#precalculus

1 messages · Page 295 of 1

agile gyro
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$G(x)=\int^x_1 f(t)dt = F(x)-F(1)$

obsidian monolithBOT
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Syst3ms

quick bluff
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Yes

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F(1) is constant

agile gyro
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And so what you did was say $G'(x) = F'(x)-F'(1)$

obsidian monolithBOT
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Syst3ms

quick bluff
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Ah

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Got it

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Thanks a lot hype

agile gyro
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one final note

quick bluff
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Sure

agile gyro
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$F : x\mapsto \int_a^x f(t)dt+\lambda$ ($\lambda$ some constant) is a sort of "canonical" look for an antiderivative

obsidian monolithBOT
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Syst3ms

sick steppe
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when you use lambda over c sully

agile gyro
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so, whenever you see something like this, you can immediately go "ah, F's an antiderivative of f, so F'=f$

quick bluff
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Okok

agile gyro
quick bluff
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I think I get it

agile gyro
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Whenever you're not sure, take an antiderivative and write the integral in terms of it

agile gyro
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np

quick bluff
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Just to verify if I got it, would F'(x) be x^2 . g(x) in the pic I gave?

agile gyro
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correct

quick bluff
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And if the upper bound had been x^2 it will become x^4 . g(x^2) . 2x?

agile gyro
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yeah, that sounds right

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proper chain rule

quick bluff
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And if the upper and lower bounds of an integration is the same it gives 0 right? kek

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Since the area at a single point will be the area of a line which is 0?

agile gyro
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yup

quick bluff
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Thanks for bearing with me, helped me a lot

agile gyro
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F(a)-F(a) = 0 in other words

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np

viscid thistle
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anyone know how to do this?

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@everyone

haughty lance
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have you tried using the quadratic formula? (if you can't right away, what's stopping you)

viscid thistle
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yeah i have

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i switched the sec squared to 1+tan squared and put it in the quadratic formula

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idk what im doing wrong tho it keeps saying im wrong

haughty lance
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did you consider the interval when taking the square root?

viscid thistle
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i didnt take a square root

haughty lance
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how did you use the quadratic formula ?

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there's a portion thats +- sqrt(b^2 - 4ac)

viscid thistle
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(tantheta-4)(tantheta+2)=0

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oooooooooooooooooh

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i understand lol

haughty lance
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and the plus minus is an issue with trig stuff

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so you could easily end up in the wrong quadrant

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if one doesn't think about it

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but also, you get multiple answers if that's what you want

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depending on context

viscid thistle
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yeah, i see

haughty lance
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and if it's electronic you might have to account for periodicity

viscid thistle
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im not that advanced yet

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lol

haughty lance
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i don't know what they want you to put

viscid thistle
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just the solution set

haughty lance
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oh no that's just the +2πn

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some systems want that

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some don't

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i think ?

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not sure lol

viscid thistle
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im pretty sure i got it now thanks bro

haughty lance
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cool

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np

viscid thistle
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can someone just tell me if they're all correct

summer knot
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Yeah, the points should be at the same location.

viscid thistle
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great thanks

fossil hazel
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Someone help evaluating this integral, exact answer not decimal and with full steps

molten garnet
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Is there an easier way of solving this

flint ingot
molten garnet
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Im currently doing the factor tree for each number

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its too long

flint ingot
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Hm,

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I'm not sure if this would be more helpful,

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but I would solve this by writing down the first few primes (1, 2, 3, 5, 7, 11, 13) and try to see which numbers are divisible by them off the bat

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1 doesn't matter,

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2 is quite obvious (even or uneven number),

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each number and its properties

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I suppose it's a little too similar,

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let me see if I can maybe find something else

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Not quite sure if there is another method, sorry

rough carbon
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Write all the numbers as prime factors
So (7x5)^4 * (7x2)^6 * …
Then you’ll find that powers of numbers will cancel

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I’m sure that’s what you were doing anyway though

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So I’m not really sure of a shorter way

uncut mulch
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Im currently doing the factor tree for each number
these numbers are relatively small and have very few factors, you shouldn't need to resort to drawing a factor tree for these

molten garnet
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hmmm

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alright

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thanks

molten garnet
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How would apply the law of indices here when there is plus n minus>

uncut mulch
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expand it like you would
(a + b)(c - d)

viscid thistle
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could someone tell me if i got any wrong

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not confident with question 19

sick steppe
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$\frac{7e^{\frac{3\pi i}{5}}}{0.5e^{\frac{\pi i}{5}}}$

obsidian monolithBOT
sick steppe
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apply power laws

viscid thistle
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so 19 is wrong

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are the others?

subtle ore
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20 is

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I think you may have read 9pi/8 as pi/8

wintry eagle
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Help?

