Algebra I
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CA Algebra I: Number Properties and Absolute Value
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CA Algebra I: Simplifying Expressions
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CA Algebra I: Simple Logical Arguments
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CA Algebra I: Graphing Inequalities
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CA Algebra I: Slope and Y-intercept
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CA Algebra I: Systems of Inequalities
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CA Algebra I: Simplifying Expressions
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CA Algebra I: Factoring Quadratics
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CA Algebra I: Completing the Square
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CA Algebra I: Quadratic Equation
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CA Algebra I: Quadratic Roots
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CA Algebra I: Rational Expressions 1
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CA Algebra I: Rational Expressions 2
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CA Algebra I: Word Problems
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CA Algebra I: More Word Problems
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CA Algebra I: Functions
CA Algebra I: Word Problems 71-74, word problems
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- We're on problem 71.
- And they ask us, which fraction is equivalent to this
- thing, 3x over 5 divided by x plus 4 over x plus 2?
- Now the best way to do this, let's simplify the denominator
- first. So that's equal to 3x over 5.
- All of that over-- let's find a common denominator.
- The common denominator is 4.
- So if we have a common denominator of 4, x/4 is of
- course x/4.
- x/2 is the same thing as 2x over 4.
- So plus 2x over 4.
- 2x over 4 is the same thing as x/2.
- And now if you divide by a certain fraction, that's the
- same thing as multiplying buy its inverse.
- So this is going to be equal to 3x over 5 times the inverse
- of this, times 4 over x plus 2x.
- Let's see what we could do.
- Well we could factor out an x here.
- So this becomes x times 1 plus 2.
- Actually, 1 plus 2 is 3.
- I should have said x plus 2x is equal to 3x.
- I should have done.
- That's silly, I shouldn't have done that.
- I should have said x plus 2x is equal to 3x.
- My bad.
- Anyway, 3x and 3x cancel out.
- Numerator and denominator.
- And you're left with 4/5.
- And that's choice C.
- It's getting late.
- Next question.
- 72.
- Let me copy and paste this.
- Word problems, it's good to parse these together.
- All right, OK, a pharmacist mixed some 10% saline solution
- with some 15% saline solution to obtain 100 milliliters of a
- 12% saline solution.
- How much of the 10% saline solution did the pharmacist
- use in the mixture?
- OK, so 10% and 15%.
- So let T equal the amount of 10% percent.
- And let's say that F is equal to the amount of 15% solution.
- I picked T for ten, F for fifteen.
- OK, so the amount of 10% solution plus the amount of
- 15% solution, that has to equal 100 milliliters.
- I'll assume we're dealing in milliliters.
- That equals 100.
- And then, how much saline are we going to end up with?
- So, if we have the 10% saline solution, however much saline
- we have, how much saline is that going to
- contribute to the whole?
- Well it's 10% saline, so we're going to contribute 10% of
- that as saline.
- So 10% times the amount of 10% solution tells me how much
- saline I'm contributing.
- Plus 15% times the amount of the 15% saline solution I'm
- contributing.
- This is the amount of saline from that solution.
- And that equals what?
- Now this is interesting.
- You end up with a 12% saline solution.
- 100 milliliters of a 12% saline solution.
- So how much saline total am I going to have when I mix it
- all together?
- I had 100 milliliters.
- It's a 12% saline solution.
- So that means 12% of this 100 milliliters is saline.
- So that equals 12.
- There's 12 milliliters of saline.
- With this much contributed from the 10% solution and that
- much contributed from the 15% solution.
- Let's see what we can do.
- So now we have two linear equations and two unknowns.
- We're ready to solve.
- What do they want to know?
- They want to know how much of the 10% saline solution.
- So the best thing, we want to cancel out the F's somehow and
- solve for T.
- So let's multiply this top equation times minus 0.15,
- because I want to cancel this out with this.
- So let's do that.
- If you multiply times minus 0.15, we end up with minus
- 0.15T minus 0.15F is equal to minus 15.
- So we add these two equations, 0.1 plus minus 0.15 that's
- minus 0.05T.
- These two cancel out.
