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: Functions, Combinatorics 60-65, functions, combinations, probability
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- We're on problem 60.
- Given the equation y is equal to x to the n where x is
- greater than 0 and n is less than 0, which
- statement is valid?
- So let's see, x is greater than 0 and n is less than 0.
- So the exponent is negative, but the base is positive.
- So I'll just tell you right now.
- Your gut reaction might be to say, oh, if I have a negative
- exponent, maybe I somehow end up with a
- negative number somehow.
- But you have to remember, a negative exponent, it just
- tells you that it's really just the inverse of the
- positive exponent.
- So this is equal to 1 over x to the minus n.
- And if n is less than 0, then minus n is
- greater than 0, right?
- So you could say this is the same thing as 1 over x to some
- other letter m, where m is minus m, but then m would be
- greater than 0, right?
- So no matter how negative this exponent is, especially if
- this base is positive, you're going to end up with a
- positive number.
- You're going to end up with very small fractions if this
- gets large enough.
- It could be 1 over x to the hundredth power or something.
- So it could be a very small number.
- It can approach zero.
- But it will always be a positive number.
- So y is always going to be greater than 0.
- Choice A.
- Problem 61.
- If x is a real number, which best describes the value of x
- for which the inequality the square root of x
- greater than 0 is true?
- So I'm assuming we're dealing with the real numbers, and if
- I get an imaginary number here, it's actually just even
- hard to compare it in terms of magnitude to 0.
- So they're actually just asking us what the domain is.
- What x's can we put in here and get an answer where we can
- make some type of sense out of the statement?
- So obviously, I could put a positive x here.
- x could be greater than 0.
- Any x greater than 0 will have a square root.
- And all of their square roots will be greater than 0.
- I mean, you could square both sides of that equation and you
- get x is greater than 0.
- You could put x is equal to 0 there, but then the inequality
- wouldn't be true, right?
- Because the square root of 0 is equal to 0.
- So if you put x is equal to 0 there, that would work, you
- can take the square root of 0, but it just won't make the
- inequality true.
- So x can't be equal to 0.
- So x can be greater than 0.
- It can't be greater than or equal to 0.
- Not all values of x work.
- I can't put a negative number in there.
- I mean, they say no values of x.
- No, sure.
- We can put any positive number there that works.
- So the answer is A.
- Next problem.
- Let me turn the page.
- Let me copy and paste.
- Which of the following conclusions is true about the
- statement below: x squared is equal to the square root of x.
- All right, so once again, let's think
- about it a little bit.
- x squared is equal to the square root of x.
- I mean, it definitely works for the number 1 and it works
- for the number 0.
- If you square both sides of that equation, you get x to
- the fourth is equal to x.
- x could be equal to 0, but if you assume that x does not
- equal 0, you could divide both sides of this equation by x
- and you get x to the third is equal to 1.
- And the only number that I know works for that
- is the number 1.
- But anyway, the statement is always true.
- No, it's not always true.
- I put a 2 there.
- Is 2 squared equal to the square root of 2?
- No, 4 does not equal the square root of 2.
- So the statement is not always true.
- The statement is true when x is negative.
- No, x can't even be negative.
- You can't take the square root of a negative number.
- You'd end up with imaginary numbers and it
- just gets very strange.
- The statement is true when x is equal to 0.
- Sure.
- It could be 0.
- The statement is never true.
- No.
- Clearly, it's true when x is equal to 0 or x is equal to 1.
- So the answer is C.
- I didn't even have to do this stuff on the side right here.
- But, hopefully, it was instructive.
- Next problem.
- I copied it now I have pasted it.
- All right.
- Let me try with the larger pen.
- Abelardo wants to create several different 7-character
- screen names.
- He wants to use arrangements of the first three letters of
- his first name, ABE-- that's why they made his name
- Abelardo-- followed by arrangements of the 4 digits
- in 1984, the year of his birth.
- How many different screen names can he
- create in this way?
- All right, so he's got 7.
- So think about it.
- Let's just think about each slot.
- So slot number 1-- that pen's too fat.
- Slot number 1, slot number 1, spot number 3, 4, 5, 6, 7.
- So slot number 1, how many different choices
- can he put in there?
- He can put in an A, a B or an E there.
- So he has three different choices for slot number 1 I'll
- do that in a different color.
- He has three different choices for slot number 1.
- Once he picks slot number 1, how many choices does he have
- left for slot number 2?
- Well, he has two letters left out of the A, B and E, right?
- So he has only two options left.
- And once he gets to slot number 3, he has to just pick
- whatever leftover letter is there in A, B and E.
