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
California Standards Test: Algebra II 1-5, absolute value equations and systems of equations
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- So we're on problem number one on the California Standards
- Test Problems for Algebra II.
- And so the theory being is if you're able to do all or at
- least understand all 80 of these questions, I think you
- have a pretty good understanding of Algebra II,
- at least from the perspective of California.
- Let's see, problem number one.
- What is the complete solution to the equation the absolute
- value of 3 minus 6x is equal to 15?
- So let's think about it.
- 3 minus 6x could clearly be 15, because the absolute value
- of 15 is 15.
- So we could say that 3 -- sorry that's a 3.
- 3 minus 6x could be equal to 15.
- But 3 minus 6x could also be equal to minus 15, right?
- Because the absolute value of minus 15 is also 15.
- So let's write that down.
- Or 3 minus 6x is equal to minus 15, right?
- If this was equal to minus 15, you'd take the absolute value,
- you'd get positive 15.
- So either of these could be true.
- So let's solve for x.
- Subtract 3 from both sides, you get minus
- 6x is equal to 12.
- Divide both sides by minus 6, you get x is equal to minus 2.
- And so here, if we subtract 3 from both sides, you get minus
- 6x is equal to minus 15 minus 3.
- That's minus 18.
- Divide both sides my minus 6, you get x is equal
- to positive 3, right?
- A negative divided by a negative , so x is equal to 3.
- So x could be minus 2 or 3, and that's choice b.
- Problem two.
- Let me switch colors to make things interesting.
- And this is a similar type of problem.
- What are the possible values of x in the absolute value of
- 12 minus 4x is equal to 2?
- Same logic.
- 12 minus 4x could be 2, right?
- The absolute value of 2 is 2.
- So you you get 12 minus 4x is equal to 2.
- Or 12 minus 4x could be minus 2, right?
- Because the absolute value of minus 2 is 2.
- So then you get 12 minus 4x could be equal to minus 2.
- So this is an or proposition.
- So let's see, subtract 12 from both sides, you get minus 4x.
- 2 minus 12 is minus 10.
- Then you get x.
- Divide both sides by minus 4 is minus 10 over minus 4.
- The negatives cancel out.
- And that's equal to 5/2, or 2 and 1/2, for 2.5.
- And the only reason I wrote 2.5 is I was looking at the
- choices, and they have everything as decimals.
- So let's look at the other possibilities.
- Subtract 12 from both sides, you get minus 4x is equal to
- minus 2 minus 12 is minus 14.
- x is equal to minus 14 over minus 4.
- Negatives cancel out, which is equal to 7/2, which is equal
- to 3 and 1/2, which is equal to 3.5.
- So x could be 2.5 or 3.5, and that is choice d.
- Problem three.
- I'll switch to maybe green.
- Problem three.
- For weddings, Sherada bought several dozen roses and
- several dozen carnations.
- The roses cost $15 per dozen, so roses are $15 per dozen,
- the carnations cost $8 per dozen, so the carnations are
- $8 per dozen.
- Sherada bought a total of 17 dozen flowers and
- paid a total $192.
- How many roses did she buy?
- All right, so let's set some variables.
- Let R equal dozens of roses.
- I hope you can read my handwriting.
- And let C equal to dozens of carnations.
- All right, so they say that she bought a
- total of 17 dozen flowers.
- So the dozens of roses she bought plus the dozens of
- carnations she bought must be equal to 17 because she bought
- a total of 17 dozens.
- It also tells us that she spent $192.
- So how do we figure out how much money she spent?
- So if she bought R dozens of roses, how much did
- she spend on roses?
- Well, each dozen of roses is $15.
- So she spent 15 times the number of dozens.
- That's how much she spent on roses.
- How much did she spend on carnations?
- Well, she bought C dozens of carnations.
- Each dozen was $8.
- She spent 8C dollars on carnations.
- So her total amount of money she spent was the amount she
- spent on roses plus the amount she spent on carnations, and
- that is equal to 192.
- Now, we have two linear equations and two unknowns.
- We can just solve them.
- We want to solve for the number of roses.
- That was the question.
- So let's try to cancel out the number of carnations.
- So if we multiply this top equation by minus 8, we'll get
- a minus 8C.
- That'll cancel out with that, so let's do that.
- So let's multiply that top equation by minus 8.
- You get minus 8R minus 8C.
- What's minus 8 times 17?
- Let's see, 17-- I'll do it in the corner here-- 17 times 8,
- 7 times 8 is 56, 1 times 8 is 13, plus 5 is 1, 1 times-- oh,
- my brain was getting ahead of me.
- 1 times 8 is 8, plus 5 is 13.
- Sorry.
- It's equal to 136.
- Oh, and we would multiply it by negative 8, so we'd put a
- negative there.
- And now we can add these two equations.
- 15 minus 8, 7R.
- 8C minus 8C is 0-- and that was the whole point behind
- multiplying this top equation by minus 8-- is equal
- to 192 minus 36.
