Monthly Archives: September 2018

week4, Pre-cal11

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this week, I learnt the new knowledge, the way to adding & subtracting the radical expression.

There are some rules in there

  1. the radicals can just be adding or subtracting between the like terms, such as,                     the 2 \sqrt{2} – 3\sqrt{2} = –\sqrt{2}. and it works when, \sqrt{2}+\sqrt{8}=3\sqrt{2},                                                                    because the \sqrt{8}=2\sqrt{2}   therefore, the first step for the adding or subtracting’s question is always to check, is there any term could be further simplify to be the same term as others obviously.

2. if the question is seem like that \sqrt{2} + \sqrt{5}, the terms cannot be further simplified because \sqrt{2} is the unlike terms with \sqrt{5}. people cannot just do the math under the radical symbol, as to do the math as 2+5=7 , and use these methods to determine the answer is \sqrt{7}.

3. there are always the track in there, such as \sqrt{2} +\sqrt[3]{2}=?   be careful with the whole radicals, not only the radicands, the index also determines two radicals are the like terms or not. thus, the questions can not have a simplified sum, it just stop there.

the radicals just could be + or – between the like terms, (same radicands and index)

COULD you do the question?

Ex: –\sqrt[3]{2}+\sqrt[3]{8}+\sqrt[3]{32}–\sqrt{4} +\sqrt{16}

 

answer:

  1. = –\sqrt[3]{2}+2 +2\sqrt[3]{3} -2 +4
  2. =-\sqrt[3]{2}+2\sqrt[3]{3}+4

 

 

Week-3 Precal 11

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  1. The absolute value of a number is the distance from zero, so both 6, or -6 are have the same absolute value, |6|=|-6|, and |x| >=0 whatever the number of Xis, positive or negative.

2. the square root and cube root,

\sqrt{x}, the result must x equal or bigger than 0, because there are no way that a factor times itself become a negative product, ex: -2latex \times{(-2)}$=4,  and 2\times{2}=4.

the cube root \sqrt[3]{x}, the x could be positive or negative, because the -2\times{(-2)} \times{(-2)}=-8

and 2\times{2} \times{2}=8.

3. the rational number and irrational number,

rational, include :whole number, fraction, natural, the decimal which can  turn to be fraction that repeating and terminating.

The decimal form of an irrational number is a non-repeating and non-terminating decimal number

 

 

 

Week 2-Pre-cal 11

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Infinite geometric series,

last week, I learnt how to calculate the series of number with same d. This week, I learnt the infinite geometric series,  is also about how to determine a long series of numbers, which continues multiplying the same ratio.

finite series, for ex: S10, S50… with very specific number

infinite series, for ex: 2, 6, 18…….with no specific number to arrive the stop, it will keep going.

S∞ is the sum of series of numbers what is infinity,

there are two situation , 1 of it is diverging, when the r >1, r<-1 that the S∞ would be no sum, just getting bigger.

another one if called converges, when 0<r<1 and -1<r<0. that the S∞ could determine by the formula, S∞=\frac{1}{1-r}   the a is the 1st term, and the r is the common ration.

( the r could be any non-zero number for geometric series)

t1=a, t2=tr, t3=tr^2,  t4=tr^3….so on,

for example, there are first term of the geo series, you have to determine the 5th term.  t1=1. t2=3, t3=9,

t5=t1\times{r^4}   =  1\times{3^4} =   1\times{81}=81

 

 

Week1-Precale 11

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Arithmetic Series

It’s  the sum of the terms in an arithmetic sequence with a definite number of terms.

for Example, the arithmetic sequence is : 1, 7, 13, 19, ….. 1 ,t1; 7, t2; 13, t3; 19, t4; etc, (ps: t=term)

then the Arithmetic Series will  be

S1= t1=1

S2=t1+t2=1+7=8

S3=t1+t2+t3=1+7+13=21

S4  1+7+13+19=40

you can do the simple math if there are just few number on the sequence, but if there are a lot, lie 30+ more term, there are two formula can be use, which can save a lot time and much easier.

Sn=n(t1+tn)/2,    or Sn=\frac{n[2t1+d(n-1)}{2}   (ps, the n is the number of the term, d is the difference between each term,)

1 . use the formula 1, the S6 for the up questions, we determine the t6 is =31， through d=6, then t6=t4+2d=19+2(6)= 31

the S6= 6(1+31)/2 = 96

use the 2nd formula , S6=6[2+6(5)]/2=98