(-2\geq 0)\) | |
| \(y_1=4/0\) | Uh oh! Dividing by zero causes an error, so this point is not graphed! |
Now, let's try plugging in 2.
| \(y_1=x^2/(x\geq0)\) | Replace every instance of x with a 2. |
| \(y_1=2^2/(2\geq 0)\) | Now, evaluate. |
| \(y_1=4/1\) | Notice how this does not cause a problem because dividing by 1 does not actually affect the output of the function. |
| \(y_1=4\) | The calculator would then plot that point. The process continues. |
The moral here is that a false statement causes a division by zero error while a true statement leaves the original function untouched.
+2441 The way I would approach this problem is to find a formula for the sum of consecutive integers 1,2,3,4 ..., n.
Let S = sum of consecutive whole numbers starting with 1
\(S=1+2+3+4+...+n \)
Now, let's reverse the sum. You will see where this is headed in a moment.
\(S= n+(n-1)+(n-2)+(n-3)+...+1\)
Both of the sums above are the same. Now, let's add them together.
| \(S= 1+\hspace{1cm}2\hspace{2mm}+\hspace{1cm}3\hspace{2mm}+\hspace{9mm}4\hspace{2mm}+...+n\\ S=n+(n-1)+(n-2)+(n-3)+...+1\) | Add these sums together. |
| \(2S=\underbrace{(n+1)+(n+1)+(n+1)+...+(n+1)}\\ \hspace{3cm}\text{n occurences of n+1}\) | Knowing that there are n-terms present, we can simplify this slightly. |
| \(2S=n(n+1)\) | Divide by 2 on both sides to figure out the sum of consecutive whole numbers up to n. |
| \(S=\frac{n(n+1)}{2}\) | |
Now that we have generated a formula (a well-known one at that), we can proceed with this problem. I have proved this sum.
We know that the sum must be equal to 666, so let's plug that in and solve.
| \(666=\frac{n(n+1)}{2}\) | Let's eliminate the fraction immediately by multiplying by 2 on both sides. | ||
| \(1332=n(n+1)\) | Expand the right hand side of the equation. | ||
| \(1332=n^2+n\) | Move everything to one side. | ||
| \(n^2+n-1332=0\) | I will tell you beforehand that this quadratic is factorable, but it is pretty difficult to factor, so let's try using the quadratic formula instead. | ||
| \(n=\frac{-1\pm\sqrt{1^2-4(1)(-1332)}}{2(1)}\) | Now, simplify | ||
| \(n=\frac{-1\pm\sqrt{5329}}{2}\) | The radicand happens to be a perfect square. | ||
| \(n=\frac{-1\pm73}{2}\) | Now, solve for both values of n. | ||
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We reject the negative answer because we only care about whole-number solutions, and -37 is not a whole number.
What this means is that \(666=1+2+3+4...+36\), so one must add consecutively 36 positive whole numbers, starting with 1, to get a sum of 666.
