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Kea QA4aGithoFrom Ihe giventapla(x)AiaOul +LHag qustionMeeting now 00o5...

Question

Kea QA4aGithoFrom Ihe giventapla(x)AiaOul +LHag qustionMeeting now 00o5

Kea QA4a Githo From Ihe giventapl a(x) Aia Oul + L Hag qustion Meeting now 00o5



Answers

The orcs of $A g$ and Cu arc concentrarct using rhcir solubiliry in (a) $\mathbf{K C N}$ (b) $11 \mathrm{NO}_{3}$ (c) $11_{2} \mathrm{SO}_{1}$ (d) $11 C l$

Okay, so we're asked his objective falling. So since one is just in to we need to borrow. So we get to and this is now 11. So 11 minus two is nine. And now, since two is us and eight, we have to borrow. So this is a five, and it turns into a 12 12. Minus eight is for Okay, So now since five minus 85 is less than five minutes borrowed, we get a one here. And in the 15 year 15 a minus eight is seven. And since when is less than seven, we again need to borrow. So this has turned into four. And now we have 11.7, which is for and in four minutes for is zero and two Do not equal to okay.

Okay, so we're asking Add the falling. So let's start with our 1st 2 terms. And actually, it's put our 43288 on top since its larger. And then we have Gates. 2651 So that's 8 to 6 Close. Eight is Fortune. 75678 four and four. And now let's add 53626789 10 11 12 13. Fortune 3456 49 h for four. So we get the following, which is equal to he. Action. Now let's not be there. H So it's, um 448 964 So that's actually de

Nothing to prove this statement. Term writer Seances de a C this larger. I'm going. That whole thing you've seen this equals two d e. Clause e A and bleed. So the smaller angle. Okay, if you can remember, this is known as the condition Post to lead. I have good was similar. Thank you. That's is this slaughter Frankel, which is a hard cred. They still 16 plus 40. So what kind of a post you later under talking here simply, I trust you.

Okay, so this is problem number 14. 14 is saying that quadrilateral A B C D. Is circumscribed about a circle discover improve our relationship between a B plus D c and a D plus B C. So if we have the circle and we have a cutting through, it's a tangent, so a envy. And then we have a d and we have d. C. And so this is See right here. We're gonna erase these guys right here. OK, so that's our picture. Our diagram. Okay, so if we know that are intersecting parts, we're going to make this one. P are our tans are where they're Tanja nap. This is gonna be cute. This one will be s and we're gonna make this one. Are some of we know that we know that, um, ap It's gonna equal to a Q, so we know that a P equals a cue. We also know that de que is gonna equal to D. R. We also know that r c it's gonna equal to see us. And we also know that, um, SB is gonna equal to B p. So we also know that a B is equal toe AP plus PB. So if we do this and we also know that, um, let me erase this here quick, we also know that, um do you see is equal to D R plus r c. We also know that a D is equal to a Q plus Q D. And we know that VC is equal to be us plus s C. So if we take a B plus d c, we know that that is gonna be the same as saying a p plus PB, and we can substitute the d r plus the R. C. Okay, so if we know like we said, a peak was a cute So let's substitute these. We could make a queue and peopie it's the same as saying SP or B s. However you want to say it and we know that d r is the same thing as D Q and R. C. It's the same of see us. So if we look at this a que plus B s, if we rearranged, um, we're going to rearrange them. So a cue plus q d plus B s plus See us. That's gonna be the same thing as saying a D because a D equals a Q plus Q D plus B C because B s plus see us equal specie. So that just proved that a B plus d C. It's gonna equal a d plus B C. And that's what they were trying to prove they wanted youto prove that relationship.


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