Math Is Fun Forum
You are not logged in.
- Topics: Active | Unanswered
#151 Re: Jokes » Star Wars joke » 2016-02-03 18:55:47
[list=*]
[*]
[/*]
[/list]
#152 Re: Help Me ! » Quadratic Equations » 2016-02-03 18:42:36
Monox's solution is wrong; that solution is only for the case when the roots are real and equal.
2. Find the set of values of k for which the roots of the equation x²+kx-k+3=0 are real and of the same sign.
For real roots we require
[list=*]
[*]
[/list]
For the roots to be of the same sign, their product must be positive. So we want
[list=*]
[*]
[/list]
Hence the set of values we want is
[list=*]
[*]
[/list]
#153 Re: Puzzles and Games » Arrange Mathematical Operation Using 0-9 to Match the Post Number » 2016-02-03 11:40:42
#154 Re: Help Me ! » Password lock » 2016-02-03 03:02:49
Sigh.
Okay, for people who still don't get it...let's just suppose the password is 133.
Person A comes along and tries to open the lock. They don't know the password but they assume the first digit is 3. So they go through all the combinations with fist digit 3: 311, 312, 313, 321, 322, 323, 331, 332, 333. The last one works. If they had tried it earlier they'd have opened the lock sooner. But these are the only combinations with first digit 3, so it is impossible to fail to open the lock after 9 tries (unless you're forgetful and repeat an already tried combination).
Person B comes along and thinks they can do better than A by assuming the first digit is 2. They go 211, 212, 213, 221, 222, 223, 231, 232, 233 – and find that they can't do better than the first person. The only way they can do it faster is by being a luckier guesser with the other two digits – but even without luck, the lock can still be opened after 9 tries at the very most.
Along comes person C and says, "I'll take the first digit to be 1." But this is just a fluke: they just happen to guess the first digit correctly. This way they can open the lock after at most four wrong tries (111, 112, 121, 122); the next one is bound to open the lock. So it is at most 5 steps for this lucky person.
This shows that by fixing the first digit and trying combinations for the other two, you can open the lock after no more than 9 steps. You can try any of 1, 2, 3 as the fixed first digit. If you're lucky and are correct with this digit, you should be able to open the lock after no more than 5 steps, but even without luck, and your guess for the first digit is wrong, you still shouldn't take more than 9 steps to open the lock.
#155 Re: Help Me ! » 2 trig problems » 2016-02-02 22:28:40
Also it seems the LaTeX package in the forum software doesn't understand the \begin{align*} command. Instead of
\begin{align*}
x &= \cos(4t),\\
y &= \sin(6t).
\end{align*}try \begin{array}:
\begin{array}{rcl}
x &=& \cos(4t),\\
y &=& \sin(6t).
\end{array}[list=*]
[*]
[/list]
The software doesn't understand \implies or \iff either, forcing the use of \Rightarrow and \Leftrightarrow instead.
#156 Re: Help Me ! » Password lock » 2016-02-02 21:56:23
I may have misunderstood something here.
Yes.
Are you saying that if the correct combination is 123 (with wild's position unknown), and you entered 111, 112, 131, 132, 211, 212, 213, 221, 222, 231, 232, 233, 311, 312, 313, 321, 322, 331, 332, and 333 (none of which will open the lock, as I understand it) before using one of the seven successful combinations you listed in post #5, that your method still holds?
No.
#157 Re: Help Me ! » Linear Equations involving floor » 2016-02-02 20:40:07
[list=*]
[*]
[/list]
#158 Re: Help Me ! » Password lock » 2016-02-02 20:23:59
Read the thread!
#159 Re: Help Me ! » Password lock » 2016-02-02 19:41:46
That's the same minimum number as in my method (post #2).
#160 Re: Help Me ! » Polynomial Factorization and Remainder/Factor Theorem » 2016-02-02 19:20:05
Also, to clarify post #2:
#161 Re: Help Me ! » 2 trig problems » 2016-02-02 18:37:42
You copied and pasted the word math from my other post. Type out the word yourself on your keyboard, don't copy and paste!
