Algebra II
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California Standards Test: Algebra II
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California Standards Test: Algebra II (Graphing Inequalities
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CA Standards: Algebra II (Algebraic Division/Multiplication)
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CA Standards: Algebra II
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Algebra II: Simplifying Polynomials
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Algebra II: Imaginary and Complex Numbers
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Algebra II: Complex numbers and conjugates
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Algebra II: Quadratics and Shifts
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Examples: Graphing and interpreting quadratics
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Algebra ||: Conic Sections
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Algebra II: Circles and Logarithms
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Algebra II: Logarithms Exponential Growth
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Algebra II: Logarithms and more
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Algebra II: Functions, Combinatorics
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Algebra II: binomial Expansion and Combinatorics
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Algebra II: Binomial Expansions, Geometric Series Sum
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Algebra II: Functions and Probability
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Algebra II: Probability and Statistics
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Algebra II: Mean and Standard Deviation
Algebra II: Logarithms Exponential Growth 49-52, logarithms, exponential growth and decay
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- We're on problem 49.
- Let me copy and paste it into the window.
- I've copied it and now let me paste it.
- There you go, problem 49.
- Let me pick a good highlighter color, magenta is good.
- A student showed the following steps in his solution to the
- equation below, but his answer was not correct.
- Then at the bottom they say, in which step did he make his
- first error?
- So let's see, log base 5 of 2x squared, minus 3x, plus 1,
- minus log base 5 of x, minus 1, plus log base 5 of 125 is
- equal to 6.
- Fair enough.
- All right, now it looks like in this first step, he kept
- this the same.
- You can't mess up there.
- Then he said log base 5 of 125 is 3.
- Yeah, that's right.
- 5 to the third power is equal to 125.
- So that is right as well.
- Let's see, here it looks like he took the 2x squared, minus
- 3x plus 1, and he factored them into 2x minus
- 1 and x minus 1.
- Let's see if that's right.
- 2x times 2x is 2x squared.
- 2x times minus 1 is minus 2x.
- Minus 1 times x is minus 3x.
- And you have a minus 1 times a minus 1, you get a plus 1.
- So you get 2x squared, minus x, minus 2x, so
- minus 3x, plus 1.
- So this is right.
- So so far so good.
- What does he do next?
- All right, in the next step he keeps this
- first term the same.
- 2x minus 1, x minus 1.
- Fair enough.
- He keeps this term the same.
- He just subtracts 3 from both sides.
- He subtracts 3 from both sides and this becomes 0 and then
- this becomes 3.
- Fair enough.
- What does he do in the next step?
- This is interesting.
- He essentially said, if I am subtracting 2 logarithms with
- the same base, that's the equivalent of
- dividing the two things.
- So he kind of did a quick step here.
- So his logic here is that log-- I don't even have to
- write the base 5-- but log of 2x, minus 1, times x minus 1,
- minus log of base 5, times x minus 1, that that would be
- equal log of 2x minus 1, times x minus 1, all of that
- over x minus 1.
- That seems fair enough.
- So that should cancel out.
- Oh, but that's where he made his mistake.
- What you end up with left over when you cancel out the 2x
- minus 1's, isn't one of the minus 1's.
- You're left with log of 2x minus 1.
- This x minus 1 and this x minus 1 cancel out.
- He should be left with just the 2x
- minus 1 in the logarithm.
- But instead, somehow he got rid of the 2x minus 1 and
- ended up with just an x minus 1 here.
- So step 3 is where he made his first mistake.
- I don't know.
- He somehow intuitively knew that you could divide x minus
- 1 into this, because of log of A minus log of B is equal to
- log of A over B.
- But I don't know, he simplified wrong or something.
- Who knows?
- Next problem.
- All right, let me copy and paste this one too.
- I'm starting to like this notion of copying and pasting.
- I don't have to rewrite the problem.
- And I can mark it up and do all sorts of things with it.
- All right, a certain radioactive element decays
- over time, according to the equation y is equal to A times
- 1/2 to the t divided by 300th power.
- Where A equals the number of grams present initially, t
- equals time in years.
- If 1,000 grams were present initially, how many grams will
- remain after 900 years?
- So we get y is equal to-- 1,000 grams were present
- initially, so that's A.
- So 1,000 times 1/2 to the-- and they say, what is time?
- How many grams will remain after 900 years?
- t is time in years.
- So 1/2 to the 900 divided by 300.
- I just substituted A and t into the
- equations they gave us.
- 1,000 is A, and 900 is t.
- So let's just simplify this thing.