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Anyone

quick bluff
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The 9th question

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The answers are given as B and C

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I'm able to prove A wrong and B correct

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But am not sure how to go about solving C and D

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I think I can say that D is incorrect as f(x) is increasing in (0, infinity) and therefore has no upper limit

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Here's what I have written so far

quick bluff
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<@&286206848099549185>

fleet yew
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Lets put C in simpler words

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For sufficiently large x, the derivative of the function is always smaller than the function

quick bluff
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thonk ok

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magnitudes right?

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oh wait f'(x) is always positive

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im still not getting any ideas sully

pliant gyro
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wait

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@quick bluff

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bro this question

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is more like u cant probably prove it with absolute method in exam

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but there is a more theoritical way

quick bluff
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what is the "theoretical way"?

pliant gyro
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look

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first of all lets not bother about that sine term

quick bluff
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in the first derivative?

pliant gyro
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it wont affect much when we take higher values

quick bluff
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or everywhere?

pliant gyro
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in every thing

quick bluff
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ok thonk

pliant gyro
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we basically have to prove f(x)>f'(x)

quick bluff
pliant gyro
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look at part C

quick bluff
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yes, cant it also be possible that f(x) is negative but of a greater magnitude?

pliant gyro
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if alpha>1 that means x>1 and modulus will open with a positve sign

quick bluff
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oh yeah

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f(x) can't be negative then

pliant gyro
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now if it depends on lnx and 1/x

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just look at the rates of increase or decrease of them

quick bluff
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ln x increases and 1/x decreases?

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so there will be point where they intersect?

pliant gyro
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yep

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precisely

quick bluff
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thinkies makes sense

pliant gyro
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and for the d part i proved it wrong

quick bluff
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im still skeptical about the "sine term removal" you did tho

pliant gyro
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nah when terms get lager they become insignificant

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the sine term integral cannot be more than 3 or 4

quick bluff
pliant gyro
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it is

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but the problem lies in the values like 20 to 80

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mb there f'(x) goes up and comes down again

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we dont know

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and u can think of it from other methods

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too

quick bluff
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thonk ok

pliant gyro
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im not sayin that my method is the only best method

quick bluff
pliant gyro
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ya i just thought of it while havin lunch

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i did not write it on paper

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you're in allen??

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@quick bluff

quick bluff
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Yes

pliant gyro
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ask ur teacher and also tell me afterwards

quick bluff
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Sure

unique pagoda
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1/sqrt(3x-1) = (3x-1)^(-1/2)

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do you see why?

knotty crescent
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Could someone help me out with this?

unique pagoda
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you can take constants out during differentiation. here you have a constant factor of 1/2

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so d/dx (3x-1)^(-1/2)/2 =
(1/2) d/dx (3x-1)^(-1/2)

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no problemo

near epoch
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can someone pls help me do this using conics?

analog ravine
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<@&268886789983436800> why is math so hard aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

somber yew
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Why are you pinging moderators for this

analog ravine
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can you make math easy?

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pls

somber yew
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I suggest you to not continue off-topic chitchat here. Move to #chill .

unborn blade
#
1951    4000000
1952    6000000
1953    9000000
1954    12000000
1955    16000000
1956    21000000
1957    26000000
1958    32000000
1959    39000000
1960    47000000
1961    56000000
1962    67000000
1963    80000000
1964    95000000
1965    112000000
1966    132000000
1967    155000000
1968    182000000
1969    214000000
1970    249000000
1971    287000000
1972    331000000
1973    382000000
1974    434000000
1975    480000000
1976    534000000
1977    593000000
1978    657000000
1979    728000000
1980    798000000
1981    870000000
1982    943000000
1983    1023000000
1984    1109000000
1985    1199000000
1986    1295000000
1987    1399000000
1988    1509000000
1989    1623000000
1990    1743000000
1991    1867000000
1992    1999000000
1993    2136000000
1994    2287000000
1995    2443000000
1996    2611000000
1997    2791000000
1998    2979000000
1999    3181000000
2000    3394000000
2001    3612000000
2002    3843000000
2003    4084000000
2004    4340000000
2005    4603000000
2006    4883000000
2007    5178000000
2008    5459000000
2009    5747000000
2010    6060000000
2011    6385000000
2012    6723000000
2013    7075000000
2014    7442000000
2015    7823000000```

how do I derive a formula based on the above data set?
i ended up with  f(x)=2.023*1.136^x, but my instructor said it was incorrect
haughty lance
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for sure it's wrong

