This is a one, so your answer is 1:16 or can be said is 16 to 1 or can be 1 to 16 left to right pretty simple concept.

Example 5.9 They have the radius of one circle is r and the radius of the second circle is 3r/5. What is the ratio of the area of the first circle to the area of the second Circle?

Well, we have A1/A2 that's pretty easy. We plug in our area for a circle which is πr^2 right there we have it written here πr^2 over πr^2. In this case, it's the second Circle which is 3r/5. This R is not representative of radius, it's just a capital R variable.

So with that being said, you can do your math and you'll end up with 9R^2/25. And that R^2 will be there and all that will be multiplied by π and then you can start to eliminate those R^2 and R^2 make a one that π and that π make a one. You end up with ultimately 1 over 9 over 25, per final answer of your ratio right of your the ratio of the areas of the first circle to the second Circle.

Last but not least we're on the 5D diagonals of rectangular solids this is also super simple concept you know the Pythagorean theorem. You're good all right. You're going to use the Pythagorean theorem twice, uh to find the formula for the length of the D diagonal in your rectangular solids.

All right so example 5.10 They tell us to find the length of the diagonal that connects Corners B and H in the rectangular solid below. I went ahead and created a little solid for you here. I colorcoded that diagonal underline with green, so you can see it pretty easy.

B to H is what ultimately they want to do. It's two-step uh two-step solution to this type of problem again we're using the Pythagorean theorem twice to figure it out. You going to find the length of BG okay, BG is down here right we're going to remove the whole base and that whole base is going to be uh rectangle fbcg and we're going to get the diagonal between B and G right because we have two given we have two legs given we're just lacking a hypotenuse which is BG.

We can use this theory to go ahead and and plug the Pythagorean theorem in to get our length of BG. You can see that bg^2 is going to be equal to BC^2 plus GC^2 right because we have two Givens we have two legs given we're just lacking a hypotenuse which is BG.

So step one would be to plug in the value, and you'll end up with BG is going to be equal to the square root of x^2 + y^2. Now that's going to be one leg and HG is going to be another leg for the hypotenuse of BH which is your ultimate goal of finding right so step two would be to plug in the BG as one of your legs to HG another known leg to find your hypotenuse.

We do that and we'll end up with the square root of x^2 + y^2 squared plus z^2 will now equal your BH squ you factor all that down and you're going to end up with BH with the length of the square root of X2 + y^2 plus z^2. We'll consider that solved.

Moving on to example 5.11 which I might be mistaken but I'm pretty sure in the problem set had the exact same question so you're going to find the length of the diagonal that connects Corners A and C in the rectangular solid below. And they change the things up put some numbers in there instead of XY variables.

But I colorcoded what you're trying to get, a to c is what you're trying to get. Step one remember we're going to move that bottom base take what's known we have a b and a c is what we're looking for. They give us a two and a three plug all that in our BC hypotenuse will be the square root of 13 and that becomes the second leg uh that you needed for another right triangle to find the hypotenuse which is your ultimate goal of AC or the diagonal of the right solid.

So we plug that in, it's the un of 13 squar. You move it down 13+ M2 AC is ultimately going to be equal to the sare root of 13 + M2 there we go

"WEBVTTKind: captionsLanguage: enhello everybody this is uh Mr Hunt speaking we're g to be covering um lesson five in video format just to see as a test if I can make video formats for classes see how you guys like them you follow along I'll try to get it out as soon as possible all right all we going to start with lesson five has four parts 5 a b c and d we're going start with 5 a and exponents and radicals see in the screen it says key things to remember about exponents and radicals right it's going to be the square root of a number factor twice is that number not too much of a hard concept I don't think set this make sure it's okay if you take a number like seven and you want the square root of that multiplied by the square root of seven ultimately you've just gotten rid of your square roots and your answer is seven easy enough all right make this thck this down a little bit all right also second Point uh we we're going to add the exponents when you're multiplying powers of the same basis so if you see if you have four to the power of whatever time four to the power whatever let's go with an example of2 four to the^ of 12 * 4 to the^ 12 You're simply just going to add these two together half and a half equal a whole P of one is going to be four uh the exponent 1/3 can be used to indicate the cube root and 1/4 to represent the fourth root and Fifth and six and seven so on and so forth right not a hard concept moving on if we were to look down here this little example of three2 the 3 * 3 to the 3 time 3 to 3 only got 33 you're going to put the whole again answer is going to be three this works so on and so forth all right example 5.1 asks us to simplify the square root of x cubed y * the fourth root of XY cubed another thing to remember over