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Weight Awav compantt that sclls wcight loss plans Ottom advertises the edfectiveness [lars bY hghlghting the stories of a (ew clicnts Kho havc cemaardinan amont ceh...

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Weight Awav compantt that sclls wcight loss plans Ottom advertises the edfectiveness [lars bY hghlghting the stories of a (ew clicnts Kho havc cemaardinan amont cehca gel 0 better indicauion the qencral cllcrtivcness the plans [email protected] asked Werht Away m inormaton Andut Eypicii clcnis Weight Away madled hrochute #tththe folloning betogcatn, which duplwy: thc wcight bxs anon Dycrtat Pit Month Weighr Away clents {Notetnat neqauve vulut eriuht ueinht qaim )Artoht kna (n puunde|t huttouatui cetrnate tha m

Weight Awav compantt that sclls wcight loss plans Ottom advertises the edfectiveness [lars bY hghlghting the stories of a (ew clicnts Kho havc cemaardinan amont cehca gel 0 better indicauion the qencral cllcrtivcness the plans [email protected] asked Werht Away m inormaton Andut Eypicii clcnis Weight Away madled hrochute #tththe folloning betogcatn, which duplwy: thc wcight bxs anon Dycrtat Pit Month Weighr Away clents {Notetnat neqauve vulut eriuht ueinht qaim ) Artoht kna (n puunde| t huttouatui cetrnate tha mean netoht Iose (pound-) Iratle drcuua nlont aud LutndYuui 4ua7 Aatcata CpTdADEA Eantanabion Oedt tt



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One proposal for an "effortless" method of losing weight is to drink large amounts of cold water. The body must provide energy in the form of heat in order to bring the temperature of the cold water to body temperature, $37^{\circ} \mathrm{C}$. This is provided by the burning of stored carbohydrates or fat. (a) How much energy in the form of heat must the body provide to warm $1.0$ liter of cold water at $4^{\circ} \mathrm{C}$ (a typical refrigerator temperature) to $37^{\circ} \mathrm{C}$ ? (b) Calculate how many grams of body fat must be burned to provide this energy. Assume that the burning of fat produces about 39 kilojoules per gram of fat. (c) How many liters of $4^{\circ} \mathrm{C}$ water must be consumed to lose $1.0$ kilogram ( $2.2 \mathrm{lbs}$ ) of body fat?

The question is asking the following. A weight loss counselor has prepared the program of diet and exercise for client. If the client sticks to the program, the weight loss can be expected to be a function off and and see to be equal to one over eight and squared minus one over Fight C plus 1 97 over eight. Where this end is the number off, 40 minutes were cast per week, and the sea is the average daily calorie intake for the week. The first part of the question is asking, How many pounds can the client expect to lose by eating an average off 1200 calories per day and participating in 4 40 minutes workouts in a week? So I am given that C is equal to 1200 N is equal to four on. I'm asked, what is the weight loss that I should expect if I eat this much? And if I exercise this much so I have a function f off N N. C. I'm just going to use the values of N. C that I'm given in the problem and plug them into that function and calculate the expected weight loss, so F off four on 1200 is going to be won over eight times. Four squared because and is equal to four minus one over five C, which is 1200 +1937 over eights. This is equal to one over a one over eight times four squared, which is two minus one over five times 1000 200. This is equal to 2 41 97 over eight is equal to 2 42.1 To find those families. I just got them by putting them into the calculator, doing this simple addition. I'm going to get 4.1 25 pounds. So to summarize, if I ordered the clients, it's 1200 calories per day and exercises for four, and he has 4 40 minute workouts. Then he should expect to lose around for 4.1 to £5. The second part of the question is asking what is partial f by portion and and to interpret it so we have our function, we want to get a partial F by portion, and so this means I'm going to have to differentiate my function f with respect to end and I'm going to treat see as a constant. So the first part of the problem, which is one over eight and square this hasn't an NSO is derivative, will be won over eight times to end, and then the remaining part of the problem does not have any ends is just a constant. So the derivative off a constant with respect to the variable end is just zero. So my portion f partial and will be equal to end over four. And what does this need? This portion of my passion and this is the rate of change in unit loss for a unit change in the number off 40 minute workouts given that I'm holding see as a constant. So this basically means how much do I expect the weight loss to change if I change the number of workouts by one unit when assuming that sea is held constant and does not change, the last part of the problem is asking, Ah, the client says the client currently average average is 1100 calories per day, and thus 3 40 minute workouts each week. What would be the approximate impact on weekly weight loss off adding 1/4 workout for weeks. So in this problem, the customer he has, he's eating a calories off 1100 is equal to three on. They're asking us what will be what? Um, I what is he? What would be the approximate impact on his weight loss by adding 1/4 workout in the week? So I'm This means he wants to get the partial at by partial and by and he wants to see how much the weight loss will be affected if I increase or decrease this rate, this number of workers. So I'm increasing this number of workers from 3 to 4. So I will substitute in my portion of my portion and that I got from part B with an end equal to three. So this is just going to be three over four pounds. So this customer, if he increases the number of workouts by one session compared to what he currently has, which is three sessions per week, So if he increases it by 1/4 Toby after have 1/4 session, he should expect to lose an extra to you over £4 per week. Notice that C or the the calories intake is no longer involved in this problem anymore because it does not affect the problem because portion of my portion and is a function off end. And it's not a function off and and see anymore. So it doesn't matter how many calories he takes. What matters is how many number of workouts ask for this formula right here. So if he increases the number of workers from three sessions to four sessions per week, then he should expect to lose on additional three over £4 for a week. Thank you so much.

