Last meeting, Mr. Marsigit showed us many videos. There were 6 videos and 4 them are related to Mathematics. Then, we’re asked to retell the contents of the videos with our own words. it is very interesting. I hope I can develop my English again. Here, I will write my story about the videos.
1. Video I
Be Different
The title of the video is Death Poet Society. In this video, there was a teacher who talk to his students in a class. He talked about what should we do as human. Shakespeare said that someone has many characters. And the characters can make he or she look at something in different point of view. So, he told that we must see something from different side. We must try to be different. Don’t be afraid to find new ground to increase our self.
2. Video II
If we believe others, we will be trusted easily. As human, we must believe that we have ability to build our self. With trust, we can create many things because we believe that we can. No matter where we come from, if we can trust someone sincerely, we will be loved by others. so, start to believe others with believing our self first.
3. Video III
What you know about math? Math is very complicated. There are a lot of kinds of math. Math is known because of calculator. If we see someone use calculator, we assume that he or she are doing something related to math. Some pieces of math are trigonometry, logarithm, exponents, limit, etc. The most popular in math is Mathematics League. Many students are attracted to join math competition. If we want to success in Mathematics, we must do math works with our self. We should not cheat others work.
4. Video IV
In this video, we can learn how to solve problems related to integral and differential.
Problem: We must get y(dependent variable) from the equation dy over dx equal to four times x square.
Solution:
· First, we move dx to right side of the equation. We multiply both sides with dx. So, we find
dx times dy over dx equal to four times x square times dx.
· Then, the equation becomes dy equal to four times x square times dx
· Then integrate the equation above. So we will find integral from dy equal to integral from four times x square times dx.
· The result is y equal to four third times x cube plus arbitrary constant.
5. Video V
A. Problem: Find the value of x from x-5=3(x minus five equal to three)
Solution: We will find the value of x from the equation x-5=3
· Add two sides by 5.
x-5+5=3+5 (x minus five plus five equal to three plus five)
x=8 (x equal to eight)
So the value of x from the equation above is eight.
B. Problem: Let 2/3 times x=8(two third times x equal to eight). Find the value of x
Solution: We will find the value of x from 2/3 times x=8
· Multiply both sides by 3/2
3/2 times 2/3 times x= 8 times 3/2 (three second times two third times x equal to eight times three second)
x=12 (x equal to twelve)
So the value of x is twelve.
C. Problem: Let 5-2x=3x+1(five minus two times x equal to three times x plus one). What is x?
Solution: We will find the value of x from the equation 5-2x=3x+1
· Subtract both sides by 3x
5-2x-3x=3x+1-3x (five minus two times x minus three times x equal to three times x plus one minus three times x)
5-5x=1 (five minus five times x equal to one)
· Then, subtract both sides by 5
5-5x-5=1-5 (five minus five times x equal to one minus five)
-5x=-4 (negative five times x equal to negative four)
· Multiply both sides by -1/5
-1/5. 5x=-4.-1/5 (negative one fifth times five times x equal to negative four times negative one fifth)
x=4/5 (x equal to four fifth)
So, the value of x is four fifth.
D. Problem: Let 3-5(2m-5)=-2. Find the value of m!
Solution: We will find the value of m from 3-5(2m-5)=-2 (3 minus 5 times open bracket 2 times m minus 5 close bracket equal to negative two)
· Multiply 5 by the number in the bracket, 2m-5
3-5(2m-5)=-2
3-10m+25=-2 (three minus ten times m plus twenty five equal to negative two)
28-10m=-2 (twenty eight minus ten times m equal to negative two)
· Then, subtract both sides by 28
28-10m-28=-2-28 (twenty eight minus ten times m minus twenty eight equal to negative two minus twenty eight)
-10m=-30 (negative ten times m equal to negative thirthy)
· Multiply both sides by -1/10
-1/10.-10m=-30.- 1/10 (negative one tenth times negative ten times m equal to negative thirty times negative one tenth)
m=3 (m equal to three)
So, the value of m is 3
E. Problem: Find the value of x from the equation x/2+1/4=x/3+5/4(x times a half plus a quarter equal to x times one third plus five fourth)!
Solution: We will find the value of x from the equation above
· Multiply this equation by the smallest union multiplication of 2,3, and 4. It is 12.
12(x/2+1/4)=12(x/3+5/4) (twelve times open bracket x times a half plus a quarter close bracket equal to twelve times open bracket x times one third plus five fourth close bracket)
6x+3=4x+15 (six times x plus three equal to four times x plus fifteen)
· Then, subtract both sides by 6x
6x+3-6x=4x+15-6x (six times x plus three minus six times x equal to four times x plus fifteen minus six times x)
3=-2x+15 (three equal to negative two times x plus fifteen)
· Multiply both sides by -1/2
-1/2.3=-1/2(-2x+15) (negative a half times three equal to negative a half open bracket negative two times x plus fifteen close bracket)
-3/2=x-15/2 (negative three second equal to x minus fifteen second)
x=-3/2+15/2 (x equal to negative three second plus fifteen second)
x=12/2 (x equal to twelve second)
x=6 (x equal to six)
So, the value of x is six.
F. Problem: Find the value of x from the equation 0,35x-0,2=0,15x+0,1(oh point three five times x minus oh point two equal to oh point one five times x plus oh point one)
Solution: We will find the value of x from the equation above
· Multiply both sides by 100/one hundred
100(0,35x-0,2)=100(0,15x+0,1) (one hundred times open bracket oh point three five times x minus oh point two close bracket equal to one hundred times open bracket oh point one five times x plus oh point one close bracket)
35x-20=15x+10 (thirty five times x minus twenty equal to fifteen times x plus ten)
· Collect the numbers which have x variable in one side
35x-15x=10+20 (thirty five times x minus fifteen times x equal to ten plus twenty)
20x=30 (twenty times x equal to thirty)
· Multiply both sides by 1/20
20x.1/20=30.1/20 (twenty times x times one twentieth equal to thirty times one twentieth)
x=30/20 (x equal to thirty twentieth)
x=3/2 (x equal to three second)
So, The value of x is 3/2 (three second).
6. Video VI
a) If logarithm base x of A equal to B, so x to the power or B equal to A.
Ø If we multiply logarithm base x of A equal to B by C, we will find
C times logarithm base x of A equal to B times C.
Ø Then, we make quadrant the equation x to the power of B equal to A by C, we will find
x to the power of open bracket B times C close bracket equal to A to the power of C.
So, logarithm base x of A to the power of C equal to B times C.
Ø From two equations above, we can conclude that C times logarithm base x of A equal to logarithm base x of A to the power of C.
b) Logarithm base x of A plus logarithm base x of B equal to log base x of A times C.
We will show the truth of that equation.
Ø Log base x of A equal to L. So, x to the power of L equal to A.
Ø Log base x of B equal to M. So, x to the power of M equal to B.
Ø Log base x of open bracket A over B close bracket equal to N, so we find that x to the power of N equal to A over B.
Ø Then, based from the first and second statement, we can conclude that x to the power of N equal to x to the power of L over x to the power of M.
Ø So, x to the power of N equal to x to the power L minus M.
Ø Then we can see that N equal to L minus M.
Ø So, we will find that log base x of A over B equal to L minus M.
Ø And finally log base x of A over B equal to log base x of A minus log base x of B is proven.
