5

15- Aldehyde or ketone reacted with water t0 formL3-diolI- Wo hydroxy compoundcarboxylic acidd-1,I-diol16- What is the product of the following reaction?CH;509 HzSO...

Question

15- Aldehyde or ketone reacted with water t0 formL3-diolI- Wo hydroxy compoundcarboxylic acidd-1,I-diol16- What is the product of the following reaction?CH;509 HzSO4Hyc-C-CH-CHz CH,CH;CH;CH3 CHj H;c C-CHz-CHzOH d: H;C CH-CH CH} H;ca-HyC-C-CH-CH; HO CH,b-Hyc-C-CHz-CH; CH,17- Which one of the following compounds is the structure Of oxime?CH;CH-NNH: b-CH;CH-NNHCSNHCHCH-NOH d-CHCH-NNHPh

15- Aldehyde or ketone reacted with water t0 form L3-diol I- Wo hydroxy compound carboxylic acid d-1,I-diol 16- What is the product of the following reaction? CH; 509 HzSO4 Hyc-C-CH-CHz CH, CH; CH; CH3 CHj H;c C-CHz-CHzOH d: H;C CH-CH CH} H;c a-HyC-C-CH-CH; HO CH, b-Hyc-C-CHz-CH; CH, 17- Which one of the following compounds is the structure Of oxime? CH;CH-NNH: b-CH;CH-NNHCSNH CHCH-NOH d-CHCH-NNHPh



Answers

The reaction of an aldehyde or ketone with a Grignard reagent (Section 12.8 ) is a nucleophilic addition to the carbon-oxygen double bond. (a) What is the nucleophile? (b) The magnesium portion of the Grignard reagent plays an important part in this reaction. What is its function? (c) What product is formed initially? (d) What product forms when water is added?

