mathgarage
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Most men and women, by birth or nature, lack the means to advance in wealth and power, but all have the ability to advance in knowledge. Get more understanding about math from this blog.
mathgarage
3w ago
The DSE (click here to see what HKDSE is) is close. The math core exam is less than 2 weeks time. I am going to share a really nice equation to find the tangent of a given point on the circle. Suppose that the general form of a circle is given by
then the equation of a tangent line at point will be as follows:
Proof:
Differentiate the equation of the circle with respect to x implicitly.
hence y’ is,
then the slope of the tangent at point would be
By point slope form of a line,
Now, since the given point satisfies the circle, we plug into the circle,
Hence, we have the following ..read more
mathgarage
1y ago
I was browsing the net couple days ago and saw this integral. At first it looks quite difficult. I have tried numerous different substitutions and they don’t work. And later I tried using complex number and it seems to work. So, here is my solution.
Solution:
By Euler’s formula, cosine function is the real part of a complex number.
Hence,
Re-arrange the exponents we’ll get
Now, since
We can use integration by parts,
By using the double angle identity of sine and cosine ..read more
mathgarage
1y ago
The actual mass of a packet of cheese is [215 g, 225 g), hence the total mass of 250 packets is within the interval [53750 g, 56250 g) or [53.75 kg, 56.25 kg). 53.6 kg correct to the nearest 0.1 kg lies outside the interval of [53.75 kg, 56.25 kg). Therefore the claim is false.
(a)
and
take the intersection of these two intervals,
(b)
-4.5 < x < 3 implies negative integer x could be -4, -3, … , -1, a total of 4 negative integers.
Let F be the number of female passengers and M be the number of male passengers. The first sentence implies
Now, 24 female passengers leave ..read more
mathgarage
1y ago
This is problem 23 of AMC 10B in 2011. So, here is the problem:
I will present two solution here. The first one uses Chinese Remainder Theorem, which is slow. And the second solution I will use the trick of using Binomial Theorem.
SLOW method:
By Chinese Remainder Theorem, we have
We have .
We need to find
Since , then
So,
To find ,
Next, we will find ,
Since , then
So,
Next, we are going to find
Therefore,
Therefore, the hundreds digit is 6.
FAST Method:
By Binomial Theorem,
Therefore the hundreds digit is 6 ..read more
mathgarage
2y ago
Saw this limit when I was watching a Korean drama, Melancholia. The drama is about a romance love (and math) story between a math prodigy and his math teacher. In the drama, the school principal and one of the math teachers had been helping out the daughter of a politician cheating in the math tests and exams so she can always be the top of the class. In one of the exam, the main actress (one of the math teachers) has modified one of the questions due to the error or typo in one question. She didn’t notify her colleagues she modified one of the questions and of course her colleague later gave ..read more
mathgarage
2y ago
Solution
In general, let
By Pythagoras Identity,
Solve this system by adding and subtracting, we get
Hence,
and
Therefore,
Let , then ..read more
mathgarage
2y ago
So here is the definite integral:
And here is the solution:
The trick here is to use Euler’s formula and the rest is kinda straight forward.
In general, for any natural number n, we have the following,
If , then
So, the integral vanishes unless or .
If n = 2k, then
Since the integral vanishes when n is odd, we can rewrite it as the following,
For example, when n = 1011, then ..read more
mathgarage
2y ago
The way of solving this problem is to recognize that this is a math olympiad type problem and it probably has an elegant solution. This equation is nice. There is a symmetry in the coefficients and this is a nice hint for us to crack this problem. My idea is coming from squaring the number 111.
But 1, 2, 3, 2, 1 is not exactly 1, 7, 14, 7, 1. However I know I am getting there. If I play around with the numbers a bit, it is not difficult to get the following.
So, here it is. The hard part is solved! The rest is trivial ..read more