- Let's see.
- We can do little borrowing, or regrouping, they call it now.
- That's 12, that's 8, so it equals 56.
- 7R is equal to 56, so R is equal to 8.
- So Sherada bought 8 dozens of roses, and that is choice C.
- Problem four.
- What is the solution to the system of
- equations shown below?
- And there's three of them with three variables, always the
- hairiest problem.
- Let's see.
- We have 2x minus y plus 3z is equal to 8.
- We have x minus 6y minus z is equal to 0.
- And then you get minus 6x plus 3y minus 9z is equal to 24.
- Now there's a bunch of ways to solve these.
- You could use matrices and inverses of matrices and
- Gauss-Jordan elimination, or you can just kind of use your
- traditional multiplying equations by each other and
- adding them so that you can kind of constrain them or take
- out variables.
- And we'll do the last one because I don't want to
- introduce any other kind of matrix manipulation here.
- So what I want to do is, let's use this equation with both of
- these, and try to cancel out the x's, right?
- Well, let's think of how we could do that.
- Well, one thing I like to do, just to simplify it, is to get
- all of the equations in what I consider lowest integer from.
- I know that's not a formal term.
- But look at all the terms in this equation.
- They're all divisible by 3, right?
- So let's divide all of them by 3.
- So this last equation would become-- if you divide
- everything by 3, you'd get minus 2x plus y minus 3z is
- equal to 8.
- So let's cancel that out.
- I just divided it by 3.
- It's the same thing.
- Remember, whatever you do to one side of an equation you
- can do to the other side of the equation.
- Anyway, so let's use this equation and this equation and
- then this equation and this equation to get two equations
- with just y's and z's.
- So essentially we want to cancel out the x's.
- So to cancel out the x's with this one and this one, we can
- just add them, right?
- Because 2x plus minus 2x, that's going to
- cancel out the x's.
- So we can just do that.
- And then when we do it with this equation, we could
- multiply this equation times minus 2 first and it'll cancel
- out the x's there.
- So let's do it.
- So let's just write it and I'll write it in green.
- So if I add this equation to this equation, 2x minus 2x is
- 0x minus y plus y, that's 0y.
- These are all zero, but I'm just writing them down to show
- that we're done.
- Well, they all cancel out.
- Plus 0z.
- These are equal to 8 plus 8 is equal to 16.
- So you get 0 is equal to 16, which makes no sense.
- So there's actually no solution here.
- And if you wanted to think about visually what's
- happening is each of these equations represents a plane
- in three dimensions.
- And the only way you get no solution or no intersection
- between this equation and this equation or this equation,
- which are essentially the same equation, the only reason why
- you can't get the intersection between them is that they must
- be parallel equations.
- You could get an infinite number of solutions if they
- were actually the same plane.
- But these are parallel planes, so they'll never intersect.
- So we saw.
- And it kind of makes sense because this part-- well, I'm
- not going to go into the details.
- We're just focused on solving problems right now.
- But it's fair enough to say that when you add this
- equation to this equation to cancel out the x's, everything
- else cancels out and you got a nonsensical statement, 0
- equals 16, so there's no solution and that's choice c.
- And those are parallel planes in three dimensions.
- Problem five.
- A restaurant manager bought 20 packages of bagels.
- Some packages contained 6 bagels each and the rest
- contain 12 bagels each.
- There were a 168 bagels in all.
- How many packets of 12 bagels did the manager buy?
- So he bought some packages of 6.
- So the packages of 6 plus the packages of 12, that's just my
- random notation that I invented.
- He bought a total of 20 packages, so that equals 20,
- and he bought a total of 168 bagels in all.
- So how many bagels did he get from the 6-packs?
- Well, each of them had 6 bagels, right?
- So he got 6 times the number of packs.
- So he got 6 times this.
- This is the number of bagels he got from the 6-packs, and
- then this is the number bagels he got from
- the 12-packs, right?
- The number of packs times 12, and then that is equal to 168.
- Two equations, two unknowns.
- They're linear equations.
- We should be able to solve this.
- So they want to know how many packages of the 12
- bagels did he buy?
- So they want to know what p sub 12, or whatever you want
- to call that is.
- So if we want to do that, let's cancel out the P6's.
- So if we multiply that top equation by minus 6, you get
- minus 6 times the 6-packs, minus 6 times the 12-packs--
- I'm just multiplying this top thing by minus 6-- is
- equal to minus 120.
- Now, we can add the two equations.
- These cancel out.
- That was the whole point.
- 12 times the 12-packs minus 6 times the 12-packs, that's
- equal to 6 times the 12-packs.
- It's equal to 168 minus 120 is 48.
- And so we get the 12-packs, there were 48
- divided by 6 is 8.
- So there were 8 12-packs.
- The manager bought 8 12-packs.
- That is choice b.
- And I'll see you in the next video because
- I'm all out of time.
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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