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but think about why though

unborn blade
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i just followed a technique i found on a khan academy vid

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how do u recommend i do it?

unborn blade
haughty lance
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i mean it's just hard to discern how the dataset and the equation was derived

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if you show what you did it's easier to point out where the misstep was

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like where does 2.023 come form

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i think it's worth revising what function you're trying to use f(x) = ab^x? some growth with a rate? how are you deriving your terms

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and then think about it in simple terms

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dont think of a huge dataset

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what text are you using, it's always good to go back to that

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which video did you use?

unborn blade
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that is the video I used

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i followed the exact procedures as outlined in that video and ended up with the formula I have

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the data set seems to show an exponential function of some sort, which is why i tried to use f(x)=ab^x form

unborn blade
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it basically just asks us to find some sort of data set online that is a function of degree 3 or higher, or an exponential function, or trig function

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and then to formulate an equation based on the data set that roughly models the data

viscid thistle
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did i do this correctly

fleet yew
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500cos(30), 500sin(30)

inland flower
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anyone know how to find angle?

toxic flame
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@inland flower apply the law of cosines

shut timber
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how would u get the solution

uncut mulch
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you could first consider the product of the constant components and eliminate some choices

shut timber
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yea but going from just the equation given

toxic flame
obsidian monolithBOT
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coycoy

shut timber
toxic flame
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factor out x^2 from the first two terms and -7 from the last two terms

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just continue factoring after that

shut timber
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oh

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wow

obsidian monolithBOT
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Jaba Teresinha

narrow lichen
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how do I solve this system of equations?

toxic flame
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multiply the top by y and the bottom by -x. then add the two equations.

obsidian crown
#

Can someone please answer my Pre-Calc question:

solemn elbow
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we cant answer, but we can help

obsidian crown
haughty lance
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this is some heavy precal...

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feels not american...

grand lotus
viscid thistle
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${(-1 + sqrt(5))/2}$

obsidian monolithBOT
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∂T/ ∂t = α ∇²T

willow bear
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did you mean $\frac{-1 + \sqrt{5}}{2}$

obsidian monolithBOT
slim echo
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I'm having trouble understanding how to get my last term

fleet yew
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@slim echo divide all terms by 5

slim echo
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ok

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I understand what you're saying but still unsure how this leads to my last term

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does the n=20 just mean that whatever term I get when when I get to the 20th term will be my last term

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nvm I was able to figure it out but thank you anyways

lament sandal
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did you mean 3x^{2}+2x+5

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o it doesnt work

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:/

viscid thistle
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did i do this correctly

dawn kestrel
viscid thistle
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this good?

willow bear
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you should not use the letter x for multiplication

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and also you didn't do what was asked:

Write a full sentence to describe the recurrence relation of the sequence

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$a_{n+1} = 3a_n + 1$ is what you meant for the first one

obsidian monolithBOT
viscid thistle
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i had a sentence but i don't think i phrased it correctly

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so i just ended up deleting it

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Every value is 3 times greater, plus one than the previous value.

wary mulch
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idk how to approach this sum can anyone help me

echo wagon
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You need to find $$\lim_{x \to 0} f(x)$$and set f(0) equal to that

obsidian monolithBOT
#

Lunasong the Supergay

echo wagon
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||hiding so no one sees, but do you know Lhospitals rule?|| @wary mulch

echo wagon
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That's probably the easiest way

wary mulch
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OK

echo wagon
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The only other way I can think of is a little ugly

wary mulch
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wats tat

ivory loom
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why not just make x=0 though?

wary mulch
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it will come as 0/0

ivory loom
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would it? my brain must not be functioning today xD

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ohh fuck

ivory loom
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ignore me, my bad

echo wagon
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Nvm what I was thinking doesn't work

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Just do ||LH|| if you know it

uncut mulch
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does that even end well here?

