SAT Prep: Test 5 Section 7 Part 3 Video

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SAT Prep: Test 5 Section 7 Part 3

SAT Prep: Test 5 Section 7 Part 3 We’re on problem number 10. Problem 10 If 2x – 2 x 2– x = 0, what are all possible values of x? If I multiply two numbers—if I multiply two expressions equal to zero that means one or both of these expressions are going to be equal to zero. And if you solved for roots of a quadratic using factor you’ll be familiar with this notion. So either 2x – 2 is zero and/or actually either of these could be equal to zero or both of them could be equal to zero or both of them could be equal to zero. So if 2x ―let’s just solve this. So add two to both sides, 2x is equal to 2, x is equal to one. In here, if we just add x to both sides we get 2 is equal to x or x is equal to 2. So x could be one or x could be 2. So choice D, one and two only and you could try them out. If x is one this becomes 0 times 1 which is of course zero. If x is 2 this becomes 2, right? 2 times 2 minus 2 is 2 times – which is zero. And zero doesn’t work. That would be in your temptation. But if you put zero in here you'll get minus 2 because this term will go away. You get minus 2 times 2 which is minus 4 so that doesn’t work. You just have to set each of these expressions to zero. Next question, Problem 11 If x3 = y9, what is x in terms of y? So what we can do here is we just want to get rid of this cube power. So how do we get rid of the cube power? Well, we could take the cube root of both sides. If that doesn’t make sense to you I’ll show you how. X3—let me give this some breathing rooms we go to the y9 I’m just writing the same thing a little bit bigger. But we do the one side of the equation. We can do the other side of the equation. If I want this exponent to be one, I just have to take it to the 1/3 power. How do I know that? Well, if I take something to an exponent and then take it to another exponent I essentially just multiply the two exponents, that’s an exponent rule and you can review that in the exponent videos. If you multiply 3 1/3 you get one so that’s why I’m raising it to the 1/3 power. And that of course is the same thing as taking the cube root. But of course if I do something to one side of the equation, I have to do it to both sides of this equation. If I think the cube root of one side for the equation the whole I have to take the cube root of the other side. So x3 to the 1/3 well that’s just x1 or just x because 3 times 1/3 is 1 so x1 is just x and that equals 1. Y9 times 1/3. But what‘s 9 times 1/3 is equal to divided 3 which is equal to 2. So x is equal is y3 power, that’s just 3. And that is choice C. Next problem—I will switch colors for variety and there’s something that I have to draw. And they’re asking us, in the x-y coordinate system above, which of the following line segments has a slope of negative 1. So what is the slope of negative 1? A slope of negative 1 means that as we move to the right one, we move down one, that’s the slope of negative 1 and if you’re familiar with slopes a lot just intuitively you know that it looks something like this. Slope of negative looks something like that. And if we look at the lines that they drew, all the choices, the ones that are moving up are definitely not our answer. This one is moving up, so this one can’t be the answer. This one moving up so this can’t be the answer. This one is moving so this can’t be answer so all of these lines, the yellow ones that I haven’t scratched up, all of this have negative slopes. And we just to figure out which one could be negative 1. If we look at the slope here between point O and point A, they mark off that this is 1, 2, 3 and this is 1. So choice for this line, we went over 1, we move over 1 but we went down 3. So this line has a slope of negative 3. And similarly, they draw—this goes out 1, 2, 3, this line we have to go 3 for to go—we have to move in the x direction positive 3 for it to move down to 1. So this has a slope of 1/3 and you can kind of tell that it goes down very gradually. The rise is negative and the run is 3, the run is just half right moving the x direction. This is—the rise is negative 1 which is right there and the run is 3. So just—so this is the answer. So just by deductive reasoning we know that this is probably going to be our answer. And it also looks like it has a slope of negative 1. And if you would actually look at it points D and C, this is about—this looks like the point if I just look at by inspection 1, 3 and this looks like 3, 1 right here. So what is the change in y over the change in x? Well, the change in y is 3 minus 1 so that equals 2 and the change in x is 1 minus 3 which is minus 2. Remember, when you could calculate slope you always have to use the first—if I use 3 the first time with the numerator I have to use 1 the first time with the denominator. So 3 minus 1over 1 minus 3 is the slope and it’s minus 1. But if you’re really good you should just able to look at it and say, well, that’s the closest thing to negative 1 because it’s not too steep and not to shallow. Next problem—so that is choice D, C, E. Problem 13 Kyle’s lock combination consists of 3, 2 digit numbers. The combination satisfies the three conditions below. One number is odd, one number is a multiple of 5 and then one month is the day of the month of Kyle’s birthday so one number day of month. So what do we know about that number? Well, it has to be—is it two digit number so that’s to be between 01 and the 31in order to be a birthday can be. If each of the numbers satisfied exactly one of the conditions, which one of the following could be the exact combination to the lock? So, one number can’t satisfy two of these conditions and another number that has base nine. Each of them has to do one of them. So let’s look at choices. Choice A is 14, 20, and 13. If—let’s see which could satisfy the odd. Well, this number is odd, so let me if this condition is satisfied by this number, it’s odd. One number is multiple of 5, so this condition—well, that’s satisfied by this number and one number is a day of the month, so this looks pretty straight forward, one number is a day of the month well, 14 could be a day of a month. And in general, in any SAT problem, I’m seeing this pattern more and more as I go through of these practice test is when they have a problem where they want you to go through every choice, the answers tend to be one of the first two choices because they don’t want make you to waste a lot of time. So you should feel pretty comfortable if you do this and you got choice A. And if you want to see the example of what work, I’m guessing B won’t work. Choice B says, 14, 25, 13, I think that doesn’t look like quite satisfying. If each number satisfies exactly one of the conditions, which of the following could be the combination to the lock, exactly, so why can’t this one be because this conditions that one number is odd. It can only be satisfied by one of the choice. Oh I’m running over, I’ll see you the next video.