Okay. This problem is asking us to put each of these compounds into a basic a quick solution and get the retro Aldo audition product. Okay, so first up, I have this to Ethel three hydroxy Hexen. All the first thing that I do to approach each of these problems is actually draw out the structure of the molecule. Okay, so whenever I have a name and I'm asked to give the structure, I always start off with the right back. So I started for the very back and I worked my way this way. So first up, I see that I have hexagonal. So I see that the all that corresponds to an Alba had and the hex corresponds to a vaccine. So six carbons within Al behind on the last carbon. So 123456 carbons. And then I have my Alba hide over here. Okay, so this is position one. This is 2345 and six. It's on position three. I have the presence of my hydroxy. So three hydroxy that's gonna correspond to an alcohol right there. Okay. And the last up, I have my two fo That's gonna be an ethic group on the second carbon. So 12 carbons. Okay, Now I can actually perform my basic acquia solution reaction in which I am going to take hydroxide, for example. And I'm gonna take off this acidic proton get when I take off at a proton. I'm gonna have to move the bond, the electrons from the bond between my auction hydrogen onto this single bond. And in the process of doing that, I'm gonna have to move the electrons from this carbon carbon bond onto this carbon right there. So that's gonna for him in this product in which I have my I'll hide like this and then my other Aldo hide just like this. Okay, so if I have lone pairs there and I just propagate that so it should pretend like we printed that in another stuff. So I'm gonna get this product keeps trying to raise just the lone pairs there. Okay, so if we get this product, we can see that we have 1234 carbons total, and 1234 carbons total. So each of these are the same product, and that will be the product of this reaction. If we reacted with a basic agree a solution. Okay, so that's that one. Let's move on to the 2nd 1 So for the 2nd 1 and again, I'm gonna start with the back and work my way to the left or right to left. So, Penton on that corresponds to a key tone. And then up the, um that's gonna be on five carbon chain. So 12345 carbons on the second carbon. So second carbon, that's the location of my Carbonell, cause that's gonna correspond to the key tone. So second, carbon is where my carbonless and I have form Ethel, that's gonna be on the fourth carbon. So I have 1234 carbons on that fourth carbon. I have my metal group. So Method group right there and then on my fourth carbon as well. I also have that hydroxy group, so that's gonna be a alcohol right there. OK, so, Avery, first thing I'm gonna do is take off that a set of car engine off of my alcohol. So it's gonna be like this. I'm gonna have my quest. Basic solution that's gonna take off that s set of hydrogen moving the electrons from this oxygen hydrogen bond onto this site right there. When I move the electrons onto that carbon single bond, I'm gonna have to move the electrons away from this carbon and onto this site right there. Okay, So that's gonna form this product in which I have my, um, key tone like that. That's corresponding to this one right here. And then the other one is going to be my deep resonated. Other version of that molecule. It's like this. Okay, so this site right here that correspondents to that key told right there, Okay. And then, of course, we just undergo a pro nation step in which we prone eight this off carbon, and we should end up with just two molecules at the same molecule. Okay, so that will be my products for that retro. A little addition direction. And next up is this one. It's a to four die cycle. Texel. Three hydroxy, beautiful. Get so get starting with back and working my way. This way. I have the al, so I'm gonna have an Alba head and then butane. That's gonna corresponds to you. Four carbons total. So 1234 carbons 1234 And then on my first carbon, I'm gonna have my old hide just right there, Okay? And then on the third carbon, I'm gonna have my hydroxy group so hydroxide Europe right there. Okay. And last up, I have my 24 die cycle Hexcel statements I'm gonna have on position two and on position for I'm gonna have a cycle hek sel, and I'm gonna have two of them total. So di cycle. Heck, so that means I'm gonna have a cycle hexane as a substitute. So just like this, And make sure that I have that, um, the carbon connected to that second carbon, for example, Don't put my cycle hek sel right there. Because otherwise I would have this cycle hek sel as a superstition went off of this stored carbon to make sure that we just put this in a correct orientation. So again, my carbon must be connected to that fourth carpet, So cycle Heck, so off of that. Okay, so that is the structure of my 24 die cycle. Hexcel three hydroxy beat. No. And now we're gonna have my basic bakery solution in which I have my hydroxide it's gonna go ahead and take off. That s a country gin moving the electrons from this oxygen hydrogen bond onto this single bond between my carbon in my oxygen. Getting. When I do that, I'm gonna have to move the electrons from this carbon onto this site right here. And I know not to move it on to this site because if I did that, then this would be an unstable um, carbon. I am right. I want to make my car ban on as stable as possible. And I'm gonna do that by moving them onto this. All right, there. That's the awful carbon. And those elections could be d localized by moving, um, onto this Carbonell. So that's why I moved them on to this site rather than this site. So with that in mind, let's do that and draw out our products. So if I were to do that reaction, I'll get my I would hide like that. And then I have my cycle, Hexcel. So just like this, and then that will be the product of this deep rotation. And then if I dip, rotate this one, I'll have the same. Exactly. If I draw that one out. I have the same thing. So another molecule of this one. And of course, we have the deep rooted carbon right there. But we're just gonna go ahead and prone eight that in the later step. So what? You just get to molecules of my same product. Okay, so next up is this five Ethel, five hydroxy four. Metal three. Happy known. So again, back to front. So happen on that corresponds to a six carbon compound with a key tone And that key tone specifically the Carbondale is gonna be located on that third carbon of my seven carbon chain. So 1234567 carbons. The third Carbon is location of Mikey Tone. It's right there. Okay. And then form Ethel. That's gonna correspond to the fourth position. I'm gonna have a method group so forth. Carbon right there. Fore position method. You're right there. And then five hydroxy that's gonna correspond to the location of my hydroxy group on the fifth carbon. It's right there. And then I'm gonna have my five Ethel, that's gonna be on the fifth Carbon as well. I'm gonna have NFL group. So just like that, Okay, Now we can go ahead, and D Protein ate that most instead of Kardashian on that alcohol. So using my equal space department, the hydrogen moving the electrons from this oxygen 100 bond onto this single bond right there to make a double bond. And when I do that, I'm gonna have to move the electrons away from this carbon, because otherwise I would have exceeded the octet of that carbon. I'm gonna move them onto this awful carbon of this Carbonell group. So it's a little bit. It might be a little bit confusing. Could this cause I drew this upside down, But this should be the product. I'll draw the left side first, and then I'll draw the right side. I'm gonna have right here in my Carbonell with this group in which I have my car ban on right there. And then if I drop out the right side of this now, I'm gonna have this product in which I have my Carbonell. And then I have my two ethyl groups attached to it like this. And then, like this and again, I'm gonna have the pro nation step to protein it this carbon on So we can just go ahead and delete that compare just to show the pronation. Okay. So again, I had the same molecules as I would basically just combined those two molecules in Aldo Addition reaction to get to the product right here. But then, if we go to undergo a retro although edition, we're gonna get those Marcos put up so that those will be on my answers.

