September 2018 – Ruben's Blog

Week 4 – Precalc 11

rubeng2016 on September 30, 2018October 3, 2018

This week in math we learned how to add and subtract mixed radicals.

example: $(6\sqrt{7}-5\sqrt{5}) (6\sqrt{7}-5\sqrt{5}) + (3\sqrt{7}+4\sqrt{5})^2$

The first thing that you would do to this equation would be to FOIL it

$(6\sqrt{7}-5\sqrt{5}) (6\sqrt{7}-5\sqrt{5}) + (3\sqrt{7}+4\sqrt{5})(3\sqrt{7}+4\sqrt{5})$

After FOILing you would end up with:

$36\sqrt{49}+30\sqrt{35}-30\sqrt{35}-25\sqrt{25}+9\sqrt{49}+12\sqrt{35}+12\sqrt{35}+16\sqrt{25}$

The next step would be to simplify where possible

$252-125+63+12\sqrt{35}+12\sqrt{35}+80$

The last step is too collect like terms

$=270+24\sqrt{35}$

Week 3 – Precalc 11

rubeng2016 on September 23, 2018September 24, 2018

This week in math we learned how to convert an entire radical into a mixed radical

$\sqrt{120}=\sqrt{4}\sqrt{30}$

This was done by dividing 120 by 4 which equals 30

$=2\sqrt{30}=\sqrt{120}$

The square root of 4 is 2, so 2 squared times 30 is equal to square root 120

Week 2 – Precalc 11

rubeng2016 on September 17, 2018

This week in math we learned how to find the sum of a geometric sequence. To do this you have to use the formula:

$s=\frac{a(r^n -1)}{r-1}$

so if you were given the geometric sequence 6, 24, 96, 384… and you want to find $s_{15}$ you would do the following

$\frac{24}{6}$ which means r= 4

once you know what r is you would use the formula shown above

$s_{15}=\frac{6(4^{15}-1)}{4-1}$

$s_{15}=\frac{6(1,073,741,824-1)}{4-1}$

$s_{15}=\frac{6(1,073,741,823)}{4-1}$

$s_{15}=\frac{6,442,450,938}{4-1}$

$s_{15}=\frac{6,442,450,938}{3}$

$s_{15}=2,147,483,646$

Week 1 – My Arithmetic Sequence

rubeng2016 on September 10, 2018September 11, 2018

the first five numbers in the sequence are 6, 13, 20, 27, 34

finding the 50th term

$t_n=t_1+(n-1)d$

$t_{50} = 6 + (49)7$

$t_{50}= 6 + 343$

$t_{50} = 349$

The sum of the first 50 terms

$s_n = \frac{n}{2} (t_{1} + t_{n})$

$s_{50} = 25 (6 +349)$

$s_{50} = 25 (355)$

$s_{50} = 8,875$

Week 1 – Precalc 11

rubeng2016 on September 9, 2018

This week in math we learned about arithmetic series and sequences. One thing that we learned that I understood very well was finding terms further along the sequence using the formula, Tn = t1 + (n – 1) d, for example. if the sequence is 7,15,23… and you want to find the 25th term you would do the following

Tn = t1 + (n – 1) d

T25= 7 + (24) (8)

T25= 7 + 192

T25 = 199