- It's equal to 12 minus 15 is minus 3.
- OK, we can multiply both sides by negative 1, you get plus
- 0.05 times T is equal to 3.
- Sorry, I'm getting messy here.
- So if you divide both sides by 0.05, you have T is equal to 3
- divided by 0.05.
- 0.05, that's the same thing is 1/20.
- So this is equal to 3 times 20, which is equal to 60.
- So we contributed 60 milliliters of the 10% percent
- saline solution.
- So T is equal to 60 and that's choice A.
- And this is a little bit confusing, because you're like
- what variable do I use for the 10% solution and how do I set
- up a second equation?
- And a good place to start is to say, oh well they want to
- know how much of the 10% saline solution did the
- pharmacist use, so let me set that is a variable.
- Let me set another variable for the
- other type of solution.
- If I add them together I have to get 100 milliliters.
- And the tricky part was just probably figuring out how much
- saline did each of the solutions contribute?
- And then when you add that up, it equals 12
- milliliters of saline.
- Anyway, next question.
- 73.
- Andy's average driving speed for a 4 hour trip was
- 45 miles per hour.
- During the first 3 hours he drove 40 miles per hour.
- What was his average speed for the last hour of his trip?
- OK, so distance is equal to rate times time.
- OK, so his total for a 4 hour trip was 45 miles per hour.
- Or you could say rate is equal to distance divided by time.
- Just divide both sides of that by time.
- So they said, the average driving speed for a 4 hour
- trip was 45 miles per hour.
- So 45 miles per hour is equal to the distance he traveled
- divided by his time.
- Divided by 4 hours.
- So we can figure out the distance right off the bat.
- So let's actually do that.
- The distance is equal to 4 times 45.
- That's what?
- It's equal to 180 miles.
- Fair enough.
- They say during the first 3 hours, he drove
- 40 miles per hour.
- What was his average speed for the last hour of his trip?
- So we just have to figure out how much ground
- did he have to cover?
- So, during the first 3 hours, he drove 40 per hour.
- So how far did he go?
- So that's a different distance, that's just the
- first 3 hours.
- So the distance over the first 3 hours is equal to the rate
- of the first 3 hours, it's equal to 40, times the time,
- times 3 hours.
- So his distance over the first 3 hours is equal to 120 miles.
- So how far did he have to travel in that last hour?
- Well, in the first 3 hours he went 120 miles.
- He went a total of 180 miles.
- So his distance in that last hour is going to
- be 180 minus 120.
- He had to go 60 miles in that last hour.
- So how fast did he go?
- Well he went 60 miles in that last hour, so he went
- 60 miles per hour.
- Choice B.
- So this might seem like a complicated problem, and it is
- a little tricky.
- But it's all about distance equals rate times time.
- And then recognizing that you could use this first statement
- to figure out the total distance he traveled.
- You could use the second statement to figure out how
- far he traveled in the first 3 hours.
- And then you know how far he had to travel in that last
- hour, or essentially how fast he had to travel.
- Next question.
- 74.
- One pipe can fill a tank in 20 minutes.
- So let's say pipe 1 its rate is equal to
- 1 tank per 20 minutes.
- That's how I like to write it.
- That's what they told us.
- One pipe can fill a tank in 20 minutes, while another takes
- 30 minutes to fill the same tank.
- So pipe 2, is equal to 1 tank per 30 minutes.
- That's its rate.
- Or 1/30 tank per minute.
- This is 1/20 tank per minute.
- Fair enough.
- How long would it take the two pipes together to fill it?
- So their combined rate is just both of these.
- If you have both pipes going in, you're going to have 1/20
- tank per minute plus 1/60 tank per minute.
- Let's see.
- 1/20 plus 1/60, common denominator is 60.
- 1/20 is 3/60 plus 1/60 is 1/60 tanks per minute.
- It's just adding fractions.
- So that equals 4/60 tanks per minute.
- 4/60, that's 1/15 tanks per minute.
- Or you view this as equal to one tank per 15 minutes.
- So how long does it take to fill one tank?
- 15 minutes.
- Choice is C.
- Oh I'm all out of time.
- I'll see 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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