The main purpose of cooking vegetables is to make them more edible, palatable, and digestible. The downside, though, is that cooking may adversely affect the levels of nutrients, especially the heat-sensitive and water-soluble ones. But even if you boil greens for 10 minutes, the drop in antioxidant capacity, for example, which is a rough proxy for a phytonutrient retention, isn't that much. Yes, there's a significant drop in each case, a 15 to 20% drop, but most of the antioxidant power is retained, even if you boil lettuce for 10 minutes. The single nutrient that drops the most is probably vitamin C, but as you can see, collards start out so vitamin C-packed that even collard greens boiled for 10 minutes have twice as much vitamin C compared to even raw broccoli. You can see the vitamin C in spinach really takes a hit. Even just blanching for 5 minutes can cut vitamin C levels more than half, with more than 90% dissolving away into the water after 15 minutes. Though most of the beta-carotene, which is fat soluble, tends to stay in the leaves. But just keeping it in a regular plastic bag, like you get in the produce aisle, can protect it. The refrigeration is important, though even in a bag, a hot day can wipe out nearly 50%. Not as bad as drying, though, which can wipe out up to 90% of the vitamin C, suggesting that something like kale chips may pale in comparison to fresh. Though vitamin C is particularly sensitive, other nutrients like beta-carotene are less affected across the board. Cooking by microwaving and steaming preserves the nutrition more than boiling. Here measured in watercress, little steaming or microwaving hardly has any effect compared to raw, though boiling even 2 minutes may cut antioxidant levels nearly in half. Watercress is a cruciferous vegetable, though a cabbage and broccoli family vegetable, so it's prized for its glucosinolate content, which turns into that magical cabbage compound sulforaphane. What does cooking do to it? Well, fresh is best, but steaming's not bad, with microwaving coming in second, and then stir-frying and boiling at the bottom of the barrel. The glucosinolates in other cruciferous vegetables are also significantly affected by boiling. The researchers conclude that red cabbage is best consumed fresh, and look, not just in salads. As I talked about how not to die, I always keep a red or purple cabbage in my crisper to cruciferize my meals, slicing off shreds and putting in basically anything. But if you aren't going to cook it, steaming may be the best bet, so as to retain the optimum benefits of the health-promoting compounds. Other nutrients we look to greens for are the eyesight-preserving nutrients like lutein, which I've talked about before, and folate, particularly important for women of child-bearing age. And vegetables are the main natural source. It's been estimated that approximately half the folate is lost during cooking, which may be true for boiling broccoli, or stir-frying spinach, or mustard greens. But the folate in stir-fried kale holds up better, only losing about a quarter, similar to steamed broccoli florets. But note that broccoli starts out so high that even boiled broccoli has more folate than raw spinach. But check how broccoli leaves. Not only do they start out with among the highest levels, the levels actually go up a bit when you cook them. No one's ever looked at the folate concentration in broccoli leaves, which ironically or commonly just cut off and thrown away, yet contributes great concentrations of this vitamin. Therefore, we should make sure to eat them. Note they also compared thinly sliced kale to kale just torn into larger pieces to determine if a larger surface area of exposure would promote greater losses of folate in kale. However, no effects were found, so slice away. Here they just looked at stir-frying. What about the effects of other cooking methods on kale? There's lots of studies on cooking cabbage and broccoli. However, very little information has been available on the queen of greens until now. First of all, fresh versus frozen. The freezing process is generally regarded as destructive to antioxidant compounds. One just assumes that frozen would have a lower antioxidant capacity compared to fresh, but kale breaks all the rules. The frozen kale showed a higher antioxidant capacity than fresh, and not just by a little. We're talking 60% more. Wow, okay. So what happens when you cook it? If you start out normalizing the starting levels at 100%, blanching and steaming actually boosts the antioxidant content, whereas microwaving or even boiling doesn't seem to do much. So you can boil kale without losing out on its antioxidant punch. I told you kale was a rule breaker, but check out that blanching and steaming. Heat can disrupt that plant cell walls and all the little subcellular compartments and spill out extra antioxidant compounds that may have been hiding. Now, that's usually counterbalanced by losses caused by high temperatures, but the kale compounds are looking pretty cruciferous and stood their ground.


Stop scrolling. Have you ever wondered if you or someone you know does not have autism? But you just do not know what exactly to look out for. Well, you are watching the right video because in the next 60 seconds I'm going to share with you five signs to tell if you or someone you know have autism. The last one is very important So keep watching if you do not want to miss out Let's get started. You're hanging out and you notice someone making great eye contact Not just a quick glance, but really looking at you. Guess what? That's sign number one Good steady eye contact could mean a lower chance of autism because often folks with autism find it hard to maintain eye contact It can make them feel super uncomfortable