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i'd probably approach it with the factorisations for a difference of two powers

echo wagon
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Oh, I was doing that, then it didn't work put, but I was dumb and only did it for the one power lol

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Yes, that's the other way, but I mean, it's hella messy with an eighth root

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Which I why recommend ||LH||

wary mulch
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i got ans as 1/5

echo wagon
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I don't think that's right

wary mulch
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is this crct

echo wagon
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You didn't use chain rule when differentiating

wary mulch
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my teacher didnt teach differentiations yet

echo wagon
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Oh, so you aren't supposed to use LH?

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You can't use LH if you haven't done differentiation

wary mulch
#

ohhh ok

echo wagon
wary mulch
#

ok

uncut mulch
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its actually not that bad

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there's heavy simplification

echo wagon
#

Multiply the numerator and denominator by
$[(256-7x)^\frac78 + 2(256-7x)^\frac68 + 2^2 (256-7x)^\frac58 + 2^3 (256-7x)^\frac48) + 2^4 (256-7x)^\frac38 + 2^5 (256-7x)^\frac28 + 2^6 (256-7x)^\frac18 + 2^7][(5x+32)^\frac45 + 2 (5x+32)^\frac35 + 2^2 (5x+32)^\frac25 + 2^3 (5x+32)^\frac15 + 2^4]$

obsidian monolithBOT
#

Lunasong the Supergay

echo wagon
#

Lol

wary mulch
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u kiddin me

echo wagon
#

You don't have to multiply out manually

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So this comes from factorization

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$x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \dots + x y^{n-2} + y^{n-1}$

obsidian monolithBOT
#

Lunasong the Supergay

echo wagon
#

So when you multiply out the numerator with the first bracket, you will just get 2^8-(256-7x)

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And in the denominator, you will just get (5x+32)-2^5

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Then with the extra pair of brackets

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You get 7x times second bracket / 5x times first bracket

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Cancelling out the x's gets rid of the 0/0 issue

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And then you can substitute in

wary mulch
#

0/0

echo wagon
#

2^8 - 256 + 7x = 7x

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5x+32-2^5 = 5x

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Cancel out the x's

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Then it won't be 0/0 anymore

wary mulch
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ohk yeaa it is

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thanks @echo wagon

echo wagon
#

👍

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This will work in general when you have 0/0 with roots like this

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It just gets uglier the higher the roots are

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8 and 5 pretty bad already lol

wary mulch
#

how did u findout with wat to multiply and divide so fast

uncut mulch
#

factorisation identities

wary mulch
#

ohhh ok

echo wagon
#

I treat the one term as x, and the other as y

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Then what I need to multiply with is that longass second bracket

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And do it for the numerator and denominator

wary mulch
#

ok

uncut mulch
#

could use sigma notation to avoid tedious writing

echo wagon
#

But then it doesn't look as crazy lmfao

uncut mulch
#

also makes evaluating the limit much easier in this case

echo wagon
#

I agreeeeee

tranquil totem
#

@uncut mulch that R is awesome... where to find it?

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the R on ur name

uncut mulch
#

old english font

tranquil totem
#

can it be found on keyboard?

uncut mulch
#

you can get it in something like microsoft word

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or search for images online

viscid thistle
#

𝔸𝔹ℂ𝔻𝔼𝔽𝔾ℍ𝕀𝕁𝕂𝕃𝕄ℕ𝕆ℙℚℝ𝕊𝕋𝕌𝕍𝕎𝕏𝕐ℤ

shut timber
#

how to find equation of expoentnial function with only one point

ivory loom
#

that doesn't seem like its the entire question, might be helpful to post all of it

fleet yew
#

You dont need the equation

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Just eyeball it

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And reverse the coordinates of the given point

shut timber
#

the answer is ln x/2 , but how?

ivory loom
#

To find the inverse of the function, you swap the x and y variables right?

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since you know its an exponential equation, the inverse would turn out to be logarithmic

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honestly the easiest way is to just imagine what the graph would look like in your head when you swap the axis and then you can easily identify which ones must not be correct

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if its easier, imagine the y=x line and mirror the graph to the other side of that line (someone else might be able to explain this better than i can)

ivory loom
#

imagine you have the function y=e^x

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swap y and x

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x=e^y

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solve for y
y=ln(x)

shut timber
#

yes

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but this doesnt have eulers number

ivory loom
#

eulers numbers?

shut timber
#

e

ivory loom
shut timber
#

yea

ivory loom
#

so then whats the problem?