here I started out is square root of a number is going to be equal to that number raised to the half power uh transversally we got the square or the CU or the X root of a number will be that number to the one over the X root that you're using so if this x is a three it'll be a third and so on and so forth step one in this process is going to be replace the radicals with parentheses and fractional exponents and then multiply the exponents where indicated so in this case I've removed the radicals here we replace those with parentheses right and plus fractional exponents so this is the square root if there's nothing here it's indicated it's automatically a square so fractional exponent will be the2 hence the repeating of this rule up here all right and this one's the fourth root so it will be raised to the power 1/4 right we're going to end up with X to 3 over2 y to one over two with X to one over4 time y to 3 over4 second step will do be rearrange the like bases and simplify by adding exponents so we'll see the X's together we're going to end up with a three Hales and 1/4 I believe over here we'll have y 12 time y 34s remember they have like bases now we're just going to add the fraction together have to get common denominators this will end up being two over four this will end up being six over four now we add those together we'll end up with X to 7 over4 y to 5 over4 for a final solution all right moving along to example 5.2 want it's to simplify this big mess all right a to X over 2 * Y 2 - x to the half over a 3X * Y -2x all right so first thing we're going to do is just need to inverse these denominators here so we can get it all into the numerator we're going to simplify and write all exponents in the numerator to do that just inverse these so it's positive three will'll turn it into a negative3 and then OB obvious here four plus the 2x here you'll do you'll multiply that over itself ultimately this thing is going to equal one so we're putting everything up into the numerator as the step requires from there we're going to have um a to the x^2 * a the -3x that's this one this one and then we'll have y to the 1 - x^ 2ar over y 2x now we did have some problems figuring out how exactly Y 2 - x to the 12 simplified down to 1 - x over2 so I wrote that down right here what we'll do is we'll zoom into this all right so we remember remember that if you have a base raised to an exponent which is raised to an exponent both your exponents will be multiply together so you'll take your y to the 2 - x over half will turn into 2 - half times2 ultimately break out to you know distribute that in half of two is one and half of negative X would be a minus X over2 hope you guys can see that all right ultimately we will go on to step two this gets a little bit involved with all this math I've got it gray out here so we can go over together see if we can't put that a lower all right so when we get our a times x over two * a to the 3x well we're going to have to add these together so we're going to put this over one to get a like denominator right 2 two we'll get A6 here and a two on the bottom that's gone that's gone all right ultimately you'll end up having your a -6x over 2 * a + x over 26 plus an X is going to equal your five x over two for the Y's we ended up with a Y with a 1 - x over two times a y to the 2x well that's going to be over one that's going to be over one we're going to add this to this right so we're going to have to get common denominators which would be two two over two gives us a 4X two over two over here gives us a two over two still a one um we're going to be Distributing this in so when you distribute that in you're going to end up with plus 4X over two ultimately Y is going to be your two plus 3x all over two it's your new exponent all right put it all together a to the -5x over 2 time Y 2 + 3x over 2 parentheses and that is done example 5.3 we're going to simplify x to the -2 + y -2 all over XY the 1 step one very easy eliminate negative exponents then add the two expressions in the numerator all right easy day here uh if you remember if we have a negative exponent you're just going to have to make it positive and put it all over one so that's what we've done here 1x^2 + 1 over y^2 over 1xy and now we've gotten rid of all of our uh negative exponents don't want those uh and then we're going to have to go ahead and add every all the Expressions into the numerator well in order to do that we're going to add all these in numerator by finding like denominators this one right here will be y^2 this one right here will be X2 ultimately we end up with y^2 + X2 over X2 y^2 that's all going to be over one over XY so step two you going to finish by multiplying above and below but the reciprocal of the denominator see if I can use my tools here this see if that captures nice all right and duplicate that down here somewhere perfect okay there all right what are we going to do all right we're going to finish by Ling above and below the reciprocal of the denominator the reciprocal of the denominator is going to be this flip easy day XY over one right and then XY over one this is my favorite part right here because we just start doing eliminations here that's gone that's gone that's gone that's gone all right and then we have an x on the top here and two x's on the bottom so that gets rid of one of the X's so this is now gone while this is gone as well this y will be gone and this Square will be gone ultimately leaving us with a final answer of y^2 plus X2 over XY and that one's done