So far, we've seen that every permutation can be written as a product of disjoint cycles, and every cycle can be written as a product of transpositions. And we have some evidence that the number of transpositions required reserves parity. But how many transpositions are required? To answer this question, let's consider the parity of the identity transformation. The advantage of looking at the identity is that we know it's going to be in every permutation group. So suppose the identity can be written as the product of n transpositions. We'll show that the number of transpositions must be... Well, let's figure it out. So suppose we could write the identity as a product of n transpositions. Let's consider the last two transpositions, which are actually at the front. Remember, we apply the transpositions from right to left. And so there are three cases. First, those last two transpositions might be the same, A, B, A, B. Second, they might be disjoint, A, B, C, D. And third, they might share one element, A, B, A, C. Let's consider those three cases separately. So since A, B is its own inverse, then this product A, B, A, B is the identity, and we can omit these last two from our product. And thus we can write the identity as a product of two fewer transpositions. The other possibility, if the last two transpositions are disjoint, because they're disjoint, we can switch them. Because the product of disjoint cycles is commutative, A, B, C, D is the same as C, D, A, B. So the last transposition does not move A. Or if the last transpositions share exactly one element, we can rewrite the product A, B, A, C as B, C, A, B. And again, the last transposition does not move A. And so this leads to a lemma, let sigma be a product of n greater than two transpositions with A, B as the last transposition. Then sigma could be rewritten as the product of n minus two transpositions, or n transpositions, where the last transposition does not move A. While we started with the last two transpositions, associativity means that this applies to any two consecutive transpositions. So remember, lather, rinse, repeat. So it appears we can move any element to the first transposition, giving us our identity, easy product, tau prime one, tau prime two, and so on, all the way up to A, x, where none of the towels move A, and x is some other element. But wait a minute, if none of the towels move A, then the identity applied to x is going to give us A, and none of the towels can restore it, so E can't be the identity. What went wrong? The only way to avoid this is to keep the transposition affecting A from reaching the front. And this can only happen if it encounters a copy of itself. And if it does, this will reduce the number of transpositions by two. And so this gives us a second lemma. Suppose the identity could be written as a product of n transpositions, where n is greater than two. Then it could be rewritten as a product of n minus two transpositions. And so now we're ready to conclude. Suppose we can write the identity as the product of an odd number of transpositions. We can reduce the number by two until we have one transposition left. But the identity can't be equal to a transposition. And consequently, we can't write the identity as a product of an odd number of transpositions. And so the identity can only be written as a product of an even number of transpositions. So now let's put it all together. Consider any other permutation sigma. And suppose we could write sigma using an odd number of transpositions, and also using an even number of transpositions. So sigma is some product of an even number of transpositions. And it's also some product of an odd number of transpositions. Now, if we find the inverse of sigma, we can invert either of these two products. And let's just invert the product of the odd number of transpositions. So the inverse will be, and consequently, when we multiply sigma inverse by sigma, we get. And then E, the identity, sigma inverse sigma, can be rewritten as the product of an odd number of transpositions. But that's impossible because the identity can only be written as a product of an even number of transpositions. And consequently, we can't write sigma as an odd number of transpositions, and also as an even number of transpositions. It's got to be one or the other. And so the parity of a permutation is invariant. Now, suppose rho and sigma are two even permutations. Their product will also be even. And so the even permutations are closed under composition. And consequently, the even permutations on n elements form a subgroup of Sn. This group is important enough that it has its own name. It's called the alternating group.


Hello and welcome to the session. Let us understand the following problem today. A, B, C, D is a cyclic quadrilateral. C figure 3.7, find the angles of the cyclic quadrilateral. Now let us look at the figure. This is a cyclic quadrilateral A, B, C, D with the given angles. Now, before writing the solution, we see that sum of the opposite angles of a cyclic quadrilateral is 180 degree. That is, angle A plus angle C is equal to 180 degree and angle B plus angle D is equal to 180 degree. Now this result is our key idea to the question. Now let us proceed with our solution. Given to us that, angle A is equal to 4y plus 20, angle B is equal to 3y minus 5, angle C is equal to minus 4x and angle D is equal to minus 7x plus 5. Now by the key idea, we know that angle A plus angle C is equal to 180 degree by key idea. So it implies angle A that is 4y plus 20 plus angle C that is minus 4x is equal to 180 degree. Which implies 4y plus 20 minus 4x is equal to 180. Which implies 4x minus 4y is equal to 20 minus 180. Which implies 4x minus 4y is equal to minus 160. Now taking 4 common we get x minus 5 is equal to minus 40. This is our equation 1. Now also, here angle B plus D is equal to 180 degree by key idea. Which implies angle B that is 3y minus 5 plus angle D that is minus 7x plus 5 is equal to 180 degree. Which implies 3y minus 5 minus 7x plus 5 is equal to 180 degree. Which implies 3y minus 7x this and this gets cancelled is equal to 180 degree. Which implies 7x minus 3y is equal to minus 180 degree. This is our equation 2. So we have got our 2 equations as x minus y is equal to minus 40 and 7x minus 3y is equal to minus 180. Equation 1 and equation 2. Now multiplying equation 1 by 3 we get here 3x minus 3y is equal to minus 120. And writing equation 2 as it is we get 7x minus 3y is equal to minus 180. Now subtracting them do not forget to change the sign while subtracting. We see that minus 3y plus 3y gets cancelled and we get 3x minus 7x is equal to minus 4x is equal to minus 120 plus 180 is equal to plus 60. Which implies x is equal to minus 60 by 4 which gets cancelled by 15. So which implies x is equal to minus 15. Now substituting x is equal to minus 15 in equation 1 that is x minus y is equal to minus 40 which implies minus 15 minus y is equal to minus 40. Which implies minus y is equal to minus 40 plus 15 which implies minus y is equal to minus 25 where minus and minus gets cancelled so which implies y is equal to 25. Therefore angle A is equal to 4y plus 20 which is equal to 4 multiplied by 25 plus 20 which is equal to 100 plus 20 therefore angle A is equal to 120 degree. Now angle B is equal to 3y minus 5 which is equal to 3 multiplied by 25 minus 5 which is equal to 75 minus 5 therefore angle B is equal to 70 degree. Now angle C is equal to minus 4x which implies minus 4 multiplied by minus 15 therefore angle C is equal to 60 degree. Now angle D is equal to minus 7x plus 15 plus 5 which is equal to minus 7 multiplied by minus 15 plus 5 which is equal to 105 plus 5 therefore angle D is equal to 110 degree. Hence angle A is equal to 120 degree, angle B is equal to 70 degree, angle C is equal to 60 degree and angle D is equal to 110 degree. This is our required answer. I hope you understood the problem. Bye and have a nice day.