+2446 I'm posting this solution as an alternate method to finding the vertex of a quadratic equation. Either method, presented by hecticlar or me, are acceptable methods.
Finding the vertex of a parabola is actually simple, or, at least, I think so. First, find the line of symmetry by using this formula:
\(\frac{-b}{2a}\)
However, we must identify what a and b stand for. Let's look at our quadratic function and analyze it. Here it is:
\(y=-\frac{1}{4}x^2+4x-19\)
As a review, a is the coefficient of the quadratic term, and b is the coefficient of the linear term. Let's plug it into the formula above, \(\frac{-b}{2a}\).
| \(\frac{-4}{2(-\frac{1}{4})}\) | Let's solve this expression by simplifying the denominator first. |
| \(\frac{-4}{-\frac{1}{2}}\) | I'll use a fraction rule that states that \(\frac{a}{\frac{b}{c}}=\frac{ac}{b}\). Let's apply it! |
| \(\frac{-4*2}{-1}\) | Simplify this |
| \(8\) | |
This is not our answer. The vertex is the point where either the minimum or maximum is on a parabola. The point we have found is the line that divides the parabola in half. To find the corresponding y-coordinate, substitute \(8\) into the function. Let's do that:
| \(y=-\frac{1}{4}x^2+4x-19\) | Anywhere you see an x substitute in an 8 in its place. |
| \(y=-\frac{1}{4}(8)^2+4(8)-19\) | According to order of operations do the exponent operations first |
| \(y=-\frac{1}{4}*64+4*8-19\) | Continue to simplify until you get the y-coordinate. |
| \(y=-16+4*8-19\) | Do 4*8 next, of course. |
| \(y=-16+32-19\) | |
| \(y=-3\) | |
After all of this, we have determined that the vertex is \((8,-3)\). This is your answer.
+2446 I am only showing this alternate solution because I think it is a nicer way of solving it than what the guest provided. His/her method, though, is sound and is perfectly fine.
| \([(1.08)(1+x)]^{\frac{1}{2}}-1=10\) | Add 1 on both sides |
| \([(1.08)(1+x)]^{\frac{1}{2}}=11\) | Simplify \((1.08)(1+x)\) by distributing the 1.08 to the 1 and the x. |
| \((1.08+1.08x)^{\frac{1}{2}}=11\) | Taking something to the power of 1/2 is the same as square root, so let's write it like that |
| \(\sqrt{1.08+1.08x}=11\) | Square both sides to get rid of the radical |
| \(1.08+1.08x=121\) | Subtract 1.08 on both sides |
| \(1.08x=119.92\) | Divide by 1.08 on both sides to isolate x. |
| \(x=\frac{119.92}{1.08}\) | Multiply the numerator and denominator by 100 to make them whole numbers |
| \(x=\frac{11992}{108}\div\frac{4}{4}\) | Divide the numerator and denominator by its GCF, 4, to put the improper fraction in simplest terms. |
| \(x=\frac{2998}{27}=111.\overline{037037}\) | You are done! This is your final answer! |
+2446 You can also solve this system algebraically, too, but it requires some logical thinking. You solve it similarly as that of a system of equations. Here are your inequalities:
1. \(2x+4y\leq12\)
2. \(3x-y<2\)
I'll use substitution. I'll solve for y on both equations. I'll solve for y on the inequality \(2x+4y\leq12\):
| \(2x+4y\leq12\) | Here is the original inequality. Subtract 2x first to begin isolating y. |
| \(4y\leq-2x+12\) | Divide by 4 on both sides of the equation |
| \(y\leq-\frac{1}{2}x+3\) | Finally, y is isolated |
Next, isolate y in the inequality 2, \(3x-y<2\):
| \(3x-y<2\) | This is the original inequality. Subtract 3x on both sides |
| \(-y<-3x+2\) | Divide by -1 on both sides. Remember that the inequality sign flips when dividing by a negative number! |
| \(y>3x-2\) | |
Our inequalities have changed to from their original to being solved for y:
This is where the logical thinking comes to play.If 3x-2 is less than y and \(-\frac{1}{2}x+3\) is equal to and greater than y, then \(3x-2<-\frac{1}{2}+3\)! Now that there is one variable in this inequality, we can solve for x.