shut timber
#

yea

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reflection across y axis

ivory loom
#

no, should be reflection across the y=x line

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right?

shut timber
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wait yea

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yea i get that

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the inverse would have 1,0 instead of 0,1

ivory loom
#

ok, now lets go through the options they provide

the first option is linear, is this a possibility for the graph u have visualised?

shut timber
#

yea, it cant be it, cuz it should be exponential

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and the second and fourth options cant be it, becaue x^2 should have a local extrema

ivory loom
#

ok, second and fourth option are quadratics, can they be a possibility?

shut timber
#

yea

ivory loom
#

ok third is logarithmic, does the function u are visualizing look logarithimic to you?

shut timber
#

yea ig

ivory loom
#

does the point they outline fit on the third option's graph?

shut timber
#

no, but the inverse does

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they give 0,1 , on ln x/2 1,0 exists

ivory loom
#

they dont give (0,1)

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look at the scales

shut timber
#

0,2*

ivory loom
#

yes

shut timber
#

2,0 exists for lnx/2

ivory loom
#

then u have your answer

shut timber
#

is there any way you can get that without looking at multiple choice

ivory loom
#

nope

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one point is not enough

shut timber
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ah, i see

ivory loom
#

you can prove it to yourself that one point is not enough to get an exponential equation, can u think of multiple exponential equations that go through that same point? use desmos graph tool and test around with some numbers

shut timber
#

Yea will do

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I spent a lot time trying to find equation for the graph, looking online

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Couldn’t find anything

ivory loom
#

quick example

shut timber
#

Yep

patent beacon
#

Acceleration is second derivative of x. Get that, find a(2)

quick bane
spark tendon
#

When a scalar ,k is multiplied with a function f(x), we represent the resultant function as (kf)(x). But when me multiply f(x) with (k+m), where m is also a scalar, which is a better representation of the resultant function : ((k+m)f)(x) or (k+m)f ?

willow bear
#

whichever you think looks better aesthetically

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you could even say (k+m)f(x) since it's unambiguous what's being talked about

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and likewise kf(x)