simplify the three to the uh time the < TK of 3es - 4 * theun of 2/3 + 2 * the < TK 24 for these types of problems they want us to change the form of radicals and use multiplication as necessary to rationalize the denominators and that's the key thing right here right let's go ahead and rationalize these denominators it's not hard to do we will multiply these denominators here by something we can get rid of and we've got a square root of two time theun of two that'll equal one so we're good over here what do you think it'll be bet it'll be this one right here yep square root of three times this over the square root of three this one right here is already good but we can see already that this is factored out the 24 isun of 4 * 6 this is a perfect root so that'll be 2 * 2 6 carry that down here all right let's move up a little bit see what we have after you rationalize these denominators right multiply as necessary you're going to end up with um rationalizing you're going to have a three times the sare < TK of 6 over what's our rule square root Times Square Root same same it's just be that number so it'll be two with no square root over here same deal all right 2 * 3 is six all over oh big fat three and then over here rationalize that out we'll just bring it down to four square roots of six okay finish by finding the common denominators and adding step number three I might want to jump up this here perfect okay so in order to do this we're going to be multiplying common denominators here in order to get this two we're going to be multiplying it by three all right this one we're going to have four squ roots of six * three we're going to multiply that thing by two over here we're going to have four six that'll be over one multiply that bad boy by six all right what's that gonna give us all right so we're gonna have nine square roots of six over sorry over six right minus we're going to have now eight square roots of six over six plus oh is that 24 24 square roots of six over six should have like denominators we got 9 - 8 is 1 1 and 24 is 25 so we will have 25 square roots times the 6 over six that can't be reduced any further so we are done with that Le 5.5 they want us to simplify that all right I don't like this one all right I can see easily we got parentheses here we're going to use the foil method first outers ERS lasts right so we have the firsts the outers the inners the lasts Square OT of two times one anything times one is itself so we're going to have the square root of two as our first term outers first outers it's going to be this minus minus theun of 2 * theun of X now we're going to have plus theun of x * 1 anything times itself so we're have plus the root of x and S otk of x times the squ RO of x uh negative s Ro of X we're going to end up a minus X getting rid of our radical all right now we're going to factor the two middle terms this will be term number one this will be term number two this is the part I don't like because it says note the coefficient of the second term is the sum of and if it's it's a sum of one minus two right how do they get that all right so this is on the right I don't like things being on the right so I put them on the left we're going to bring this common term these are the common terms or the commonalities between the two middle terms so one will come out and then Factor inside in order to get our answer and in order for this to happen we'll end up being times one and then a minus two all right we'll have root X time theun 2un x when factored out right that's then factored out so we'll end up with the square root of two minus one 2 x- x okay clear as mud moving along less than five at 5B complex numbers and in math the letter i is going to be used for the positive square root of nea1 so I equals the square < TK of1 I the shorthand symbol that's used uh first used by the Swift Swiss mathematician uler and said he lived from 1707 to 1783 square roots of a negative number can be written as a product of a real number and I you can see down here we've gotun of5 can be Rewritten Asun 15 times1 or theun of 15 is equal times the of1 will be equal to 15 I so I equaling the of1 we just Factor any negative out all right we're coming up here we have since I is equal to the square root of negative 1 then I2 is equal to negative 1 because remember our rule Rewritten here anything square rooted times itself will be itself with gone so i^2 equal 1 because the of NE 1 * of1 equals guessed it1 complex number uh is made up of two real numbers a and b and the letter I so a plus b i complex number said to be in standard form and complex numbers have a real component a and an imaginary component B I've color coded these here so a is your real component B is your imaginary component you can see it real clearly though the imaginary component usually has that little I in there right for an imaginary number I put standard form in green that'll be important later because your answer can be one wanting those to be in standard form so if we tracking over here we've got all of these are considered complex numbers in standard form we have 4 plus 2 I 3.01 minus the2 over 3 I still following the same format right which is a plus bi you look at uh C and D however you just got 4.06 well they rationalize that by saying it can be imagined that 4.06 plus a zero times I and then over here on the on the transverse you're can have Z Plus 3 I so that's your real component uh 3 I is going to be your imaginary component so theoretically anything could be an imaginary component example 