- So we get this is equal to 1,000 times 1/2 to the--
- what's 900 divided by 300?
- It's just 3.
- So that equals 1,000.
- What's 1/2 to the third power?
- It's 1/8.
- It's 1/2 times 1/2 times 1/2.
- So we get 1,000 divided by 8, which is 125.
- 125 grams are left in 900 years.
- That is choice C.
- Next problem.
- All right, let me see if I have to copy
- and paste this one.
- Yeah, sure, they have a chart, even better.
- I've copied it, and there we go.
- Look at that.
- All right.
- Bacteria in a culture are growing exponentially with
- time as shown in the table below.
- Which of the following equations expresses the number
- of bacteria y present at any time t.
- OK, so growing exponentially tells us that if I say, the
- amount of bacteria after any days, so let's say y would be
- the amount of bacteria, exponential form tends to be
- you'd have your initial amount-- like A in that
- previous example-- times some number that is kind of the
- exponent factor, I guess you could call it.
- The last one it was 1/2.
- When you do continuous compound interest, it'll be e.
- It'll be how fast something is either growing or decaying.
- I don't know, that'll be x.
- I'm just making up a letter.
- It's usually x to some power t that's usually a function of
- time, some constant times time.
- This is the general form of any kind of
- exponential growth or decay.
- You might see different letters than that, but that's
- the general idea.
- So anyway, if we look at all of these choices, well, if you
- just look at the thing, you're seeing it's
- doubling every day.
- The amount of bacteria is doubling every day.
- So that tells us that's going to be some power of 2.
- So y is going to be-- the amount of bacteria after any
- day is going to be equal to some initial amount.
- Actually, we could just think about it, right?
- We have some initial amount times 2 to some constant times
- the day's power.
- But I think if we just take 2 to the day's
- power-- let's see.
- At 0 days you have a 100 bacteria.
- So y is equal to 100-- I think this gets you there; and
- instead of writing days, let's write time--
- times 2 to the time.
- Sometimes there's a constant here, but this is
- straightforward enough.
- When t is equal to 0, this 2 to the t becomes 1, and you
- have 100 here.
- When t is 1, then this becomes 2.
- 2 times 100 is 200.
- You could say this is t and this is y.
- When t is 0, y is 100.
- When t is 1, y is 200.
- When t is 2, y is 400.
- So it is choice B.
- Probably the easiest way is just to look at these.
- And say, wow, there's only one-- well, two of these
- choices, well, three of them have kind of
- an exponential form.
- We're just like, oh, well, this doesn't take into account
- the initial condition.
- At day 0 I have 100 bacteria.
- This one, I don't know, it says at day 0, I have 200
- bacteria, which is just wrong.
- I have 100.
- You take t equals 0, and you're left with what
- you have at day 0.
- So depending on how you look at it, you
- get to B either way.
- Next question.
- I'm running out of space.
- Oh, there you go.
- All right, next question.
- They write, I'll write the whole thing here.
- If the equation y to the 2x is graphed, which of the
- following values of x would produce a point
- closest to the x-axis?
- So if you think about it, let's just think about how
- that graph would even look.
- Let me draw it here actually.
- I'll draw it in a darker color.
- I can do it right on top of the white.
- Let's just get an intuition.
- What's 2 to the 0?
- 2 to the 0 is 1, right?
- Its y-intercept would be 1 at x equals 0.
- 2 squared, then 2 to the 2 power.
- And you were looking at 4, we'll go there.
- Sorry, let's say if this is 1, this would be 2, 2 to the
- first power.
- Then 2 to the second power would be 4.
- Then 2 to the third power would be 8.
- Actually, 4 would be like.
- So it's going to look something like that.
- We don't have to worry.
- It's going to go up really fast.
- Now what happens as you go into the negative powers?
- So 2 to the negative 1 power is 1/2.
- You're going to get right there.
- 2 to the negative 2 is 1/4.
- So you're just going to get smaller and it's just going to
- asymptote to 0.
- So that's what the graph is going to look like.
- I didn't even have to worry about negatives, because there
- are no negative numbers here.
- So really, the smallest of these are going to produce a
- point closest to the x-axis.
- The smallest of these values, you want to
- get as small as possible.
- Let's see, 3/4 is bigger than 1/4.
- So that's not right.
- 5/3 is bigger than 1/4.
- 8/3 is bigger than 1/4.
- So the answer is 1/4, A.
- 2 to the 1/4 power is going to be a little bit more than 1.
- It's going to be something like that.
- Anyway, next question.
- Oh, I'm all out of time.
- See you in the next video.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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