- So he only has one option.
- But then when he gets to fourth digit, now he's trying
- to pick from 1984.
- For the fourth digit, once he has four options, he can pick
- the 1, the 9, the 8 or the 4.
- He has four options.
- Then on the fifth digit, he would have picked one of the
- four, so he has three of them left.
- By the sixth digit, he only has two of them left.
- By the seventh digit, he's forced to pick whatever's
- leftover, what he hasn't picked yet.
- He has one digit left.
- So the combinations, you just multiply all of them.
- 3 times 2 is 6 times 1 is 6.
- 6 times 4 is 24.
- 24 times 3 is 72.
- 72 times 2 is 144 times 1 is 144.
- So there's 144 different screen names he can
- create in this way.
- Next problem.
- 64.
- I've copied it.
- And now I'm pasting it.
- A train is made up of a locomotive, seven different
- cars and a caboose.
- If the locomotive must be first and the caboose must be
- last, how many different ways can the train be ordered?
- So we don't even have to worry about the
- locomotive and the caboose.
- They're always going to be in the same position.
- The locomotive's first, caboose is last. So we just
- have to worry about the seven different cars, the order that
- they're in.
- It's really the same problem as the last one.
- In some ways, it's a little bit easier.
- So let's just say that's the position.
- Position 1, 2, 3, 4, 5, 6, 7.
- And let's say we just have to decide which car we put in
- each position, or how many different options do we have.
- So let's say we are filling the slots starting with
- position 1.
- Well, we have seven cars to choose from.
- The seven is different from them.
- That's not including the locomotive and the caboose, so
- we have seven cars to pick from.
- We've picked one of them.
- Then in slot 2, we have six cars to pick from.
- Slot 3, we have five cars to pick from because we've
- already picked two.
- And in slot 4, four cars to pick from,
- three cars, two cars.
- And then in slot 7, we've already picked six of the
- cars, so whatever's leftover, we have to pick.
- So let's see what this number is.
- 1 times 2 is 2.
- 2 times 3 is 6.
- 6 times 4 is 24.
- 24 times 5 is 120.
- 120 times 6 is 720.
- 720 times 7-- I'm just going to have to do a little bit
- multiplication here on the side-- 720 times 7.
- 7 times 0 is 0.
- 7 times 2 is 14.
- 7 times 7 is 49 plus 1 is 50.
- 5,040.
- So there's 5,040 different ways that the
- train can be ordered.
- Let's see.
- Next problem.
- Maybe I'll just clear all of this.
- All right.
- Let me copy the-- all right.
- Teresa and Julia are among 10 students who have applied for
- a trip to Washington, D.C.
- Two students from the group will be selected at
- random for the trip.
- What is the probability that Teresa and Julia will be the
- two students selected?
- So let's think about this.
- There are 10 students who have applied for the trip.
- Two students from the group will be selected at
- random for the trip.
- What is the probability Teresa and Julia will be the two
- students selected?
- All right, so let's think about student 1 selected and
- student 2 selected.
- What's the probability that Teresa is the first student?
- So what's the probably that we get Teresa and then we get
- Julia, that Teresa is the first student selected and
- that Julia is the second student selected?
- Well, the probability that Teresa is the first-- well,
- let me take a step back.
- There's two ways that what they want to happen could
- happen, that Teresa and Julia will be the
- two students selected.
- Either Teresa is the first student selected and then
- Julia is the second; or Julia is the first student selected
- and Teresa is the second.
- So if we figure out the probability of this happening
- and the probability of this happening, then we can just
- add those probabilities together and then we have the
- probability of either of those happening.
- All right, so let's do that.
- The probability of Teresa happening is-- let me do this
- in a different color.
- There's 10 students, so there's a probability that we
- pick Teresa first. And then once we've picked Teresa
- first, the probability that we pick Julia next, well, now
- there's 9 students left, right?
- Because you've already picked Teresa, so there's a 1 in 9
- chance you pick Julia.
- So that is equal to 1 in 90.
- So there's a 1 in 90 chance of this happening.
- There's a 1 in 90 chance of that happening.
- And then what's the chance of this happening?
- What's the chance of that happening?
- Well, it's the same thing.
- This is 1 in 10 and this is 1 in 90, right?
- And so you have a 1 in 90 chance of it happening.
- And so, what is the probability of either of these
- things happening?
- Of this combination or this combination?
- Well, you just add them together, and you get 2 out of
- 90, which is equal to 1 out of 45, which is choice A.
- And I'm out of time.
- See you in the next video.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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