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SAT Prep: Test 5 Section 7 Part 3

We’re on problem number 10.

Problem 10

If 2x – 2 x 2– x = 0, what are all possible values of x?

If I multiply two numbers—if I multiply two expressions equal to zero that means one or both of these expressions are going to be equal to zero. And if you solved for roots of a quadratic using factor you’ll be familiar with this notion. So either 2x – 2 is zero and/or actually either of these could be equal to zero or both of them could be equal to zero or both of them could be equal to zero. So if 2x ―let’s just solve this. So add two to both sides, 2x is equal to 2, x is equal to one.

In here, if we just add x to both sides we get 2 is equal to x or x is equal to 2. So x could be one or x could be 2. So choice D, one and two only and you could try them out. If x is one this becomes 0 times 1 which is of course zero. If x is 2 this becomes 2, right? 2 times 2 minus 2 is 2 times – which is zero. And zero doesn’t work. That would be in your temptation. But if you put zero in here you'll get minus 2 because this term will go away. You get minus 2 times 2 which is minus 4 so that doesn’t work. You just have to set each of these expressions to zero.

Next question, Problem 11

If x3 = y9, what is x in terms of y?

So what we can do here is we just want to get rid of this cube power. So how do we get rid of the cube power? Well, we could take the cube root of both sides. If that doesn’t make sense to you I’ll show you how. X3—let me give this some breathing rooms we go to the y9 I’m just writing the same thing a little bit bigger. But we do the one side of the equation. We can do the other side of the equation.

If I want this exponent to be one, I just have to take it to the 1/3 power. How do I know that? Well, if I take something to an exponent and then take it to another exponent I essentially just multiply the two exponents, that’s an exponent rule and you can review that in the exponent videos.

