if
0 = 720,500(1.022)x - 1,548,800
how do I find X?
help is greatly appreciated
-N
0 = 720,500(1.022)x - 1,548,800
how do I find X?
help is greatly appreciated
-N
If its to the power x, take logs is always the key.Noobpatty wrote:
if
0 = 720,500(1.022)x - 1,548,800
how do I find X?
help is greatly appreciated
-N
x = (ln(1,548,800/720,500))/ln(1.022)
edit: lol sad face
double edit: forgot a ln and messed up a digit.
Last edited by presidentsheep (2010-01-26 16:37:55)
I'd type my pc specs out all fancy again but teh mods would remove it. Again.
0 = 720,500(1.022)x - 1,548,800Noobpatty wrote:
if
0 = 720,500(1.022)x - 1,548,800
how do I find X?
help is greatly appreciated
-N
add 1,548,800 to both sides
1,548,800 = 720,500(1.022)x
divide both sides by 720,500
2.1496 = 1.022x
take logs
log1.0222.1496 = x
x = 35.17
Well I kinda made a wrong move,presidentsheep wrote:
If its to the power x, take logs is always the key.Noobpatty wrote:
if
0 = 720,500(1.022)x - 1,548,800
how do I find X?
help is greatly appreciated
-N
x = (ln(1,548,800/720,500))/ln(1.022)
edit: lol sad face
double edit: forgot a ln and messed up a digit.
1548800= 720500(1.022)x
so then
2.14961..=1.022^x
and then
log1.0222.14 = x
right?
right.Noobpatty wrote:
Well I kinda made a wrong move,presidentsheep wrote:
If its to the power x, take logs is always the key.Noobpatty wrote:
if
0 = 720,500(1.022)x - 1,548,800
how do I find X?
help is greatly appreciated
-N
x = (ln(1,548,800/720,500))/ln(1.022)
edit: lol sad face
double edit: forgot a ln and messed up a digit.
1548800= 720500(1.022)x
so then
2.14961..=1.022^x
and then
log1.0222.14 = x
right?
I'd type my pc specs out all fancy again but teh mods would remove it. Again.
eek
got about 20 minutes, anyone fluent in calc?
ive got 2 problems involving l'hospital's rule.
47. lim (x -> 1) ((x/(x-1)) - (1/ln(x)))
I somehow reduced this one to (1/x)/ln(x) + 1 - (1/x), but this leaves me with 1/0, which I don't want. I need 0/0. If I were to derive once more, I'd get the correct answer of 1/2.
64. lim (x -> inf) ((2x-3)/(2x+5))^(2x+1)
Got this to (2x+1)*(ln(2x-3) - ln(2x+5)) which is also no bueno. need lny = -8 so that y = e^-8.
Wolfram is no help on either of them. Thoughts?
got about 20 minutes, anyone fluent in calc?
ive got 2 problems involving l'hospital's rule.
47. lim (x -> 1) ((x/(x-1)) - (1/ln(x)))
I somehow reduced this one to (1/x)/ln(x) + 1 - (1/x), but this leaves me with 1/0, which I don't want. I need 0/0. If I were to derive once more, I'd get the correct answer of 1/2.
64. lim (x -> inf) ((2x-3)/(2x+5))^(2x+1)
Got this to (2x+1)*(ln(2x-3) - ln(2x+5)) which is also no bueno. need lny = -8 so that y = e^-8.
Wolfram is no help on either of them. Thoughts?
@Bevo
If you're reading this then you're probably in class already ... sorry, I got bored and started watching House M.D instead
If you're reading this then you're probably in class already ... sorry, I got bored and started watching House M.D instead
Np. I figured out the first one and got close on the second. For the first one I accidently derived the top and distributed the bottom without deriving both. The second... Well I'm at a loss.
lol, fucking hellBevo wrote:
Np. I figured out the first one and got close on the second. For the first one I accidently derived the top and distributed the bottom without deriving both. The second... Well I'm at a loss.
He asked at the beginning of class if we had any questions over the HW. I said 64. He did the first half of the problem exactly like I did, and he was like "you should be able to finish it"
I just worked out what I could and put down an incorrect answer. He's a terrible TA.
Fuck, I hate calculus. Limits were the first thing we learned, and I still don't understand what the frig I'm doing.Bevo wrote:
eek
got about 20 minutes, anyone fluent in calc?
ive got 2 problems involving l'hospital's rule.
47. lim (x -> 1) ((x/(x-1)) - (1/ln(x)))
I somehow reduced this one to (1/x)/ln(x) + 1 - (1/x), but this leaves me with 1/0, which I don't want. I need 0/0. If I were to derive once more, I'd get the correct answer of 1/2.
64. lim (x -> inf) ((2x-3)/(2x+5))^(2x+1)
Got this to (2x+1)*(ln(2x-3) - ln(2x+5)) which is also no bueno. need lny = -8 so that y = e^-8.
Wolfram is no help on either of them. Thoughts?
Well erm this is Calc 2, so don't expect to know how to do these.Ryan wrote:
Fuck, I hate calculus. Limits were the first thing we learned, and I still don't understand what the frig I'm doing.Bevo wrote:
eek
got about 20 minutes, anyone fluent in calc?
ive got 2 problems involving l'hospital's rule.
47. lim (x -> 1) ((x/(x-1)) - (1/ln(x)))
I somehow reduced this one to (1/x)/ln(x) + 1 - (1/x), but this leaves me with 1/0, which I don't want. I need 0/0. If I were to derive once more, I'd get the correct answer of 1/2.