5.6 it wants to simplify 3 I cubed + 2 I 5us 3 I + 2 I 2 this is pretty simple here we're just going to go ahead and expand and pair the I terms right so you've got three I's here I I I well okay so it's three times I and then you pair up two of them right so 3 I * I I and then 2 I and then we have five total here so two pairs two and two and one make five so they're all paired up and you can turn all of these double ey pairs into negative ones later on that's the second step replace each pair of I's with negative one you'll end up with 3 I * 1 + 2 I * 1 * 1 - 3 I + 2 * 1 you simplify all of that out right and you write the results in standard form there's that green color again your standard form you have3 I plus 2 IUS 3 I minus 2 to equal -2 minus 4 I but is that in standard form no oh no it is on to example 5.7 a little bit lengthy but you can see what happened here we take all this negative3 stuffs and we're we're going to translate that into three times a square root of negative one which ultimately will Factor down to an I so when doing these you do have to realize that be on the lookout for something like a three and an I that this right here is not above right that's a nogo they're usually on the outside of your radical so you use the I notation right so once you get rid of all of these and do your simplification you'll end up with all of this right sare < TK of 3 I * 2 plus uh 3 * theun of 2 I * 3 i+ 4 I + 4 we simplify the 2 > 3 I to minus 22 + 4 I + 4 and finish by grouping real and imaginary Parts doing that you can see that we had um 3 I * 2 + 3 I * < TK of 2 * I time 3 I plus 4 I + 4 this reduces down this two comes over to the left I can't write this change my color up right two over here 3 I over here we're going to end up with n time theun two and then you're going to have your eyes which will turn into one pair pair of I equals 1 so that's going to change your sign don't forget to change your sign and plus your 4 I four simplify all this down you'll end up with four minus 9un two plus the sare root of or two * the of 3 + 4 and you can how they just factored your like terms right here your eyes they got factored in because they both had one common so two or 2 * 3 + 4 alltimes I works out to be the same ultimately it's just this making it inverse so it'll be a negative at the end of the day whatever the number turns out to be all right moving on to Five Point C areas of sim geometric figures you see that the ratios of the areas of similar two-dimensional geometric figures equals the square of the scale factor between the figures the book will tell you this is easy to say but it's hard to conceptualize or visualize whatever I disagree it's pretty easy to see you have example 5.8 right here it tells you the scale factor between two rectangles is four what is the ratio of the areas well let me let bring this up into view here you see you have one little pink box here and then you have 16 pink boxes here right because there's only one here now there's 16 because this was just cut and paste four times up and four times to the right so you end up with a 4x4 which is 16 very simply 1 to 16 Ratio or 16 to1 the formula they give us is just ASAS to understand you have a w a width and you have a length and you have a scale factor given to you as a four so it's four W's and then four L's so the area one will be length times width the area two will be four L * 4 W and that'll be equal to 16 lws you put area one over area two you'll have LW over 16 lws this is a one so your answer is 116 so ratio can be be said is 16 to one or can be 1 to 16 left to right pretty simple concept uh example 5.9 they have the radius of one circle is r and the radius of the second circle is 3r over five so what is the ratio of the area of the first circle to the area of the second Circle okay well we have A1 over A2 that's pretty easy and then we just plug in our area for a circle that's pi r s right we have have that written here Pi r^ 2 over Pi * R 2 in this case it's the second Circle which is 3 R over 5 this R is not representative of radius it's just a capital R variable all right so with that being said you can do your math and you'll end up with uh I already crossed this out here for you but you'll have N9 R over 25 and that R squ will be there and all that will be multiplied by and then you'll have up on top you'll have a pi r squ and then you can start to eliminate to that R squ and R squ make a one that pi and that Pi make a one and you end up with ultimately one over 9 over 25 and to get rid of this we just inverse these numerator denominator ultimately you'll end up with 25 over 99 per final answer of your ratio right of your the ratio of the areas of the first circle to the second Circle okay last but not least we're on the 5D diagonals of rectangular solids this is also super simple concept you know the Pythagorean theorem you're good all right you're going to use the Pythagorean theorem uh twice uh to find the formula for the length of the D diagonal in your rectangular solids all right so example 5.10 they tell us to find the length of the diagonal that connects Corners B and H in the rectangular solid below I went ahead and created a little solid for you here I colorcoded that diagonal underline with green so you can see it pretty easy B to H is what ultimately they want to do so it's two-step uh two-step solution to this type of problem again we're to Pagan theorem twice to figure it out you going to find the length of BG okay BG is down here right we're going to remove the whole