Okay. This problem is asking us to propose a mechanism for a manic reaction. Given the following, the regions have been given a second exit on just a six Governor in with a carbon on top. I'm also giving a nitrogen with hydrogen and two month groups and then also in a seat, childhood like that. And finally, that is all in an acid medium. So acid catalyst okay, and should result in the following compound that looks something like this. Something cycle hex unknown. And then in the awful position I have this, so just connection with my foot gets cut off. Okay, so how do we make this? Well, let's go through the mechanism. So first off, I see that I'm in an acid catalyst in an acid medium. So I'm going to protein ate my most basic compound, and my most basic compound is actually this and mean. So I'm going to go ahead and put it that I mean, with my acid resulting in the following compound. So I just have my propagated. I mean, Okay, next up, I'm going to utilize my album head and get my other had propagated so a little bit about why I'm using my all the head is opposed, Mikey. Tone first is because my album head is considered to be more reactive and general than Mikey Tone. And that's due to Sterk effects because this hydrogen is smaller than this group of carpets. And then also because of the inductive effects. Basically, my, my carbons are pushing electron density onto that carbon, making that carbon considerably less electric folk. Okay, so my childhood is considered more Electra Filic, which is why we're going to you utilize this album Hide in the first reaction. Okay, so I'm gonna first off Protein ate my oxygen without 100 moving the electrons onto that night. This After that, I should end up with my natural with two methyl groups and then my appropriated other hide. Okay, so now that I have my opponent adulthood, this album head is considered to be activated so I can more easily attack it with a nuclear file. So I'm going to use the power of my main group and make that my new profile attack my Electra Philip Carbon, and moved the electrons of to the auction to make an alcohol. Okay, So the result of that should be this in which I have my alcohol over here. And then I have my group, which has my extra pro tone and then my two methyl groups. So that means gonna be having a posit charge and then my hydrogen. Okay, so now I know that I'm I want to make some type of, um I mean, our enemy perhaps. So I need to kick off this alcohol. But obviously we know that alcohol is not the best leaving groups, so we want to make it into a better leaving group. So I want to make it water. So because I have this extra pro time, I'm going to move that proton onto this oxygen, and I'm going to represent that using a proton transfer. So just PT protect transfer will result in the following molecule. I just have water with a positive charge. And then I have my I mean with my two methyl groups and then my hydrogen. Okay, so my water is a much better leaving group now, I could make that water leave, and I can bring the elections down from that estrogen onto that single bond to make a double bomb case that I should result in the following molecule. It just ends up quickly. My nitrogen double bonded to my carbon and then my hydrogen there. So this nitrogen has a positive charge. Okay. And then of course, I have water down here. Okay, so now I have to. That's basically the first part of my of my reaction. The second part of my reaction is used utilizing my cycle accident. My cycle hex unknown, is also going to be in an acidic medium. Okay, so we're gonna have my acid, and we're going to propagate that oxygen resulting in just a protein, a taquito. Okay, so that probably the key tone is considered to be activated now. But it has this acidic proton. So the aesthetic proton is in the awful position right here. Okay, So if I have water, which I do and I have water because in my previous direction with the nitrogen, I formed water at a side product over here. And also because it let's assume that I'm in hydro Liam. If this was from a hydro Nehemiah such as age three plus, then after I appropriated that auction, then I would result in the formation of water. So this is we have two options of why we have water. So I'm going to move the electrons onto the hydrogen pro native, my water and in the process, moving the electrons onto this single bond. Okay, then simultaneously moving the electrons onto that oxygen. Okay, so after that, I should have the following compound in which I have my alcohol here and then my all keen here so that all cane is going to be considered to be nuclear folk in an eventual reaction with my, um, enemy compound. Okay, so what I'm gonna dio is bring this and I mean, over here on the right side of my screen, I'm gonna bring it down, and then I'm gonna have my vinyl alcohol attacked the base of that, um, enemy structure. So this is someone so carbon. Also imagine that this is an oxygen. I'm just going to co go ahead and attack the base of that carbon. Get so after I attacked that carbon, I'm gonna have to move the electrons up to this nitrogen to relieve of its positive charge. So that should result in the following. Oh, and by the way, simultaneously as I'm attacking those carbon with the elections from this double bond, I'm going to move the elections down from this oxygen to create a double bond. Okay, so I should end up with this. So auction with a hydrogen. So that means I have a positive charge there. And then this is considered to be my nuclear Philip Carbon. So I'm attacking my nitrogen. So I'm attacking the carbon. Sorry. So this carbon corresponds to this Corbyn attached to that carbon. We have my nitrogen, which has a single bond. And then we also have that those two month grips attached to it. And then, of course, to this central carbon, because this one we have that metal group is Well, okay, so now we have to consider What do we need to end up with? We need to end up with this molecule. And how do we end up with at all? We have to do Is deep protein ate my, um my protein a taquito. So I'm just gonna utilize water. So you're using water. I'm going to go ahead and deep, rotate that pro native key tone and move the elections back onto the auction, so we should end up with the following molecules. Just my product. So cyclo, hexane, own and then attached to that awful carbon. We have this. Okay, so this would be my final answer. And this whole thing is my mechanism.

