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Dt (7 (t) xv(t)) = 0Select one:TrueFalse...

Question

Dt (7 (t) xv(t)) = 0Select one:TrueFalse

dt (7 (t) xv(t)) = 0 Select one: True False



Answers

True or false: $(t+7)(t-7)=(t-7)(t+7) ?$

Phillips whereby reality which fantasies? So we have the loving minus something there has four is for equal to pull This statement is true. You have to and flies Gates dispirited and girl This is also true. So since both of these are true are conditional is

This problem lost. Calculate the last lap last transform of the piece wise to find function at 50 which is equal to a warm Linus T when tea is between 01 and zero when tea is greater than want. In this case, if we just plug in f into the lap last transform will find that it's integral from zero to infinity of each of the negative esty times the function integrated, respected T masses f vanishes When tea is greater than one. You can actually simplify this integral as the integral between zero and one of each of the negative s t times one minus t d t to integrate this, I'm gonna do an integration by pops. I'm gonna differentiate this side anti different, shooting this side differentiating the right hand side. If we do this, then we obtain that this is equal to negative one of s each of the negative s t one minus t evaluated between zero and one plus the integral from their toe one of each of the negative esty. Now the value here at one vanishes because one minus 10 for the value of zero, it doesn't match the value here is long. So you obtain that this is equal to, um, one of the s, um plus Thea, anti derivative of here. Evaluated of one and zero, which in this case, is negative. One of the S s we get. This is Eagle two plus one over s squared Times E to the negative. S t evaluated at zero and evaluated at one. So we get this. Integral is one of the S plus one of the S squared, each of the negative s minus one.

Head over there. So for this exercise we're going to have a statement and we need to say if it is true or false, uh we're going to consider the set of all the positive. All the possible is mid curves and all the values of T. For which the derivative is different from the zero vector. So here we have that. Basically what we're saying here is that the derivative of the curve is perpendicular to the curve at any point. And well, clearly this is false because we know that the first derivative of a curve, it's going to be tangent to the curve. So it's not it's not necessarily true that for all the points and for all the possibles mitt curbs these, there's going to be the origin. So we are going to have a factor pointing to to our point of interest. And here we are going to have a derivative. Right? So it's not a general result that the doctor, the velocity factor, it's going to be perpendicular to the curve. This happened sometimes. Well, in the circular motion where the position vector, it's going to be perpendicular to the velocity vector. But it's a special case is not a general result in this case this statement is false.

Hello there. In this case we need to say if the statement is true or false. Uh And we need to verify the statement for all the positive, all the possible is mid curve and all the values of T. Where the for which the derivative is different from the zero factor. So we need to say if the following statement that say that the first derivative trust product with the it's mid curve at some point T. Is equal to the zero vector. And in this case this is false. Why? Because the only way that across product is equal to the zero vector is that the zero vector is more Is Part one of these two. Uh huh vectors. He is the zero vector. Okay. And this is only possible if but this is not possible actually is a contradiction because we are considering all these positive smooth curves. So not all. Okay. The 00 curve is one special case, but it's not part of the set. So there is not true. And we have specified that the derivative at that point is different from the zero vector. So this is also not true. So it is impossible that this cross product to be equal to zero. So the statement here is false.


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