base and that whole base is going to be uh rectangle fbcg and we're going to get the diagonal between B and G right because we have two Givens we have two legs given we're just lacking a hypotenuse which is BG we can use this Theory to go ahead and and plug the Pythagorean theorem in to get our length of BG you can see that bg^ s is going to be equal to bc2 plus GC SAR factor that all down that'll give you a BG is going to be equal to the square root of x^2 + y^2 now you have found the length of BG now that's going to be one leg and HG is going to be another leg for the hypotenuse of BH which is your ultimate goal of finding right so step two would be to plug in the BG as one of your legs to HG another known leg to find your hypotenuse we do that and we'll end up with the square root of x^2 + y^2 squared plus z^2 will now equal your BH squ you factor all that down and you're going to end up with BH with the length of the square root of X2 + y^2 plus z^2 and we'll consider that solved moving on to example 5.11 which I might be mistaken but I'm pretty sure in the problem set had the exact same question so you're going to find the length of the diagonal that connects Corners A and C in the rectangular solid below and they change the things up put some numbers in there instead of XY variables but I colorcoded what you're trying to get a to c is what you're trying to get step one remember we're going to move that bottom base take what's known we have a b and a c is what we're looking for they're given us a two and a three plug all that in our BC hypotenuse will be the square root of 13 and that becomes the second leg uh that you needed for another right triangle to find the hypotenuse which is your ultimate goal of AC or the diagonal of the right solid so you plug that in it's theun of 13 squar you move it down 13+ M2 AC is ultimately going to be equal to the sare root of 13 + M2 there we gohello everybody this is uh Mr Hunt speaking we're g to be covering um lesson five in video format just to see as a test if I can make video formats for classes see how you guys like them you follow along I'll try to get it out as soon as possible all right all we going to start with lesson five has four parts 5 a b c and d we're going start with 5 a and exponents and radicals see in the screen it says key things to remember about exponents and radicals right it's going to be the square root of a number factor twice is that number not too much of a hard concept I don't think set this make sure it's okay if you take a number like seven and you want the square root of that multiplied by the square root of seven ultimately you've just gotten rid of your square roots and your answer is seven easy enough all right make this thck this down a little bit all right also second Point uh we we're going to add the exponents when you're multiplying powers of the same basis so if you see if you have four to the power of whatever time four to the power whatever let's go with an example of2 four to the^ of 12 * 4 to the^ 12 You're simply just going to add these two together half and a half equal a whole P of one is going to be four uh the exponent 1/3 can be used to indicate the cube root and 1/4 to represent the fourth root and Fifth and six and seven so on and so forth right not a hard concept moving on if we were to look down here this little example of three2 the 3 * 3 to the 3 time 3 to 3 only got 33 you're going to put the whole again answer is going to be three this works so on and so forth all right example 5.1 asks us to simplify the square root of x cubed y * the fourth root of XY cubed another thing to remember over here I started out is square root of a number is going to be equal to that number raised to the half power uh transversally we got the square or the CU or the X root of a number will be that number to the one over the X root that you're using so if this x is a three it'll be a third and so on and so forth step one in this process is going to be replace the radicals with parentheses and fractional exponents and then multiply the exponents where indicated so in this case I've removed the radicals here we replace those with parentheses right and plus fractional exponents so this is the square root if there's nothing here it's indicated it's automatically a square so fractional exponent will be the2 hence the repeating of this rule up here all right and this one's the fourth root so it will be raised to the power 1/4 right we're going to end up with X to 3 over2 y to one over two with X to one over4 time y to 3 over4 second step will do be rearrange the like bases and simplify by adding exponents so we'll see the X's together we're going to end up with a three Hales and 1/4 I believe over here we'll have y 12 time y 34s remember they have like bases now we're just going to add the fraction together have to get common denominators this will end up being two over four this will end up being six over four now we add those together we'll end up with X to 7 over4 y to 5 over4 for a final solution all right moving along to example 5.2 want it's to simplify this big mess all right a to X over 2 * Y 2 - x to the half over a 3X * Y -2x all right so first thing we're going to do is just need to inverse these denominators here so we can get it all into the numerator we're going to simplify and write all exponents in the numerator to do that just inverse these so it's positive three will'll turn it into a negative3 and then OB obvious here four plus the 2x here you'll do you'll multiply that over itself