Reflection Of Videos

by silfia yulianis 09.50 0 komentar

1^st video : Dead Poets Society

This video tells us about look at something in a different way. We shouldn't look something in a common way because we can't be creative. We must look our own way.

2^nd video : Believe

In this video, there is a boy. That boy said on the stage. First, he asked audiences, "Do you believe in me?" And the audiences answered,"Yes!!". Then, the little boy replied, "Because I believe in me." This video tell us to believe in ourselves. Because we can do anything, say anything, think anything, and become anything if we can believe in ourselves.

3^rd video : What You Know about Math

This video's contents is a song about mathematics. This song is funny because the singers (two men with rapper style) give us a question that is what you know about math. Then, the singer tell us little about mathematics. I just can say that this song is very interest.

4^th video : Solving Differential Equation

Let dy over dx is equal to four times x square. To find variable y, first, we can multiply both sides by dx. So, we get dy is equal to four times x square times dx. Now, we must integrate both sides. We get integral dy is equal to integral four times x square times dx. Now, we can solve it. We get that y is equal to three fourth times x cube plus c. c is constant.

5^th video : Solving Linear Equation with One Variable

1. Let x – 5 = 3, Find the value of x!

To get the value of x, we can add both sides with five. It will be x minus five plus five is equal to three plus five. So we get x is equal to eight.

2. Let 7 = 4a – 1, Find the value of a!

We can add both sides with one. It will be seven plus one is equal to four a minus one plus one. We get eight is equal to four a. Then, we divided both sides by four. So, we get two is equal to a.

3. Let 2/3 x = 8, Find the value of x!

To get the value of x, we can multiply both sides by three second. It becomes three second times two third x is equal to eight times three second. So, we get the value of x is eight times three second, that is twelve.

4. Let 5 – 2x = 3x + 1, Find the value of x!

First, we can subtract both sides by three x. It becomes five minus two x minus three x is equal to three x plus one minus three x. We get five minus five x is equal to one. Then, we can subtract both sides by five. That equation becomes negative five x is equal to negative four. Now, we can divide both sides by negative five. So, we get x is equal to four fifth.

5. Let 3 – 5(2m – 5) = -2, find the value of m!

First, we should multiply negative five by two m minus five. The equation becomes three minus ten m plus twenty five is equal to negative two. We get twenty eight minus ten m is equal to negative two. Then, we can add both sides by negative twenty eight. We get negative ten m is equal to negative thirty. Last, we can divide both sides by negative ten. So we get m is equal to three.

6. Let ½ x + ¼ = 1/3 x + 5/4, find the value of x!

First, we should subtract both sides by a quarter. It becomes a half x is equal to one third x plus one. Then, we can subtract both sides by one third x. We get one sixth x is equal to one. Last, we can multiply both sides by six. So we get x is equal to six.

7. Let 0,35 x - 0,2 = 0,15 x + 0,1, find the value of x!

First, we can add both sides by negative nought point one. We get nought point three five x minus nought point three is equal to nought point one five x. Then, we add both sides by negative nought point one five x and we get the equation is nought point two x minus nought point three is equal to nought. Then, we add both sides by nought point three. We get nought point two x is equal to nought point three. Last, we can divide both sides by nought point two. And we get x is equals to one point five

6^th video : Proof Log Base x of A is Equal to Log Base x Minus Log Base x of B

Let say that logarithm base x of A is equal to B (log_x A = B). We can say that it is same to x to the B is equal to A (x^B =A). If we multiply log base x of A with C, we get C times log base x of A is equal to B times C (C log_x A = BC).

Go back to x to the B equal to A. If we rise x to the B to the power of C, we get x to the B to C power is equal to A to the C. We can say that x to the BC is equal to A to the C (x^BC = A^C).

Then, write x to the BC is equal to A to the C in logarithm expression. So, it becomes logarithm base x of A to the C is equal to BC (log_x A^C = BC).

Now, we can see that C times logarithm base x of A is equal to logarithm base x of A to the C (C log_x A = log_x A^C).

We know that C times logarithm base x of A is equal to logarithm of A to the C and we just learn that logarithm base x of A plus logarithm base x of B is equal to logarithm base x of A times B (log_x A + log_x B = log_x AB).

What happens if we change the addition with a subtraction?

Let say that:

1. Log base x of A is equal to l, it says that x to the l is equal A.

2. Log base x of B is equal to m, it says that x to the m is equal B

3. Log base x of A over B is equal to n, it says that x to the n is equal to A over B.

Now, we change A over B with x to the l over x to the m. We can write that x to the l over x to the m is equal to x to the l times x to the negative m or that also equal to x to the l minus m. We can see that x to the n is equal to x to the l minus m. So, n is equal to l minus m.

Then, log base x of A over B is equal to l minus m. That also equal to log base x of A minus log base x of B (because log base x of A is equal to l and log base x of B is equal to m). So, log base x of A over B is equal to log base x of A minus log base x of B.