viscid thistle
#

${\frac{x/x}}$

obsidian monolithBOT
#

∂T/ ∂t = α ∇²T

${\frac{x/x}}$
```Compilation error:```! Argument of \frac  has an extra }.
<inserted text> 
                \par 
l.55 ${\frac{x/x}}
                  $
I've run across a `}' that doesn't seem to match anything.
For example, `\def\a#1{...}' and `\a}' would produce
this error. If you simply proceed now, the `\par' that
I've just inserted will cause me to report a runaway
argument that might be the root of the problem. But if
your `}' was spurious, just type `2' and it will go away.```
viscid thistle
#

$\frac{x_x}$

spark tendon
#

suppose f is a real valued function, then kf(x) is a real number while (kf)(x) is a function ,right?

obsidian monolithBOT
#

∂T/ ∂t = α ∇²T
Compile Error! Click the errors reaction for more information.
(You may edit your message to recompile.)

willow bear
viscid thistle
#

thank you

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didnt know how to use it

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😦

willow bear
#

@viscid thistle there is a cheat sheet pinned in #latex-testing if you want to familiarize yourself with the basics of latex for using the bot with

viscid thistle
#

ohh k ty

spark tendon
willow bear
#

(kf)(x) and kf(x) both refer to a number tho

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that is if you refuse to acknowledge that the former is defined as the latter

viscid thistle
#

Write an equation for the parabola with focus (−1,2)and vertex (3,2).

#

for this problem, would I use the horizontal or vertical equation of a parabola

haughty lance
bitter delta
#

can anyone help me out with this question, no clue what to do here

viral juniper
#

Hello.
Help, thanks!

#

<@&286206848099549185>

tranquil lava
#

You ca start by evaluating dy/dx

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This will give the turning point

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@viral juniper

viral juniper
#

solved it 😓

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-5, -11 from what i remember

tranquil lava
#

okay cool

left sleet
#

Hey, can someone help solve this?A rocket makes a one dimensional motion/movement.

g is the gravitational acceleration,
x is the vertical coordinate,
m is the mass,
t is the time.

Assuming that there is no air resistance/friction etc and only gravity affects the motion of the rocket, the force (F) is
F = –mg.
In this situation, the equation of motion (F= ma where a is acceleration) is (attached as a photo).

(1) Find and show the rocket's velocity as a function of time. However, when t = 0, its velocity is v0.
(2) Find and show the rocket’s position as a function of time. However, when t = 0, its position is x0.
(3) When v0 is much greater/a very large number, describe the rocket's motion.

How would you solve this problem? I'm thinking integrate once to get velocity, and twice to get position. It would be great if you could help out!

sand dune
#

don't post in multiple channels to get more attention

viral juniper
signal vine
#

Anyone able to help me figure out how to go about solving this?

gritty echo
#

What does the derivative tells you about a function?

viscid plank
#

When de function decrease and increase and the maxims and minimums

sick steppe
signal vine
heavy finch
#

If you'd like I can help you with the question

#

it would be helpful if you posted the function itself but if you only want the knowledge, and then to apply it yourself thats better!

heavy finch
#

yep i didn't read it fully sorry

worn crest
#

Could someone walk me through these? Going through my review and haven't done these in awhile

#

Synthetic division and stuff

worn crest
#

I went back to try and solve it, I got this far with part a but I can’t remember how to go from here, or if I did something wrong

#

I used the rational zero theorem, do I have to finish it by using quadratic formula?

#

Alright I did quadratic formula and figured out another zero but I need one more, am I supposed to use that zero I just got and do synthetic division again?

gritty echo
signal vine
#

Yeah

gritty echo
#

So, there you know that if x<2 the derivative is negative, that means that with each step the function goes down right?

wary mulch
#

im getting a=1,b=0 but ans is a=1,b=1/2 can anyone help me

willow bear
#

show work?

#

@wary mulch

wary mulch
#

fully mind calculation

willow bear
#

then do it on paper

wary mulch
#

ok

#

now im getting the ans

haughty lance
#

mind calculation is cool for when you're walking without paper. it's almost always helpful to write some stuff out. even if it's easy, you might see some other properties that you wouldn't notice if it's all calculated by head

jolly raven
#

Wouldn't the answer be a > 0 and b = 0

north cloud
#

Can anyone help me solve this problem?

rough carbon
#

Work out the modulus of the vector, which is 5
We want a vector with modulus 1, so our vector will be 1/5 of v given

north cloud
#

What do you mean by modulus?

#

Do you mean as in this "%"?

viscid thistle
north cloud
#

Why do you need to use modulus in calculating unit vector direction?

sick steppe
#

modulus of a vector is the length/norm

viscid thistle
#

ahh too used to programming lol

sick steppe
#

clearly modular arithmetic has nothing to do with vectors

#

bar Z_p scalar fields

viscid thistle
#

why not just say magnitude or length, why modulus 😠

#

but yea that makes more sense now that I think about it lol

sick steppe