Okay. This problem has ah, weight loss function as a function off and and see where and his number of workouts and see is your daily caloric intake. So this problem has three parts A, B and C A asks how much a person is expected to lose and wait if they eat on average, 1200 calories per day and they do for workouts per week. So they're asking you to solve for the function off. Four because they're four workouts and 1200 calories. So 400 well, 1206 1200 calories. So if we were to plug in, we're plugging these values into our function would get 18 times and square, so be 16 foursquare in 16 by this 1/5 time. See which is told 100 1200 plus 193 7/8 and this turns out to be. This turns out to be 4.1 to 5 pounds lost, so this person would lose approximately or 4.1 to £5 if they eat 1200 calories a day and do for 40 minute workouts a week. Now what those part B ask you doing part b ask you to find an interpret the partial derivative with respect to so t f. Yeah. So if we took the derivative with respect again, we just have to use power rule. Bring on the to Teoh 1/4 and and this is your partial derivative with respect to run. And they are They also ask you to interpret the meaning off your partial derivatives. So we're gonna go ahead and type that out here. Uh, this, uh, number 1/4. And it is a great race of change. Ready to change of weight loss her unit for you. Change. You can change in workouts since, and the invariable means the number workouts for changing bad thing, changing the number of workouts with respect. And so that's your answer for part B. Now hard See parts. He asks you if a person is currently eating 1100 calories per day and does three workouts each week. Well, what happened if he would if you were to increase the number of workhouse from 3 to 4 while keeping the 1100 calorie caloric intake. So what this problem is asking you to do is to find the parts of there it is oh, off function at the 0.3 and 1100 calories because we're increasing the number of workouts from fetal four. So we need to find the partial derivative with respect to the number of workouts. So we already saw for the partial derivative in Part B. It's all we have to do is plug in. Only have to do is plug in are variable, so it's only three. So it just turns out to be three fours or zero point stand. If I so if the person did an additional work out, he would lose an additional your £0.70.0.75 pounds a week.

Okay, so the person is uh initial weight is £164, and the gold weight is £128. Okay? Okay, assume that the number of weeks by X. Okay. And the weight loss after X weeks our goal is two lost £36 of weight. Okay? So that this person uh there's a loss of at least £1.. week. So let's um writer inequality to represent This. Okay, so I have 1.5 times x is less than or equal £36. Okay, so let's solve for X. So X is less than or equals 24. So the maximum autograph weeks taken by the person to attain his goal weight is 24. Okay.

Hey, everyone, this is question number 70 from chapter 14. And this problem were It's about the body. We're told that one pound of fat equals 35 100 food calories. Not very comforting. We're told that 80% of metabolized food goes to heat, and we're we're get a situation that you're going to run and running. Produces a metabolic rate of 1,290 watts of metabolic power. Excuse me. Whereas, to find the hours to burn one pound of fat, how much heat produced by burning one pound of fat. And how many leaders of sweat you would need to evaporate to get rid of that heat. Okay, so part A, um, we're as to find out hours. We need to run to burn one pound of fat. So first we need to figure out how much heat this pound of fat actually is. We can convert food calories to Jules. You could find that conversion in your book, but essentially one food calorie, I'm just gonna put FC is equal. Tio 4.186 I'm standing on 1/3 Jules. Okay, So the energy released with that 35 100 is just 35 100 times that amount 4.186 times 10 of third Jules. And that gives us 1.47 times 10 to the seventh Jules. And then we are trying to solve. For time we have energy. We have a power which were given in Watts, and we know that power equals Q over T energy or, in this case, heat overtime. So we rearrange this equation and Saul for tea t equals Q over. Pete. This part A T equals Q. We just found it. 1.47 times 10 10 of the seventh. Jules over 1,290 watts we know is a jewel per second. Jules, cancel your left. The seconds equals 1.13 times 10 to the fourth seconds and where has to find an hours. So you divide by 3,600 you get 3.2 hours. So this is not really, really realistic. For a normal port person, you'd have to be very, very in shape. Okay, so now we move on apart. Beat? Yeah, and for part B, we are looking for the heat produced, and we're told we just solved for you the energy release, which was 1.47 times 10 to the seventh. Jules, and we're told that 80% of that is get 80% of that goes to heat in the body. So we just multiply this by 0.8, and that gives us 1.18 times 10 to the seventh. Jules. That part's pretty simple. That's what actually is heat instead of just energy that's answered Part B, and then we move on apart. See if you had Teo get rid of all this heat sweated out, essentially evaporated. How much water, or how much sweat would you need to How much sweat with your body need to get rid of and were given a heat of vaporization? How much sweat you need, but you need to evaporate. You were given a here, vaporization for water in the body, and we know the equation. Q equals M LV, so we rearrange and solve for mass. M equals Q over LV. Plug in our numbers m equals Q. We just saw for at 1.18 times 10 to the seventh Jules over l V were given the problem 2.42 times 10 to the sixth Jules per kilogram, and that equals 4.88 kilograms. But we're asked how much water is in leaders, and we also know that one kilogram equals one leader of water said. This is going to be 4.8 leaders of water that is a lot of water.


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