This is the answer to Chapter 24. Problem number five from the Smith Organic Chemistry textbook. This problem asks what Alba ha dorky tone is needed to prepare each compound by an algal reaction. Okay. Ah. And so, um, in order to determine that we need to locate um, the Alfa Carbon of the carbon eel. In each of these molecules, we need to locate the Alfa Carbon that is bonded to the beta carbon with the alcohol on it. Or in the case of see the Alfa Carbon that is participating in the double bond If the actual Al Gore product has been dehydrated. Okay. S o for a, um, the Alfa Carbon. Ah, and the beta carbon are going to be Ah, here will be the beta. Ah, And here will be the Alfa Carbon. And so basically, we cut between them on we cut between them. Um, and that is gonna be our two pieces. So we have one piece that's gonna be this Alba hide. That would be the left Feats that the piece that had data Corbin Ah, and then the peace with the Alfa Carbon. Oh, so actually, um right, yes. So actually there Exactly the same. Okay. All right. S o, that is part, eh? Ah, for part B. Um, we could do the same thing s o the Alfa Carbon. Mmm. Try that again. Uh, the alpha carbon will be this one. The beta carbon will be this one. And so if we cut between them, we get to pieces. Uh, that looked like this. So again, these air the same molecule. Um, So it's just the one molecule that that these out our products are coming from, Um, and then for C ah. This time we have the dehydrated product, so we can still identify. This is the Alfa Carbon. And then this would be the beta carbon, and so we can cut right through that. That double bond. All right. And that will give us again to equivalents of just one molecule, Eso you know, to equal once of it. Yeah. Okay. Ah, And that is the answer to Chapter 24. Problem number five

