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12 ) J 7-8+25 5 13) Jccsct-) siittsa< 14) Givem fex) _ X-6x+7 Ha eoate 4+ Exe Io #" cuwe e Po)...

Question

12 ) J 7-8+25 5 13) Jccsct-) siittsa< 14) Givem fex) _ X-6x+7 Ha eoate 4+ Exe Io #" cuwe e Po)

12 ) J 7-8+25 5 13) Jccsct-) siittsa< 14) Givem fex) _ X-6x+7 Ha eoate 4+ Exe Io #" cuwe e Po)



Answers

$$\frac{3^{2} \cdot 4^{5} \cdot 5^{3}}{5^{3} \cdot 3^{3} \cdot 4^{6}}=$$ (F). \frac{1}{60} (G). \frac{1}{12} (H).\frac{3}{4} (J)$$12$$

Okay. We won't walk through the process of finding, um, the center of Mass, um, of a given system of point masses. And what I'm gonna do is I'm gonna go ahead and put these in the corn. It plain. So here is the cornet playing and make sure I have enough here, okay? And so the first point mass is at two three. So there is my first point mass. And that point mass is, um, a value of 12 and is at 23 Then I have one at negative one five. And so that point mass is six. And he's that negative. 15 Then I have 1/3 1 at 68 and I believe eight was a here. And that point Mass is four and 1/2 and occurs at 68 And then I have a point mass of 15 at two. Negative too. So there are my four point masses, and, um so just start finding the center of mass. I need the total mass, so we'll just take each of the masses and add them together. And so I get a total of 37.5, Then I want to find the moment about the why axes. So I want how, um, each of these point masses or distributed about the vertical axis, so that's gonna be dependent upon how far they are from the Y axis in the ex direction. So, um, the moment in the Y axis to pin him at the Y axis is each of the point masses times their respective X values. So we take each of the point masses and multiply and by the respective X values. And when we do that, we get a value, uh, 75. Then we're gonna find the moment about the X axis. And so, if I'm looking about at the moment about the X axis, I'm trying to see how these point masses are arranged about this X axis. So that's gonna be dependent upon their why value. And so my take each of the masses and multiplying by the respective why values? Um, because that's gonna be their distance away from the X axis. And then when we do that, we get a total or combined of 72. And so to find the, um X coordinate at the point mass, it is going to be the moment about the Y axis over the total mass. Um, so we get 75 over 37.5, which gives me a two to find. The Y coordinate is the moment about the X axis to buy by the total mass, which is going to give me 72 over 37.5, which becomes 48 over 25. So my center of mass is going to be two comma, 48 over 25. And so let's go ahead and taught him in. And so that means 48 over 25. So it's at two comma, just a little over two would be 50. So it's just a little under two, so right, right there.

In this video would go through the answer to question number 15 from chapter 9.4 where were asked to find the values off T um, for which thes Baptist A and B are miliary dependence on Will you invest it? So Okay, let's have a look at this. So a can be written as each the tea times by minus three minus 15 at times by minus the bird. Right? Because this guy, it is just the back to warm five. What's this? This is just minus 1/3 times by B and we haven't specified what by your tea we have yet. So for any value of tea, then you can write a year. Someone you think should be so. Therefore, they are many early independent afternoon off the realign, so 40 linearly independent for tea and empty set. There's no value

For J Plus 12.54 equals 18.12 I'm just gonna work with the decimals here because although it's easier, I'm going to subtract 12.54 from both sides and I'm left with four. J equals 5.58 Divide by four Divide by four J. Walls one point 39 lied.

In this video, we're gonna go through the answer to question of a 13 from chapter nine. Fight forthe where we asked a show in order to determine whether thes to vector functions are linearly dependent or linearly independence on the integral. Uh, t between minus infinity. Infinity. Okay, so we're basically looking for it. Ah, value of tea for which at some multiple off b is equal to So some what? Some constant Alfa B is equal to a for that given by your tea. Okay, so let's have a look to that. So just just by observation, we can see that, Um, if t is equal to 12 then a is gonna be equal. Thio 12 three, which is equal to three times a day s. So therefore, when t is equal to 12 then A and B are linearly dependent because you could write a has some linear multiple off B for any of the value of t. You can't do this. They're gonna be at linearly independence. So we'll say that they are linearly independence t in the sets minus infinity too infinity without about you 12


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