#162 Re: Help Me ! » Polynomial Factorization and Remainder/Factor Theorem » 2016-02-02 18:33:30
Please give detailed answer and don't skip any step.
If this is a homework question, you should do it yourself.
#163 Re: Help Me ! » Some couple of questions » 2016-01-31 21:21:16
Just question three [3]
the answer is at the back of the book and is 62 or 122 but need the steps
I had to work backwards from the answers to figure out the mistake in your post.
(3) the sum of the first 3 terms of a G.P is 14. If the first term is 2, find the possible values of the sum of the first 5 terms
If you seriously want help, you have to make an effort yourself. How is anyone to help you if you don't bother to post your question correctly? I was able to work out your typo because I happened to have time to do that. You must bear in mind that most people are busy and have things to do in life, so they won't be as blessed with free time as I am!
Anyway.
Now you carry on from there.
#164 Re: Help Me ! » Some couple of questions » 2016-01-31 08:16:53
Please-- how about the others?
How about you putting in some of your own effort?
#165 Re: Help Me ! » Password lock » 2016-01-31 07:26:42
In other words, if the password is 123, then all of the following will open the lock, right?
[list=*]
[*]123, 223, 323, 113, 133, 121, 122[/*]
[/list]
My question was whether the wildcard could be in any position, or whether it was in a definite (but unknown) position. In the former case, the lock will open for any of the above seven combinations; in the latter case, then, for example, if the wildcard is in the centre, only 113, 123 and 133 will open the lock.
#166 Re: Help Me ! » Password lock » 2016-01-31 03:24:59
For example, if the password is 123, the lock opens when you try 1*3, or *23, or 12*.
Do you mean any one of 1*3, *23, 12* can open the lock, or just one of them will?
If any one of them works,
#167 Re: Help Me ! » Some couple of questions » 2016-01-31 03:06:04
(1)
Substitute from [1] and solve.
#168 Re: Help Me ! » 2 trig problems » 2016-01-29 20:31:04
Hi Hate Number Theory.
Please use [math]...[/math] tags, not dollar signs.
Points
, , and are on the circumference of a circle with radius 2 such that and . Find the area of .In convex quadrilateral
and The perimeter of is Find
#170 Re: Help Me ! » Urgent system of linear equations » 2016-01-29 07:14:22
In 4 years' time his son will be x+4 years old, not x+36 years old.
#171 Re: Help Me ! » Urgent system of linear equations » 2016-01-29 06:07:07
y=x+32
y+4=5(x+4)
#172 Re: Help Me ! » 3 Trig Problems » 2016-01-26 20:18:50
1. In , we have , , and . Find the sum of all possible values of .2. In , we have , , and . Find the sum of all possible values of .3. Let be positive real numbers. Prove that[list=*]
[*][/*]
[/list]Under what conditions does equality occur? That is, for what values of , , and are the two sides equal?
3. Rewrite
[list=*]
[*]
[/*]
[/list]
Let
[list=*]
[*]
[/list]
and consider the complex numbers
[list=*]
[*]
[/list]
Then the given inequality follows from the triangle inequality for complex numbers:
[list=*]
[*]
[/list]
Equality occurs if and only if the two complex numbers and the origin are collinear with the origin not in between, i.e. if and only if
[list=*]
[*]
[/list]
#173 Re: Help Me ! » Geometry question - rectangle » 2016-01-26 06:45:36
You might want to try this additional problem:
#174 Re: Help Me ! » Geometry question - rectangle » 2016-01-25 18:39:17
What is the formula to use to find the area of a rectangle using the diagonals, not the length and breadth?
You can't determine the area from just the diagonals alone. Two different rectangles may have different areas but the same diagonal length. For example, a rectangle with sides 3 and 4 and one with sides 2 and √21 both have diagonals of length 5 but their areas are clearly different.
Are the diagonals of a rectangle always the same length?
Yes.
You can however determine the area of a square from its diagonal. Maybe this is what you are thinking of?
#175 Re: Help Me ! » Critical Numbers-Help! » 2016-01-22 21:58:52
The last question should be "Maybe".