Okay. This problem is asking us. How do we make these compounds from an algal conversation? So what I'm given is given a benzene first connected to an eventual keen connected to I never knew. And then something depending on the site. Okay, so how do we make this compound use interactive? It's in an Al Bell conversation. Luckily for Alba conversations, I have a trick, and my trick is as follows. First up, we locate the position of my, um, alphabet on saturated system. We make sure that we do, in fact, have one. Because in alittle conversation reactions, we have the addition of an alphabet on Saturday system characterized by the carbon era and then my all cane in the Alfa Beta position case without to my alpha beta and stature, The system next up, I'm located my all keen. So this my cocaine, once I locate my oking identified the carbon that is closest to my that is not representative of, um, induction. I locate the carbon that is closest to my carbonell. So this carmiel, this carbon is closest to it. So once identified that I draw a schoolgirl line down that carbon, ensuring that I have nothing attached to, um, this side of my Okay, So here's my skin. I have nothing attached to this side of it. Okay, So once I dropped that school squiggly line, then I proceed. That squiggly line is representative of a breaking of bonds. Okay, so I'm just going to break that bond like this, like that, and then I have on the other side, the remainder of my compound. Okay, so this is obviously on the mechanism. We're not breaking that bond, or we're not adding this all keen to right here. We're doing, in fact, something completely different. But this is just for the trip. And then for the trick, Once I locates, herbal tea, I make this compound the one that has the arcane. All I have to do is add an oxygen where it was connected to this carbon. Okay, so that is my those directions. That's what I have. I have a benz alta hide racking with this compound right there. Okay, so they're not necessarily the same. Okay, next up, I have this. I'm given cycle Pantene that and then I have connected to it a key tone. Okay, so how do we make this molecule. Well, Santana's before first up, I make sure that I do, in fact, have my alphabet and saturated system. And in this case, I do because I have my carbon deal. And then in the Alfa Beta position, have my okay. Okay. So once I identify that, I do have those things. I locate my talking, and then I located the carbon that is closest to my Carbondale. Okay, So here's my Carbonell. Here is my carbon that is closest to it. And then I draw a squiggly line to ensure that nothing is attached to this al cane. So I make sure that nothing is attached to this out of my okay. All I would have if are two separate. This would be Hodgins. Okay, so that might be a little bit confusing, but all I would have to do is stretch this molecule out, represented by the breaking of this bond. Kates out. Have this. Oops. I won't draw the oxen yet. I have that. And then I have this this this that and then, Of course, I don't have, um, this bond any longer, So I would have had to draw this like this like that. But I'm breaking this bond so I don't have I don't have that any longer. So I said I just have this. Okay, so that is where I'm at right now. And then, of course, all I have to do is add my oxygen to the former bond between this cocaine and this carbon. Okay? And then that's it. I'm done with that. That's why I started with that would undergo inaudible conversation to result in this product. Okay, Next up, I have this one in which I'm given Hey, Benzie and then attached to the Benzie. And I have a cycle Pantene like that and then attached to that. I have my mother. Okay, so how would we make or how would we figure out the Rechnitz needed for this album conversation? Same thing is before this is all according to the trick. So the trick. Make sure I have my Alfa Beta on Saturday system. Once I locate that, I locate my O'Kane and then I find the carbon that is closest to my Carbonell. Once I find that carbon, I draw a squiggly line to ensure that nothing is attached. Two This hulking sulking, right here. Nothing will be attached to it. If I were to separate these, all would have would be Hodgins. Okay, So what I'm gonna do is I'm going to stretch this out. So Quincy unaffected. And then I have my former cycle painting. But now it is just and OK, okay. And then I have the remainder my compound like this. Okay, So all I would have to do to make this molecule, um, performing all organization is at an auction to that alky. Okay. And of course, obviously in the mechanism were not We're not like sandwiching this auction into this position. We're not forcing that combine. It's completely different. But this is just according to the trick. Okay, Next up, we have this one in which were given this compound with carbon on top. And then we have Elkins here and attached to each of those four carbons. We have fennel groups, so fennel is representative of benzene as a substitute. Okay, but we just write down pH because it's easier to write than ready down entire benzene. Okay, So what I'm gonna do is same thing. I identify that I do have a alpha beta and sexual system. So I have one right here. But wait, I have one right here that will. So the trick might not be completely the same on this one, but I'm going to continue. So I have this debate and saturated system. I locate my Arquin, identify the carbon that is closest to my Carbonell. And then I just draw excluding line to make sure that have nothing attached to this out of my cocaine. And then I proceed. So it's cool to line. And then I have this. So benzene. Sorry, Mikey. Talk connected to this. Connected to this with my former Al Keen fennel fennel. And then I have fun over here as well. It's all would have to do is add an oxygen to that spot right there. And then we could put proceed with unknowable conversation. But if we wanted to break this this down even more into its other components, we could also could have formed this from an album. Conversations. Well, because once again, we do have an Alfa Beta on Saturday system. Here we have my Carbonell, and then we also have my double bond on the Alfa Beta carbons. So Santana's before I'm gonna draw my school squiggly line right there. So I'm separating this semi looking from this side of my molecule. So if our to draw this out, I'll just have hydrogen associated with each of those carbons. Okay, so I e I drive my squiggly line and then I proceed. I just represent that. Is it breaking the bonds like that? I have my key tone, and then I have Oops. Sorry about that. Zem backing. Okay. And then I have this set of direction. Should have my Carneal, my fennel, and then it has to right here I have my cocaine in a federal, and then, once again, I'll have to do is at an auction to that get. So, if this were in basic conditions with sodium hydroxide, my sodium hydroxide would be probably the most instead of country in which would be right here that would reacts with that Carbonell producing this compound. Okay. And then I would undergo, um, more basic deep rotation of my acidic 100 right here. And then we end up with the attacking of this carbon to form this product. Okay. And then that would be it.


