5

03310.0 pointsFinal le th-lerivalive. fl}(} olfir)f"(_)r! (6+#) In |4)fo() ({-1)"x' (6 + r)-(a+l}ff"(~)T)i6 | ~}ff")(=)(v+ 1J (G+1)-6"...

Question

03310.0 pointsFinal le th-lerivalive. fl}(} olfir)f"(_)r! (6+#) In |4)fo() ({-1)"x' (6 + r)-(a+l}ff"(~)T)i6 | ~}ff")(=)(v+ 1J (G+1)-6"=1)r(w) (x) (-1) "(n 191(6 +%)-"fin) (r)(-)"(n 1)(6 | ~)

033 10.0 points Final le th-lerivalive. fl}(} ol fir) f"(_) r! (6+#) In |4) fo() ({-1)"x' (6 + r)-(a+l} ff"(~) T)i6 | ~} ff")(=) (v+ 1J (G+1)-6"=1) r(w) (x) (-1) "(n 191(6 +%)-" fin) (r) (-)"(n 1)(6 | ~)



Answers

Redraw Figure $10-6$ with the goal of adding one $E c o \mathrm{RI}$ end and one $X h o \mathrm{I}$ end. Below is the Xhol recognition sequence.

So for this question, we need to remember that slope is the measure of the rise over the run or the change in the Y direction divided by the change in the X direction. So for a the line that we would have would look something roughly like this. So okay, our slope is going to be positive and it's going to be somewhat steep, So out of the possible options that we have, um, the positive options are only Sarah 0.5 or is there a 0.85? So we know that it's a fairly steep lines or slope is going to be 0.85 for be the line that we would have. It's going to look something like this. We see that it's still a positive line, so it will have a positive slope, but it's a little bit less steep this time, so that leaves us with our other option for a slope of 0.5 for C. Our line is going to look something like this and we see that there is no direction to this line, so the slope is therefore going to be zero, since it's not positive or a negative. It's not a positive or a negative slope. Therefore, the slope zero for a D R line would look something like this, and we realize that we're we've got a negative slope that is somewhat steep, which means that our option that we choose is going to be negative 0.45 since that wine was going to be steeper than the one we would get for E, which would look something like this. So that leaves us with 0.30 as our option for E.

For this exercise not to use the following when we use the value t cause one half we have. If zero off one half is we'll have a P zero plus one half off B one on this expression represents the middle point between P zero and P one, which you can see now here in this scheme on the same for if one. So if one of one half is one half off p one plus one half off B two, we choose the middle point between p one and P two aunt the same freaky off one half. So this is one half off if serial off one half plus one half of if one of one half on this expression represents the middle point between F 01 half and if one of one half. So this is the picture that we were asked for.

Destroyed the electron configuration for the group E elements, um one A all the way down two eight A. And we're gonna use the NSX n p Y notation. So for group one a, we have n. s. one To a BNS two three A would be NS. two. and he won For a would be NS two, NP 2 five A&S. 2, and P 36 AMS two, N p fourth 7, 2, NP five and 8 ASS two, N p six. Uh huh.

Okay, so this problem wants you to take the Grady in of the function I have written here at the 0.111 So to calculate the grading of a function, all we have to do is to write a vector comprises of its partial, the routers with respect to X Why and see. So let's go ahead and calculate them. So first part of their derivative that we have think of the function is X and there are two terms here and here. These two terms are have the X variable in them. So we have differentiate those terms. Everything else is constant. So those go away. If we differentiated, we get negative. Six eggs e plus the are tendon functions. So if we differentiated, that would end up with Z in the numerator. And in the normally we would get XY squared, plus one. All right, so that's the derivative with respect to X. Now, if we took a derivative with respect, why, well, we have to differentiate is the middle term since that's the only term with wine it and that would just simply give us native six. Why is he And now we have to finally take the derivative with respect to Z. And since all the terms have Xena and we have to take the way to differentiate all the terms and this would give us this would give us 60 squared for insert for the first term minus three times X circles Weiss Word for the second term and for the final term, which is the Ark tendon term again. Well, give us exit the numerator. I don't x'd squared plus one in the nominator. And this is our radiant in terms of X, y and Z. So all we have to do now is plug in the point 111 into our Grady in to find the green at the point p. So we were to plug in 111 into our great and we would get No, you're six. Close 1/2 for X component. We get just native six for a white component and then finally, Farsi component, we get six minus six post 1/2. And if we summed it all together and we would get a vector off native 11 half, I had minus six j hat plus 1/2 que hat. And this is your final answer


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