Recent content by gmmstr827

Nvm, I figured it out. Thanks anyway.

Thanks.

Alright, because the professor I have right now thinks those are two completely different things. He reads it as "there exists an x such that there exists a y (for that specific x)." This is the second time I had the class (it didn't transfer colleges) and my previous professor said they're the...

Is saying \existsx, \existsy the same as saying \existsy, \existsx?

Okay, thanks. I was wondering about the inverted image, though the result seemed better.

Okay, thank you!

d_obj = do = 25 ft = 300 inches h_img = hi = -3 inches h_obj = ho = 6 ft = 72 inches hi/ho = -di/do -3/72 = -di/300 di=12.5 inches 1/do + 1/di = 2/r 1/300 + 1/12.5 = 2/r r=24 inches So the radius of the mirror is precisely 24 inches? That still seems like a rather large convenience store mirror.

The problem: "A convex spherical mirror is 25 ft from the door of a convenience store. The clerk needs to see a 6 ft. person entering the store at least 3 inches tall in the mirror to identify them. What is the radius of the mirror?" d_obj = do = 25 ft = 300 inches h_img = hi = 3 inches...

Yep. I just didn't realize how to change the expression until I looked at examples enough and finally noticed you simply substitute >.< Thanks for your help.

1+2(2+3+4+···+k+k+1)+(k+1+1) 1+2(2+3+4+···+k)+2(k+1)+(k+1)+1 1+2(2+3+4+···+k)+(k+1)+2(k+1)+1 Recall: 1+2(2+3+4+···+k)+(k+1)=(k+1)2-1 (k+1)2-1+2k+2+1 k2+2k+1-1+2k+2+1 k2+4k+3 k2+4k+4-1 (k+2)(k+2)-1 (k+2)2-1 LHS=RHS Therefore, 1+2(2+3+4+···+n)+(n+1)=(n+1)2-1 for all n within N.

Wait, you just substitute the part of the LHS that existed before adding +1 to both sides for the RHS before adding the +1? If so, this is so much less complicated that I thought. I was under the impression you had to do some kind of summation or use a formula to convert it into the values you...

The problem is that I have no idea where to find formulas to convert the expression 2(2+3+4+···+k) into a term. I know that I need to end up with k^2+4k+3 in order to show the RHS, but it seems like I'll be getting too many values. I've been looking through my book and trying to research it...

Would 2(2+3+4+···+k+k+1) = k(k+2)+k+1? So it would all equal 1+k(k+2)+k+1+k+1+1 =k^2+4k+4 However, I need it to equal k^2+4k+3 Which I can get if I don't add an extra 1. So I use: 1+2(2+3+4+···+k+k+1)+(k+1)=(k+1+1)2-1 instead of 1+2(2+3+4+···+k+k+1)+(k+1+1)=(k+1+1)2-1

First, by using n=2 for the basis, I confirmed that the equation is true. Then, Assume true for n=k 1+2(2+3+4+···+k)+(k+1)=(k+1)2-1 for all k within N Show true for n=k+1 1+2(2+3+4+···+k+k+1)+(k+1)+(k+1+1)=(k+1+1)2-1 I'm trying to contradict this somehow, and went through with it unsuccessfully...

That's what I was thinking. So since 1 is a natural number, and the problem states that the equation is true for all n in N-the set of natural numbers, should it be seen as 1 not being included in n, and therefore not being in N?