#

cause modulus is technically correct since C is iso to R^2, ie you can express complex numbers as tuples of real numbers

#

and modulus of a complex number is identical to norm of a vector in R^2

#

since both are just pythagorean

viscid thistle
#

I see-ish

weak goblet
#

Can someone help

#

<@&286206848099549185>

gritty echo
#

Still need help? What have you tried/where are you stuck

pliant wadi
#

My doubt is if cos x = -1/2 , then x could be -60°

uncut mulch
#

there are infinitely many solutions to cos(x) = -1/2
but -60° isn't one of them

inland flower
#

anyone know how to graph this

uncut mulch
#

seems that ← is supposed to be <

#

which part of graphing are you stuck on?

inland flower
#

like i dont know where to start/plot

uncut mulch
#

do you know how to graph straight lines?

inland flower
#

sort of

uncut mulch
#

e.g would you be able to graph
y = x/2 + 4?

inland flower
#

yes

uncut mulch
#

start by doing that

#

seems that ← is supposed to be <
actually its supposed to be < -

#

and restrict the domain for that piece to (-inf, -4)

#

since f(x) only behaves that way for x < -4

inland flower
#

so would graph this y = x/2 + 4

uncut mulch
#

until where x is -4

#

since those are the conditions being given

#

f(x) = x/2 + 4 if x < -4

haughty lance
#

as mosh stated. also useful for dividing complex nums

viscid thistle
#

Can commutative and associative laws over sum and product of integers be proved?

pulsar geyser
#

help

#

how do i get the answer to blank 5 using the information i already put there

trim hemlock
#

Is this an exam or some marked quiz?

pulsar geyser
#

no its a practice quiz for the finals its not graded

trim hemlock
#

Hmm 5points is kind if sketchy

#

Anyway

#

Try to draw the tiles out

#

Draw the hexagon and squares out, fill them in

#

And see whats missing

#

And what kind of shape you can fill in that gap

pulsar geyser
#

im pretty sure i have to use the exterior angle of the green shape to get the number opf sides

#

cuz 360/the measure of one exterior angle = the number of sides in a figure

#

i think its 12nvm idk

trim hemlock
#

Tho there is still one aspect of the question i dont quite understand, is the green shaded area supposed to be a whole shape?

pulsar geyser
#

ok 12 is right

#

i am pro

#

ge

#

yes

#

i think so

trim hemlock
#

Hmmm then why was the top corners coloured purple

#

It seems pike the square are supposed to be attaches to the top two sides of the hexagon

#

Thats where im kind of unclear about the problem

pulsar geyser
#

its a

#

tessellation

trim hemlock
#

Tho but if the green shaded area is supposed to be a whole shape then yeah, 12 seems correct

viscid thistle
patent beacon
#

@weak goblet
Define important variables. Hint: there's almost always 2.

Write the important equations - they give you a perimeter, so what's the equation for perimeter? They want an area, what's the equation for area?

Then, write area in terms of one variable. Differentiate. Set to 0.

weak goblet
#

I still dont understand, is it possible if you could do it for me? That just helps me better if I get a visual or something

willow bear
#

we don't do problems for you

#

@weak goblet do you still need help?

weak goblet
#

Yes I do

willow bear
#

are you ok with me taking you through this by having you do it step by step?

weak goblet
#

Yes please

willow bear
#

okay, the first thing you should ask yourself is: what does the problem ask you to find?

#

i.e. what will the answer look like?

weak goblet
#

To find the area of the rectangle?

willow bear
#

no

#

you're asked to find the dimensions of the rectangle, i.e. its length and width

weak goblet
#

Oh okay

willow bear
#

this should have been clear from a careful reading of the problem

#

your answer will look like:

The area of the rectangle is maximized when its length is ___ and its width is ___; this gives it an area of ___.
#

once we fill in the blanks for length and width, the area is (hopefully) easy to find.

#

therefore the quantities we care about are the length and width.

#

do you follow?

weak goblet
#

So we have to find the max length and max width?

willow bear
#

no

#

it's area that is maximized. not either dimension on its own.

#

we didn't get there yet.

weak goblet
#

Oh okay I understand now

#

Whats the next step

willow bear
#

when solving math problems, typically you take the quantities you care about and assign them names, yes?

weak goblet
#

Yes

willow bear
#

so why don't you do that now with the length and width?

#

which names would you like to call them?

weak goblet
#

L and W

willow bear
#

alright

#

you should write at the beginning of your work:

L := length of the rectangle, in cm
W := width of the rectangle, in cm
weak goblet
#

Done ✅

willow bear
#

do you understand why i'm having you write this?

weak goblet
#

Yes so we can get a variable

#

And then solve for that

willow bear
#

no, it's so we can define our variables.

weak goblet
#

Thats what I was thinking lol

willow bear
#

then you didn't word it properly.

weak goblet
#

Yea sorry so whats next?

willow bear
#

anyway, ok, now we can get to the actual problem.

weak goblet
#

Okay

willow bear
#

we care about the area of the rectangle.

#

so we will need to express it in terms of our variables.

weak goblet
#

How will we do that?

willow bear
#

well do you know how to find the area of a rectangle?

weak goblet
#

L•W

#

Length x width

willow bear
#

length times width. don't use the letter x for multiplication.