ultimately this thing is going to equal one so we're putting everything up into the numerator as the step requires from there we're going to have um a to the x^2 * a the -3x that's this one this one and then we'll have y to the 1 - x^ 2ar over y 2x now we did have some problems figuring out how exactly Y 2 - x to the 12 simplified down to 1 - x over2 so I wrote that down right here what we'll do is we'll zoom into this all right so we remember remember that if you have a base raised to an exponent which is raised to an exponent both your exponents will be multiply together so you'll take your y to the 2 - x over half will turn into 2 - half times2 ultimately break out to you know distribute that in half of two is one and half of negative X would be a minus X over2 hope you guys can see that all right ultimately we will go on to step two this gets a little bit involved with all this math I've got it gray out here so we can go over together see if we can't put that a lower all right so when we get our a times x over two * a to the 3x well we're going to have to add these together so we're going to put this over one to get a like denominator right 2 two we'll get A6 here and a two on the bottom that's gone that's gone all right ultimately you'll end up having your a -6x over 2 * a + x over 26 plus an X is going to equal your five x over two for the Y's we ended up with a Y with a 1 - x over two times a y to the 2x well that's going to be over one that's going to be over one we're going to add this to this right so we're going to have to get common denominators which would be two two over two gives us a 4X two over two over here gives us a two over two still a one um we're going to be Distributing this in so when you distribute that in you're going to end up with plus 4X over two ultimately Y is going to be your two plus 3x all over two it's your new exponent all right put it all together a to the -5x over 2 time Y 2 + 3x over 2 parentheses and that is done example 5.3 we're going to simplify x to the -2 + y -2 all over XY the 1 step one very easy eliminate negative exponents then add the two expressions in the numerator all right easy day here uh if you remember if we have a negative exponent you're just going to have to make it positive and put it all over one so that's what we've done here 1x^2 + 1 over y^2 over 1xy and now we've gotten rid of all of our uh negative exponents don't want those uh and then we're going to have to go ahead and add every all the Expressions into the numerator well in order to do that we're going to add all these in numerator by finding like denominators this one right here will be y^2 this one right here will be X2 ultimately we end up with y^2 + X2 over X2 y^2 that's all going to be over one over XY so step two you going to finish by multiplying above and below but the reciprocal of the denominator see if I can use my tools here this see if that captures nice all right and duplicate that down here somewhere perfect okay there all right what are we going to do all right we're going to finish by Ling above and below the reciprocal of the denominator the reciprocal of the denominator is going to be this flip easy day XY over one right and then XY over one this is my favorite part right here because we just start doing eliminations here that's gone that's gone that's gone that's gone all right and then we have an x on the top here and two x's on the bottom so that gets rid of one of the X's so this is now gone while this is gone as well this y will be gone and this Square will be gone ultimately leaving us with a final answer of y^2 plus X2 over XY and that one's done simplify the three to the uh time the < TK of 3es - 4 * theun of 2/3 + 2 * the < TK 24 for these types of problems they want us to change the form of radicals and use multiplication as necessary to rationalize the denominators and that's the key thing right here right let's go ahead and rationalize these denominators it's not hard to do we will multiply these denominators here by something we can get rid of and we've got a square root of two time theun of two that'll equal one so we're good over here what do you think it'll be bet it'll be this one right here yep square root of three times this over the square root of three this one right here is already good but we can see already that this is factored out the 24 isun of 4 * 6 this is a perfect root so that'll be 2 * 2 6 carry that down here all right let's move up a little bit see what we have after you rationalize these denominators right multiply as necessary you're going to end up with um rationalizing you're going to have a three times the sare < TK of 6 over what's our rule square root Times Square Root same same it's just be that number so it'll be two with no square root over here same deal all right 2 * 3 is six all over oh big fat three and then over here rationalize that out we'll just bring it down to four square roots of six okay finish by finding the common denominators and adding step number three I might want to jump up this here perfect okay so in order to do this we're going to be multiplying common denominators here in order to get this two we're going to be multiplying it by three all right this one we're going to have four squ roots of six * three we're going to multiply that thing by two over here we're going to have four six that'll be over one multiply that bad boy by six all right what's that gonna give us all right so we're gonna have nine square roots of six over sorry over six right minus we're going to have now