Video 1
Someone has many character. Everyone has positive side and negative side. We should look at someone from many different point of view. So we can understand the character of someone. Start look at something in our environment from many different point of view.
Video 2
A faith is very important and influence self of someone. If we believe someone and someone believe me so we can do everything better. So faith is very interest. If someone believe me we must keep the faith by doing everything which is our responsibility. So someone will believe me.
Video 3
What you know about math
There are a lot of knowledges in mathematics. Everything in our environment contains mathematics. By learn mathematics we can do everything in our life easier. In mathematics we can learn algebra, trigonometry, curve, logic, line, linier equation, etc.
Video 4
Solving deferential
We will find dy over dx equal four times x square to get the equation y.
• Dy over dx equal four times x square
• Dy equal four times x square dx
• Integral two side, so integral of dy equal integral of four times x square dx
• Integral of dy is y and integral of four times x square dx is four third times x cubeb plus Constanta.
• So y equal four third times x cubeb plus Constanta.
Video 5
Problem solving:
1. We will find the value of x in equation x minus five equal three.
• X minus five equal three
• Add two side with five, so we get x minus five plus five equal three plus five
• x equal three plus five
• so the value of x in equation x minus five equal three is eight.
2. we will find the value of a in equation seven equal four times a minus one
• seven equal four times a minus one
• add two side with one, so we get seven plus one equal four times a minus one plus one
• eight equal four times a
• multiply two side with a quarter , so we get a quarter times eight equal a quarter times four times a
• a quarter times eight is equal two, and a quarter times four times a is equal a
• so the value of a in equation seven equal four times a minus one is two
3. we will find the value of x in equation two third times x equal eight
• two third times x equal eight
• multiply two side with three second, so we get three second times two third times x equal three second times eight
• three second times two third times x is equal x, and three second times eight is equal twelve
• so the value of x in equation two third times x equal eight is twelve
4. we will find the value of x in equation five minus two times x equal three times x plus one.
• Add two side with negative three times x, so we get five minus five times x equal one
• Add two side with negative five, so we get negative five times equal negative four
• Multiply two side with negative one fifth, negative one fifth times negative five times x is equal x, and negative one fifth times negative four is equal four fifth.
• So the value of x in equation five minus two times x equal three times x plus one is four fifth.
5. We will find the value of m in equation three minus five times open bracket two times m minus five close bracket equal negative two
• three minus five times open bracket two times m minus five close bracket equal negative two
• three minus ten times m plus twenty five equal negative two
• so negative ten times m plus twenty eight equal negative two
• add two side with negative twenty eight, we get negative ten times m equal negative thirty
• divide two side with negative ten , we get negative ten times m over negative ten equal negative thirty over negative ten
• negative ten times m over negative ten is equal m and negative thirty over negative ten is equal three
• so the value of m in equation three minus five times open bracket two times m minus five close bracket equal negative two is three.
6. We will find the value of x in equation a half times x plus a quarter equal one third times x plus five forth
• Multiply two side with twelve, so twelve times a half times x plus a quarter in bracket equal one third times x plus five forth in bracket times twelve
• We get six times x plus three equal four times x plus fifteen
• Add two side with negative four times x, we get two times x plus three equal fifteen
• Add two side with negative three, so two times x equal twelve
• Divide two side with two, so two times x over two is equal x and twelve over two is equal six
• So the value of x in equation a half times x plus a quarter equal one third times x plus five forth
7. We will find the value of x in equation oh point three five times x minus oh point two equal oh point one five times x plus oh point one
• Multiply two side with one hundred, we get thirty five times x minus twenty equal fifteen times x plus ten
• Add two sides with twenty minus fifteen times x, we get twenty times x equal thirty
• Divide two side with twenty, twenty times x over twenty is equal x and thirty over twenty is equal three second
• So the value of x in equation oh point three five times x minus oh point two equal oh point one five times x plus oh point one is equal three second
Video 6
We will prove that C times logarithm base x of A equal logarithm base x of A to the C in bracket.
• C times logarithm base x of A equal logarithm base x of A to the C in bracket
• We know that if logarithm base x of A equal B, so X to the B equal A,
• So x to the B in bracket to the C equal A to the C
• X to the B times C in bracket equal A to the C, so logarithm base x of A to the C equal B times C. its second equation
• Substitution second equation in first equation, we get C times logarithm base x of A equal logarithm base x of A to the C in bracket. Its shown
We will prove that logarithm base x of A minus logarithm base x of B equal logarithm base x of A over B in bracket
• If logarithm base x of A equal m, so x to the m equal A
• If logarithm base x of B equal n, so x to the n equal B
• If Logarithm base x of A over B in bracket equal p so x to the p equal A over B
• We know that A equal x to the m and B equal x to the n, so x to the p equal x to the m over x to the n
• X to the p equal x to the m minus n in bracket, we get p equal m minus n
• So logarithm base x of A over B equal logarithm base x of A minus logarithm base x of B. its shown.
