Wednesday, October 28, 2009
last work period before bi-weekly test #3
in today's class we were given a work period to tie p any lose ends and concerns we may have about tomorrows test. today class i should have used my time more wisely but i didn't, which sucks.
Tuesday, October 27, 2009
more double angle identity examples.
today we did more double angle identities. i learned that when your asked to prove an identity, it means make the left side look like the right side. yaaaaaaay. :)
Monday, October 26, 2009
Work period, Hurray
today was a work period in which i spent the whole class doing accelerated math and i mastered 2 objectives. this weeks bi-weekly test has been moved to Thursday.
Thursday, October 22, 2009
more trig identities double angles
today in class we learned how to use the double angle identities that are on our formula sheet. and the best part was that it does not even seem to hard.
exercise 17, 1-12 and exercise 18, 1-18
exercise 17, 1-12 and exercise 18, 1-18
Wednesday, October 21, 2009
more trig identities (Sum / Difference & such)
today we learned how to find the exact value of 7 pi/12 and similar questions by using sum/difference identities and the best part is i understand it :)
exercise 16 1-15 except 3 and 5 because you'll get :(
exercise 16 1-15 except 3 and 5 because you'll get :(
Tuesday, October 20, 2009
more trig identities
trig identities are like puzzles and at the moment these puzzles have me curfuffled. when I'm in class watching the examples, i get it , but when i go home and try to do the work i get messed up and can't get it.
Monday, October 19, 2009
Friday, October 16, 2009
Trig Identities
Identity-an equality (in this case, trigonometric) that evaluates as true for any value of input; that is, both sides of the trig equations are true for ALL possible variables.
*NOTE: trig equations that are Not identities are conditional equations.
*NOTE: graphing is not adequate to prove "identity-ness", graphing is definitely enough to disprove identity status.
exercise 14 questions 1-11
*NOTE: trig equations that are Not identities are conditional equations.
*NOTE: graphing is not adequate to prove "identity-ness", graphing is definitely enough to disprove identity status.
exercise 14 questions 1-11
Thursday, October 15, 2009
WORK PERIOD!!!
Today we were given a work period because of the number of students missing.
we were assigned 1-11 on exercise 12, 1-10 on 13 due monday
we were assigned 1-11 on exercise 12, 1-10 on 13 due monday
Yesterdays math test #2
Yesterday was yet another math test.
the test consisted of questions from exercises 1 through 12.
have a good day
the test consisted of questions from exercises 1 through 12.
have a good day
Tuesday, October 13, 2009
Bi-Weekly test #2 prep
Good Afternoon, today we prepared for tomorrows test.
we also had a homework check for lessons 1-10
We also worked on accelerated math.
good luck on tomorrows test every one :)
we also had a homework check for lessons 1-10
We also worked on accelerated math.
good luck on tomorrows test every one :)
Thursday, October 8, 2009
absolute value functions (more transformations)
in todays class we did some more examples.
first we graphed f(x)=x^2-1
then f(x)= abs(x^2-1)
then f(x)=1/abs(x^2-1)
and these were caled peice-wise functions
I kinda get this stuff
first we graphed f(x)=x^2-1
then f(x)= abs(x^2-1)
then f(x)=1/abs(x^2-1)
and these were caled peice-wise functions
I kinda get this stuff
Wednesday, October 7, 2009
reciprocal functions
Rational expressions
f(x) becomes 1/f(x)
the x intercept in f(x) is an asymotpoe in 1/f(x)
i kind of understode todays lesson
f(x) becomes 1/f(x)
the x intercept in f(x) is an asymotpoe in 1/f(x)
i kind of understode todays lesson
Friday, October 2, 2009
transformations (flips, Slides)
in todays class we learned how to shift a function either left or right, or up and down. lets say y=x^2 is our reference, to shift it 3 units left it would be y=(x+3)^2 and right would be y=(x-3)^2.
moving y=x^2 5 units up would be y=x^2+5 and down would be y=x^2-5
moving y=x^2 5 units up would be y=x^2+5 and down would be y=x^2-5
yesterdays math test
Yesterday we had a math test on circular functions. i thought i would do well but i dont think i did.
the test was harder than i thought it would be and i obviously did not properly prepare.
the test was harder than i thought it would be and i obviously did not properly prepare.
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