eight square roots of six over six plus oh is that 24 24 square roots of six over six should have like denominators we got 9 - 8 is 1 1 and 24 is 25 so we will have 25 square roots times the 6 over six that can't be reduced any further so we are done with that Le 5.5 they want us to simplify that all right I don't like this one all right I can see easily we got parentheses here we're going to use the foil method first outers ERS lasts right so we have the firsts the outers the inners the lasts Square OT of two times one anything times one is itself so we're going to have the square root of two as our first term outers first outers it's going to be this minus minus theun of 2 * theun of X now we're going to have plus theun of x * 1 anything times itself so we're have plus the root of x and S otk of x times the squ RO of x uh negative s Ro of X we're going to end up a minus X getting rid of our radical all right now we're going to factor the two middle terms this will be term number one this will be term number two this is the part I don't like because it says note the coefficient of the second term is the sum of and if it's it's a sum of one minus two right how do they get that all right so this is on the right I don't like things being on the right so I put them on the left we're going to bring this common term these are the common terms or the commonalities between the two middle terms so one will come out and then Factor inside in order to get our answer and in order for this to happen we'll end up being times one and then a minus two all right we'll have root X time theun 2un x when factored out right that's then factored out so we'll end up with the square root of two minus one 2 x- x okay clear as mud moving along less than five at 5B complex numbers and in math the letter i is going to be used for the positive square root of nea1 so I equals the square < TK of1 I the shorthand symbol that's used uh first used by the Swift Swiss mathematician uler and said he lived from 1707 to 1783 square roots of a negative number can be written as a product of a real number and I you can see down here we've gotun of5 can be Rewritten Asun 15 times1 or theun of 15 is equal times the of1 will be equal to 15 I so I equaling the of1 we just Factor any negative out all right we're coming up here we have since I is equal to the square root of negative 1 then I2 is equal to negative 1 because remember our rule Rewritten here anything square rooted times itself will be itself with gone so i^2 equal 1 because the of NE 1 * of1 equals guessed it1 complex number uh is made up of two real numbers a and b and the letter I so a plus b i complex number said to be in standard form and complex numbers have a real component a and an imaginary component B I've color coded these here so a is your real component B is your imaginary component you can see it real clearly though the imaginary component usually has that little I in there right for an imaginary number I put standard form in green that'll be important later because your answer can be one wanting those to be in standard form so if we tracking over here we've got all of these are considered complex numbers in standard form we have 4 plus 2 I 3.01 minus the2 over 3 I still following the same format right which is a plus bi you look at uh C and D however you just got 4.06 well they rationalize that by saying it can be imagined that 4.06 plus a zero times I and then over here on the on the transverse you're can have Z Plus 3 I so that's your real component uh 3 I is going to be your imaginary component so theoretically anything could be an imaginary component example 5.6 it wants to simplify 3 I cubed + 2 I 5us 3 I + 2 I 2 this is pretty simple here we're just going to go ahead and expand and pair the I terms right so you've got three I's here I I I well okay so it's three times I and then you pair up two of them right so 3 I * I I and then 2 I and then we have five total here so two pairs two and two and one make five so they're all paired up and you can turn all of these double ey pairs into negative ones later on that's the second step replace each pair of I's with negative one you'll end up with 3 I * 1 + 2 I * 1 * 1 - 3 I + 2 * 1 you simplify all of that out right and you write the results in standard form there's that green color again your standard form you have3 I plus 2 IUS 3 I minus 2 to equal -2 minus 4 I but is that in standard form no oh no it is on to example 5.7 a little bit lengthy but you can see what happened here we take all this negative3 stuffs and we're we're going to translate that into three times a square root of negative one which ultimately will Factor down to an I so when doing these you do have to realize that be on the lookout for something like a three and an I that this right here is not above right that's a nogo they're usually on the outside of your radical so you use the I notation right so once you get rid of all of these and do your simplification you'll end up with all of this right sare < TK of 3 I * 2 plus uh 3 * theun of 2 I * 3 i+ 4 I + 4 we simplify the 2 > 3 I to minus 22 + 4 I + 4 and finish by grouping real and imaginary Parts doing that you can see that we had um 3 I * 2 + 3 I * < TK of 2 * I time 3 I plus 4 I + 4 this reduces down this two comes over to the left I can't write this change my color up right two over here 3 I over here we're going to end up with n time theun two and then you're going to have your eyes which will turn into one pair pair of I equals 1 so that's going to change your sign don't