| \(3x-2<-\frac{1}{2}x+3\) | This is what we concluded above. Add 2 on both sides |
| \(3x<-\frac{1}{2}x+5\) | Add (1/2)x on both sides |
| \(\frac{7}{2}x<5\) | Multiply by 2/7 to isolate x. |
| \(x<\frac{10}{7}<1\frac{3}{7}\) | |
Yet again, it requires a bit of logic again. Plug in 10/7 into both inequalities! Yes, you must plug it into both inequalities! You'll see why after the calculations are made. First, I'll plug x into the first and second equation
| \(y\leq\bf{-\frac{1}{2}*\frac{10}{7}}+3\) | Simplify the right hand side by evaluating \(-\frac{1}{2}*\frac{10}{7}\)first. |
| \(y\leq-\frac{5}{7}+3\) | Change 3 to an improper fraction so you can add the fractions together! |
| \(y\leq-\frac{5}{7}+\frac{21}{7}\) | Add the fractions now because of the common denominators! |
| \(y\leq\frac{16}{7}\) | |
Great! Let's plug in x for the second equation and see what we get:
| \(y>3(\frac{10}{7})-2\) | Multiply 3 by 10/7 first |
| \(y>\frac{30}{7}-2\) | Convert the 2 to an improper fraction |
| \(y>\frac{30}{7}-\frac{14}{7}\) | Subtract the fractions to get the solution set for the other equation! |
| \(y>\frac{16}{7}\) | |
Now that we know our x- and y-values, let's write them out as a solution set:
\(((x<\frac{10}{7},\frac{16}{7}
I guess you could stop here and move on to the next problem, but do you notice how y is really equal to every number? Currently, the solution set is saying that y is greater than 16/7, equal to 16/7 and less than 16/7. That's a verbose way of saying that y can be any of the real numbers. I'll write it out for you. Sorry, I had a hard time making the real-number symbol in LaTeX:
\((x\in{\rm I\!R}| x<\frac{16}{7},y\in{\rm I\!R})\)
You are done now because you have identified the possible values for x and y! You can verify this solution set by graphing. Luckily, Cphill has already done that! You can check it out, if you'd like.
One last note before you go!
I spent around 3 hours in total (with a slight respite in between) to generate this response. I truly attempted to provide a complete answer with comprehensible explanations. I hope it helped! Also, if anyone sees a typo, please tell me. I'm 95% confident that I missed one or two.
+2446 I'm pretty sure that you mean what does \(0.\overline{9999}\) equal? I'll use some algebra to show the real value here. This method is well-known, but here it goes anyway:
| \(0.\overline{9999}=x\) | I'm going to set this answer equal to a variable. I'll multiply 10 on both sides |
| \(9.\overline{9999}=10x\) | This is probably the trickiest step to understand. Subtract \(0.\overline{9999}\) on the left hand side and \(x\) on the right. I can do this because of the first statement I made |
| \(9=9x\) | Divide by 9 on both sides |
| \(x=1\) | |
Therefore, \(0.\overline{9999}=1=x\).
Now, I have a challenge for you.
\(...9999999=x\)
Using the same algebra I utilized, what does this equal? You should get a bizarre answer
+2446 I think I have an explanation that is simple to understand! In advance, I have not referenced any of Hecticlar's sources before writing this explanation, so mine could be similar or different. To start, I am going to make a table of powers that we can calculate:
| \(2^n\) | Written-out | Result |
| \(2^5\) | 2*2*2*2*2 | 32 |
| \(2^4\) | 2*2*2*2 | 16 |
| \(2^3\) | 2*2*2 | 8 |
| \(2^2\) | 2*2 | 4 |
| \(2^1\) | 2 | 2 |
| \(2^0\) | ? | ? |
| \(2^{-1}\) | 1/(2) | 1/2 |
| \(2^{-2}\) | 1/(2*2) | 1/4 |
| \(2^{-3}\) | 1/(2*2*2) | 1/8 |
| \(2^{-4}\) | 1/(2*2*2*2) | 1/16 |
| \(2^{-5}\) | 1/(2*2*2*2*2) | 1/32 |
Do you notice a pattern? I do. As you go down the list, you can divide by 2 to get the next number in the sequence! FIrst, I'll generalize this statement:
\(\frac{2^n}{2}=2^{n-1}\)
What I have done here is manipulate the powers so that I can circumvent raising to the power of 0. If I make n=1, I will raise to the power of zero and get a result of what that answer should be. Let's try it!
| \(\frac{2^1}{2}=2^{1-1}\) | Let's simplify the right hand side first by doing 1-1 |
| \(\frac{2^1}{2}=2^0\) | Woah! Evaluate the left hand side to figure out what 2^0 truly equals. |
| \(\frac{2}{2}=2^0\) | |
| \(1=2^0\) | |
We can generalize this further to say that any number raised to the power of zero is 1 using some algebra:
| \(1=\frac{x^n}{x^n}\hspace{1mm},x\neq0\) | This statement is true because any number divided by itself is one! I'll use an exponent rule that says that \(\frac{x^n}{x^n}=x^{n-n}\) |
| \(1=x^{n-n}\) | n-n=0, so let's simplify that |
| \(1=x^0\hspace{1mm},x\neq0\) | This is saying that any number to the power of zero is one. |