Similar Solved Questions

5 answers
Evalunte the Iine intewal v 21*+0 #)t+( 2r") dr Mhcre € h formed by the Intcrstclton ol Uhe Eyllndct r' +V muth thepe t"+
Evalunte the Iine intewal v 21*+0 #)t+( 2r") dr Mhcre € h formed by the Intcrstclton ol Uhe Eyllndct r' +V muth thepe t"+...
5 answers
Determine which of the following subsets of V = R? are subspaces Prove or give counter example; W = {(a,0,0) |a e R} W = {(a,1,1) |ae R} W = {(a, b, c) | b=a +€,anda,b,c € R} W = {(a, b,c) | b=a+c+1, Jnd a,b,c
Determine which of the following subsets of V = R? are subspaces Prove or give counter example; W = {(a,0,0) |a e R} W = {(a,1,1) |ae R} W = {(a, b, c) | b=a +€,anda,b,c € R} W = {(a, b,c) | b=a+c+1, Jnd a,b,c...
5 answers
Value ot_ wheekhai conversion van that oriqinally cost 541,4J0 depreciates s0 that Eech year (0) Fina moje for Vit) , che value the van Jller years. vtt)Wvotthvalue for the previous VenrDeterming the vuluc 0f the vjn veats aiter was purchused, (Round Your answicrdacnNoad Holp?QoT
value ot_ wheekhai conversion van that oriqinally cost 541,4J0 depreciates s0 that Eech year (0) Fina moje for Vit) , che value the van Jller years. vtt) Wvotth value for the previous Venr Determing the vuluc 0f the vjn veats aiter was purchused, (Round Your answicr dacn Noad Holp? QoT...
5 answers
Point) Circlc all of the following Integrals that are improper: No justification needed_Inlz - Jdr(2 points) Doerdr converge or divcrge? Explain your answr .
point) Circlc all of the following Integrals that are improper: No justification needed_ Inlz - Jdr (2 points) Doer dr converge or divcrge? Explain your answr ....
5 answers
Question 8An unknown compound produces the IR spectrum below: What FUNCTIONAL GROUP does the compouat contain? Brielly explain what peaks lead vou t0 this conclusion:Edit View Insert Format Tools Table12ptParagraph4 4 ~ 2~ Tr~ |0
Question 8 An unknown compound produces the IR spectrum below: What FUNCTIONAL GROUP does the compouat contain? Brielly explain what peaks lead vou t0 this conclusion: Edit View Insert Format Tools Table 12pt Paragraph 4 4 ~ 2~ Tr~ | 0...
4 answers
Design trapezoidal-shaped waterway with a 4:1 side slope to carry0.6 m3/s from a terraced field where the soil is resistant to erosion,and the channel slope is 4%. The vegetation choice is Bromegrass: Thechannel must be occasionally crossed with equipment.
Design trapezoidal-shaped waterway with a 4:1 side slope to carry 0.6 m3/s from a terraced field where the soil is resistant to erosion, and the channel slope is 4%. The vegetation choice is Bromegrass: The channel must be occasionally crossed with equipment....
5 answers
5.25mertically upward with velocily of Am/s from height ot 5 25m abote tlle Abal nhrown ound The time taken (n sec) for the ball to reach the ground is: 0
5.25m ertically upward with velocily of Am/s from height ot 5 25m abote tlle Abal nhrown ound The time taken (n sec) for the ball to reach the ground is: 0...
5 answers
Solve the equation Ax = b by using the LU factorization given for A3 -43 | A = -9 10 -40 0-4 3-3-2 5b =136 -62212Let Ly =b and Ux =y. Solve for x and yy
Solve the equation Ax = b by using the LU factorization given for A 3 -4 3 | A = -9 10 -4 0 0 -4 3 -3 -2 5 b = 13 6 -6 2 2 12 Let Ly =b and Ux =y. Solve for x and y y...
5 answers
Perform the operation: 2 ~ (-3)A) 5 B) -[ C) 1 D) -
Perform the operation: 2 ~ (-3) A) 5 B) -[ C) 1 D) -...
5 answers
3hours8/15QuEsTIONPDINTThie Iigure above shows "ghe triangle velth dmensions Uncnes not Inclucle unts In your answer)37 What Is the Fengtn afy the nearest inch? (DoOnlintProvide youn anover DelovrWNM6commandoplion
3hours 8/15 QuEsTION PDINT Thie Iigure above shows "ghe triangle velth dmensions Uncnes not Inclucle unts In your answer) 37 What Is the Fengtn afy the nearest inch? (Do Onlint Provide youn anover Delovr W N M 6 command oplion...
5 answers
U 1 1 winuly +90 1 ua F
U 1 1 winuly +90 1 ua F...
5 answers
Question #9 (10 points) Explain the major functions (pleural) of' the pentose phosphate pathway (pentose shunt). (Don give the pathway)
Question #9 (10 points) Explain the major functions (pleural) of' the pentose phosphate pathway (pentose shunt). (Don give the pathway)...
5 answers
Consider Ax = b with A(10 points) Find the least squares soltion using the normal equations_points) Form thc matrices Q_ and R such thatQR;
Consider Ax = b with A (10 points) Find the least squares soltion using the normal equations_ points) Form thc matrices Q_ and R such that QR;...

-- 0.020719--