# Week 4 – Precalc 11

Radical expression: the term a $\sqrt[b]{c}$

Radical: a root sign (represented by √)

Coefficient: the number outside the radical, it is being multiplied by the radical

When adding or subtracting radical expressions, the radicals (the radicand and the root) must be like terms. Some radicals can be simplified to achieve like terms. The coefficients are then added or subtracted, and the radical remains the same.

Example: $\sqrt{2}$ + $\sqrt{8}$

= $\sqrt{2}$ + $\sqrt{4}$ $\sqrt{2}$

= $\sqrt{2}$ + 2 $\sqrt{2}$

= 3 $\sqrt{2}$

# Week 3 – Precalc 11

This week in Precalculus 11, we started the Absolute Value and Radicals unit. We first learned about the absolute value of a real number.

Absolute value of a real number: its distance from zero on a number line/the principal square root of its square

Principal square root: the positive (+) square root

Example: $\sqrt{1}$ = ±1     +1 = principal square root     -1 = negative (-) square root

The symbols “||” represent absolute value.

Example: <-0-1-2-3-4-5->     |0| = 0

|0|

= $\sqrt{0^2}$

= 0

This is a new concept I had not yet learned about and can add to my knowledge of Pre-Calculus.

# Week 2 – Precalc 11

This week in Precalculus 11, we learned about geometric sequences.

Geometric sequence: a list of terms in which each term is multiplied by a common ratio to equal the next term

Term: a number (represented by “ $t_n$“, “n” being the term’s place in the geometric sequence)

Common ratio: a common number multiplied by each term to equal the next term (represented by “r”)

Example: 1, 2, 4, 8, 16     r = 2

To find the value of a term in a geometric sequence, we use the formula $t_n$ = $t_1$ · $r^{n - 1}$.

Example: Find $t_{10}$: 1, 2, 4, 8, 16 $t_{10}$ = ? $t_1$ = 1     r = 2     n = 10 $t_n$ = $t_1$ · $r^{n - 1}$

( $t_{10}$) = (1) $(2)^{(10) - 1}$ $t_{10}$ = 1 · $2^9$ $t_{10}$ = 1 · 512 $t_{10}$ = 512

This method of finding the value of terms in an geometric sequence is faster than extending the list.

# 6 Kingdoms

Eubacteria

Bacillus anthracis https://en.m.wikipedia.org/wiki/Bacillus_anthracis

Escherichia coli https://en.m.wikipedia.org/wiki/Escherichia_coli

Bacillus anthracis and Escherichia coli are members of the Eubacteria Kingdom because they have prokaryotic cells, are unicellular, and their cells have a cell wall of peptidoglycan.

Archaebacteria

Halobacterium salinarum http://rachelyscientist2.blogspot.com/2008/02/archaebacteria-halobacterium-salinarum.html

Sulfolobus acidocaldarius https://microbewiki.kenyon.edu/index.php/Sulfolobus_acidocaldarius

Halobacterium salinarium and Sulfolobus acidocaldarius are members of the Archaebacteria Kingdom because they have prokaryotic cells, are unicellular, and their cells have a cell wall containing uncommon lipids.

Protista

Aegagropila linneai https://en.m.wikipedia.org/wiki/Marimo

Undaria pinnatifida https://en.m.wikipedia.org/wiki/Wakame

The Aegagropila linnaei and the Undaria pinnatifida are members of the Protista Kingdom because they have eukaryotic cells, and their cell wall is cellulose.

Fungi

Hericium erinaceus https://en.m.wikipedia.org/wiki/Hericium_erinaceus

Hydnellum peckii

https://en.m.wikipedia.org/wiki/Hydnellum_peckii

The Hericium erinaceus and the Hydnellum peckii are members of the Fungi Kingdom because they have eukaryotic cells, are multicellular, are heterotrophs, and their cell walls are chitin.

Plantae

Hydnora africana https://en.m.wikipedia.org/wiki/Hydnora_africana

Rafflesia arnoldii https://en.m.wikipedia.org/wiki/Rafflesia

The Hydnora africana and the Rafflesia arnoldii are members of the Plantae Kingdom because they have eukaryotic cells, are multicellular, are autotrophs, and their cell walls are cellulose.

Animalia

Hydropotes inermis https://en.m.wikipedia.org/wiki/Water_deer https://en.m.wikipedia.org/wiki/Purple_frog

The Hydropotes inermis and the Nasikabatrachus sahyadrensis are members of the Animalia Kingdom because they have eukaryotic cells, are multicellular, are heterotrophs, and have no cell wall.

# Week 1 – My Arithmetic Sequence

13, 26, 39, 52, 65… $t_n$ = $t_1$ + d(n – 1)

( $t_{50}$) = (13) + (13)[(50) – 1] $t_{50}$ = 13 + 13 · 49 $t_{50}$ = 13 + 637 $t_{50}$ = 650 $S_n$ = $\frac{n}{2}$( $t_1$ + $t_n$)

( $S_{50}$) = $\frac{(50)}{2}$[(13) + (650)] $S_{50}$ = 25 · 663 $S_{50}$ = 16575

# Week 1 – Precalc 11

This week in Pre-Calculus 11, we started the Sequences and Series unit. We first learned about arithmetic sequences.

Arithmetic sequence: a list of terms in which a common difference is added to each term to equal the next term

Term: a number (represented by “ $t_n$“, “n” being the terms place in the arithmetic sequence)

Common difference: a common number added to each term to equal the next term (represented by “d”)

Example: 0, 1, 2, 3, 4     d = +1

t₁  t₂  t₃  t₄  t₅

To find the value of a term in an arithmetic sequence, we use the formula $t_n$ = t₁ + d(n – 1).

Example: Find t₁₀: 0, 1, 2, 3, 4 $t_{10}$ = ? $t_1$ = 0     d = +1     n = 10 $t_n$ = t₁ + d(n – 1)

(t₁₀) = (0) + (1)[(10) – 1]

(t₁₀) = 0 + 1 · 9

(t₁₀) = 0 + 9

(t₁₀) = 9

This method of finding the values of terms in an arithmetic sequence is faster than extending the list.