forget to change your sign and plus your 4 I four simplify all this down you'll end up with four minus 9un two plus the sare root of or two * the of 3 + 4 and you can how they just factored your like terms right here your eyes they got factored in because they both had one common so two or 2 * 3 + 4 alltimes I works out to be the same ultimately it's just this making it inverse so it'll be a negative at the end of the day whatever the number turns out to be all right moving on to Five Point C areas of sim geometric figures you see that the ratios of the areas of similar two-dimensional geometric figures equals the square of the scale factor between the figures the book will tell you this is easy to say but it's hard to conceptualize or visualize whatever I disagree it's pretty easy to see you have example 5.8 right here it tells you the scale factor between two rectangles is four what is the ratio of the areas well let me let bring this up into view here you see you have one little pink box here and then you have 16 pink boxes here right because there's only one here now there's 16 because this was just cut and paste four times up and four times to the right so you end up with a 4x4 which is 16 very simply 1 to 16 Ratio or 16 to1 the formula they give us is just ASAS to understand you have a w a width and you have a length and you have a scale factor given to you as a four so it's four W's and then four L's so the area one will be length times width the area two will be four L * 4 W and that'll be equal to 16 lws you put area one over area two you'll have LW over 16 lws this is a one so your answer is 116 so ratio can be be said is 16 to one or can be 1 to 16 left to right pretty simple concept uh example 5.9 they have the radius of one circle is r and the radius of the second circle is 3r over five so what is the ratio of the area of the first circle to the area of the second Circle okay well we have A1 over A2 that's pretty easy and then we just plug in our area for a circle that's pi r s right we have have that written here Pi r^ 2 over Pi * R 2 in this case it's the second Circle which is 3 R over 5 this R is not representative of radius it's just a capital R variable all right so with that being said you can do your math and you'll end up with uh I already crossed this out here for you but you'll have N9 R over 25 and that R squ will be there and all that will be multiplied by and then you'll have up on top you'll have a pi r squ and then you can start to eliminate to that R squ and R squ make a one that pi and that Pi make a one and you end up with ultimately one over 9 over 25 and to get rid of this we just inverse these numerator denominator ultimately you'll end up with 25 over 99 per final answer of your ratio right of your the ratio of the areas of the first circle to the second Circle okay last but not least we're on the 5D diagonals of rectangular solids this is also super simple concept you know the Pythagorean theorem you're good all right you're going to use the Pythagorean theorem uh twice uh to find the formula for the length of the D diagonal in your rectangular solids all right so example 5.10 they tell us to find the length of the diagonal that connects Corners B and H in the rectangular solid below I went ahead and created a little solid for you here I colorcoded that diagonal underline with green so you can see it pretty easy B to H is what ultimately they want to do so it's two-step uh two-step solution to this type of problem again we're to Pagan theorem twice to figure it out you going to find the length of BG okay BG is down here right we're going to remove the whole base and that whole base is going to be uh rectangle fbcg and we're going to get the diagonal between B and G right because we have two Givens we have two legs given we're just lacking a hypotenuse which is BG we can use this Theory to go ahead and and plug the Pythagorean theorem in to get our length of BG you can see that bg^ s is going to be equal to bc2 plus GC SAR factor that all down that'll give you a BG is going to be equal to the square root of x^2 + y^2 now you have found the length of BG now that's going to be one leg and HG is going to be another leg for the hypotenuse of BH which is your ultimate goal of finding right so step two would be to plug in the BG as one of your legs to HG another known leg to find your hypotenuse we do that and we'll end up with the square root of x^2 + y^2 squared plus z^2 will now equal your BH squ you factor all that down and you're going to end up with BH with the length of the square root of X2 + y^2 plus z^2 and we'll consider that solved moving on to example 5.11 which I might be mistaken but I'm pretty sure in the problem set had the exact same question so you're going to find the length of the diagonal that connects Corners A and C in the rectangular solid below and they change the things up put some numbers in there instead of XY variables but I colorcoded what you're trying to get a to c is what you're trying to get step one remember we're going to move that bottom base take what's known we have a b and a c is what we're looking for they're given us a two and a three plug all that in our BC hypotenuse will be the square root of 13 and that becomes the second leg uh that you needed for another right triangle to find the hypotenuse which is your ultimate goal of AC or the diagonal of the right solid so you plug that in it's theun of 13 squar you move it down 13+ M2 AC is ultimately going to be equal to the sare root of 13 + M2 there we go\n"