64. lim (x -> inf) ((2x-3)/(2x+5))^(2x+1)
Got this to (2x+1)*(ln(2x-3) - ln(2x+5)) which is also no bueno. need lny = -8 so that y = e^-8.
Wolfram is no help on either of them. Thoughts?
Limits are annoying as fuck, indeed. We did 3 limit problems today in about 45 minutes. The last 25 minutes the TA spent writing and explaining the last problem... half of the class was laughing towards the end because we had no fucking idea what he was doing. I got lost somewhere after 15 minutes and just tuned him out completely.
@Bevo Do you know what the answer to the second one is meant to be?
e^-8, wolfram saysliquidat0r wrote:
@Bevo Do you know what the answer to the second one is meant to be?
I got inf/8 once, and then when I re-did it I got 1/1.
Yeah I keep getting 1/1
Well I'm in Calculus AP, and we are starting this shit in the next semester (which starts Monday).Bevo wrote:
Well erm this is Calc 2, so don't expect to know how to do these.Ryan wrote:
Fuck, I hate calculus. Limits were the first thing we learned, and I still don't understand what the frig I'm doing.Bevo wrote:
eek
got about 20 minutes, anyone fluent in calc?
ive got 2 problems involving l'hospital's rule.
47. lim (x -> 1) ((x/(x-1)) - (1/ln(x)))
I somehow reduced this one to (1/x)/ln(x) + 1 - (1/x), but this leaves me with 1/0, which I don't want. I need 0/0. If I were to derive once more, I'd get the correct answer of 1/2.
64. lim (x -> inf) ((2x-3)/(2x+5))^(2x+1)
Got this to (2x+1)*(ln(2x-3) - ln(2x+5)) which is also no bueno. need lny = -8 so that y = e^-8.
Wolfram is no help on either of them. Thoughts?
Limits are annoying as fuck, indeed. We did 3 limit problems today in about 45 minutes. The last 25 minutes the TA spent writing and explaining the last problem... half of the class was laughing towards the end because we had no fucking idea what he was doing. I got lost somewhere after 15 minutes and just tuned him out completely.
I got 32 out of 50 on the midterm, but our teacher lowered the total from 60 to 50, otherwise everyone would have pretty much failed. She's a fucking shit teacher.
Calc AP =/= Calc 1 in collegeRyan wrote:
Well I'm in Calculus AP, and we are starting this shit in the next semester (which starts Monday).
I got 32 out of 50 on the midterm, but our teacher lowered the total from 60 to 50, otherwise everyone would have pretty much failed. She's a fucking shit teacher.
You may get the credit, but my friends who took Calc AP and passed the test fine went and took calc 1 in college because they said calc 2 was far too hard. It's not really the same thing.
Don't worry, it's difficult. I breezed through every area of math before calc and I barely squeaked by with a B last semester in calc 1.
can someone help me with binary, specifically conversion when the number is decimal.
something like something this.
1234.789
i got this so far
10011010010.???
something like something this.
1234.789
i got this so far
10011010010.???
fuck i hated binary...lxcpikiman wrote:
can someone help me with binary, specifically conversion when the number is decimal.
something like something this.
1234.789
i got this so far
10011010010.???
I'll look in my old textbook to see if i can remember anything bout decimals.
I'd type my pc specs out all fancy again but teh mods would remove it. Again.
102 101 100 . 10-1 10-2 etc.
.25 = 2(10-1) + 5(10-2)
23 22 21 20 . 2-1 2-2 etc.
.75 in binary is .11 because it =
1(2-1) +1(2-2) = 1(1/2) + 1(1/4)
.25 = 2(10-1) + 5(10-2)
23 22 21 20 . 2-1 2-2 etc.
.75 in binary is .11 because it =
1(2-1) +1(2-2) = 1(1/2) + 1(1/4)
Google ftwlxcpikiman wrote:
can someone help me with binary, specifically conversion when the number is decimal.
something like something this.
1234.789
i got this so far
10011010010.???
http://acc6.its.brooklyn.cuny.edu/~gurw … 2tool.html
1234.789 = 10011010010.110010011111101111100111011
so 1234.789=10011010010.1100101??Flaming_Maniac wrote:
23 22 21 20 . 2-1 2-2 etc.
well i think that =1234.7890125 but i guess youre rounding
edit:please recheck as that was in my head
edit2:ignore this, the unrounded answer is about 2 inches above it
Last edited by mkxiii (2010-02-09 11:58:44)
yeah you got the idea, faggot teacher that gave you a stupid pattern instead of something that actually terminates after a reasonable amount of numbers after the binary point...
thats what i was thinking, i figured if it was just to test whether you could do decimals it would surely terminate after 5 or 6Flaming_Maniac wrote:
yeah you got the idea, faggot teacher that gave you a stupid pattern instead of something that actually terminates after a reasonable amount of numbers after the binary point...
yeah i could have used a binary converter but i wanted to learn how to do it.
andi think i got it now.
andi think i got it now.
Calling all liquidators.
Doing a series sum (n = 1 -> inf) e^n / 3^(n-1)
Ive figured out that the terms approach zero (e/3 < 1 and raised to the power of some n), but it has to approach 0 and be decreasing both to be convergent. The problem is, I have no idea what the difference is. The terms decrease... towards zero, just like the sum of 1/n would. But that's divergent.
what the hell is the difference?
Doing a series sum (n = 1 -> inf) e^n / 3^(n-1)
Ive figured out that the terms approach zero (e/3 < 1 and raised to the power of some n), but it has to approach 0 and be decreasing both to be convergent. The problem is, I have no idea what the difference is. The terms decrease... towards zero, just like the sum of 1/n would. But that's divergent.
what the hell is the difference?