If you multiply 3 1/3 you get one so that’s why I’m raising it to the 1/3 power. And that of course is the same thing as taking the cube root. But of course if I do something to one side of the equation, I have to do it to both sides of this equation. If I think the cube root of one side for the equation the whole I have to take the cube root of the other side. So x3 to the 1/3 well that’s just x1 or just x because 3 times 1/3 is 1 so x1 is just x and that equals 1. Y9 times 1/3. But what‘s 9 times 1/3 is equal to divided 3 which is equal to 2. So x is equal is y3 power, that’s just 3. And that is choice C.

Next problem—I will switch colors for variety and there’s something that I have to draw. And they’re asking us, in the x-y coordinate system above, which of the following line segments has a slope of negative 1. So what is the slope of negative 1?

A slope of negative 1 means that as we move to the right one, we move down one, that’s the slope of negative 1 and if you’re familiar with slopes a lot just intuitively you know that it looks something like this. Slope of negative looks something like that. And if we look at the lines that they drew, all the choices, the ones that are moving up are definitely not our answer.

This one is moving up, so this one can’t be the answer. This one moving up so this can’t be the answer. This one is moving so this can’t be answer so all of these lines, the yellow ones that I haven’t scratched up, all of this have negative slopes. And we just to figure out which one could be negative 1. If we look at the slope here between point O and point A, they mark off that this is 1, 2, 3 and this is 1.

So choice for this line, we went over 1, we move over 1 but we went down 3. So this line has a slope of negative 3. And similarly, they draw—this goes out 1, 2, 3, this line we have to go 3 for to go—we have to move in the x direction positive 3 for it to move down to 1. So this has a slope of 1/3 and you can kind of tell that it goes down very gradually. The rise is negative and the run is 3, the run is just half right moving the x direction.

This is—the rise is negative 1 which is right there and the run is 3. So just—so this is the answer. So just by deductive reasoning we know that this is probably going to be our answer. And it also looks like it has a slope of negative 1. And if you would actually look at it points D and C, this is about—this looks like the point if I just look at by inspection 1, 3 and this looks like 3, 1 right here. So what is the change in y over the change in x?

Well, the change in y is 3 minus 1 so that equals 2 and the change in x is 1 minus 3 which is minus 2. Remember, when you could calculate slope you always have to use the first—if I use 3 the first time with the numerator I have to use 1 the first time with the denominator. So 3 minus 1over 1 minus 3 is the slope and it’s minus 1. But if you’re really good you should just able to look at it and say, well, that’s the closest thing to negative 1 because it’s not too steep and not to shallow.

Next problem—so that is choice D, C, E.

Problem 13

Kyle’s lock combination consists of 3, 2 digit numbers. The combination satisfies the three conditions below.

One number is odd, one number is a multiple of 5 and then one month is the day of the month of Kyle’s birthday so one number day of month. So what do we know about that number? Well, it has to be—is it two digit number so that’s to be between 01 and the 31in order to be a birthday can be.

If each of the numbers satisfied exactly one of the conditions, which one of the following could be the exact combination to the lock? So, one number can’t satisfy two of these conditions and another number that has base nine. Each of them has to do one of them. So let’s look at choices.

Choice A is 14, 20, and 13. If—let’s see which could satisfy the odd. Well, this number is odd, so let me if this condition is satisfied by this number, it’s odd. One number is multiple of 5, so this condition—well, that’s satisfied by this number and one number is a day of the month, so this looks pretty straight forward, one number is a day of the month well, 14 could be a day of a month.

And in general, in any SAT problem, I’m seeing this pattern more and more as I go through of these practice test is when they have a problem where they want you to go through every choice, the answers tend to be one of the first two choices because they don’t want make you to waste a lot of time. So you should feel pretty comfortable if you do this and you got choice A. And if you want to see the example of what work, I’m guessing B won’t work. Choice B says, 14, 25, 13, I think that doesn’t look like quite satisfying.

If each number satisfies exactly one of the conditions, which of the following could be the combination to the lock, exactly, so why can’t this one be because this conditions that one number is odd. It can only be satisfied by one of the choice. Oh I’m running over, I’ll see you the next video.

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