weak goblet
#

Okay

willow bear
#

but yes. so you just did exactly what was needed. why did you try to ask me?

weak goblet
#

What do you mean?

willow bear
#

why did you ask me "How will we do that?" as if you didn't know?

weak goblet
#

I guess I just didnt think of it and I mightve been doubting myself

willow bear
#

mkay...

#

so we have this:

Area of rectangle = LW
weak goblet
#

Yes

willow bear
#

there is another piece of info we care about

#

the rectangle is enclosed by a border whose length is 36 cm.

weak goblet
#

Yes 36cm

willow bear
#

look at the diagram to see the shape of the border, and translate the sentence "The length of the border is 36cm" into an equation.

weak goblet
#

L=36cm???

willow bear
#

no

#

you're saying the length of the rectangle is 36 cm

#

that's not what you're told here

weak goblet
#

Oh the total length of 3 sides is 36cm

willow bear
#

perhaps i should have said "The total length of the border is 36cm" so as to prevent you from getting confused.

#

yes, so translate that into an equation, will you?

weak goblet
#

Total length of border = 36cm

willow bear
#

no. make that into an equation in L and W.

weak goblet
#

L•w=36cm

willow bear
#

what's lowercase w?

#

did you mean L*W = 36cm?

weak goblet
#

Yea

willow bear
#

this is nonsense.

#

what does L*W represent?

weak goblet
#

Length times width

willow bear
#

there is a name for that.

#

it's called area.

weak goblet
#

I sorry im just really confused

willow bear
#

which is not what we're talking about right now.

weak goblet
#

Oh okay

willow bear
#

you yourself correctly responded with "L*W" when i asked you how to find the area of a rectangle

weak goblet
#

Yea I did

willow bear
#

so then why do you say L*W again when prompted to express the total border length?

weak goblet
#

Oh right oops

willow bear
#

try again.

weak goblet
#

Would it be a quadratic function?

willow bear
#

you're overthinking it.

#

the words "quadratic function" here are extraneous, and should be ignored for now.

#

again

weak goblet
#

W^2 *L

willow bear
#

width squared, times length???

weak goblet
#

I think so??

willow bear
#

what kind of physical quantity do you get when you multiply three lengths together?

weak goblet
#

L^3

willow bear
#

i'm looking for a word here.

weak goblet
#

Length cubed

willow bear
#

that's called volume.

weak goblet
#

Yea

willow bear
#

i asked you for the total border length and you reply with a volume?

#

does this not strike you as odd?

weak goblet
#

Yea im sorry again, so the total border is 36cm how would I write that as an equation

willow bear
#

look at your diagram again, here

#

what is the total length of the solid edges?

weak goblet
#

There is 1 length

#

And 2 widths

willow bear
#

answer my question.

#

what is the total length of those?

weak goblet
#

36

willow bear
#

no, as an expression

weak goblet
#

L=36

willow bear
#

no!

#

why are you insisting that L = 36? L isn't the entire border, is it now?

weak goblet
#

36 = L*W^2

willow bear
#

why on earth are you MULTIPLYING them!?!?!?!?!?!

#

oka y

#

look

weak goblet
#

W^2+L=36

#

Is that right?

willow bear
#

why are you insisting on squaring the width?

#

now you're adding an area to a length! this makes even less sense!

#

look. do you claim the perimeter of this triangle is (5 * 7 * 9) feet? because that's what it should be by your logic.

weak goblet
#

Im not sure what im supposed to do, sorry about that please help I will listen to what you say now

willow bear
#

you should have listened to what i say from the very beginning. that you say you will do so now implies you didn't before, which is a bit disconcerting.

#

okay, so how do you find the perimeter of this triangle? of my triangle?

weak goblet
#

Yea It would be 9+7+5

willow bear
#

yeah, so you would ADD the sides

#

not multiply them

#

not multiply one side by another and add the third

#

you would add them

#

now back to this

weak goblet
#

Yes I would so in my case, I would do width plus width plus length

willow bear
#

yes

#

W + W + L

#

now can you simplify this?

#

simplify W + W + L

weak goblet
#

2W + L

willow bear
#

okay

#

great

#

so after all this blood, toil, tears and sweat, we have:

#
Area = LW
2W + L = 36
Goal: max Area
weak goblet
#

Yes!

willow bear
#

the thing about our expression for area is that it has two variables, which makes it hard to maximize. maximizing functions of two or more variables is much harder than those of only one.

#

luckily, we have an equation relating our two variables.

#

this will let us isolate one of the variables, and plug the result into the expression for area. which in turn will make the expression for area contain one variable only.

#

do you understand this?

weak goblet
#

Yes I understand this part is clear to me

willow bear
#

okay, great.

#

now would you be so kind as to follow up on this?

#

take our equation (2W + L = 36) and isolate one of the variables in it.

weak goblet
#

So like subtract either W or L

willow bear
#

isolate either W or L.

weak goblet
#

Okay. Subtract L from both sides

willow bear
#

so you've chosen to isolate W?

weak goblet
#

Yes

willow bear
#

okay

#

so you choose to subtract L from both sides. what does this give you?

weak goblet
#

2W=36-L

willow bear
#

okay

#

keep going

weak goblet
#

What would I do now

fading cedar
#

divide by both sides

#

by the coefficient in front of w

weak goblet
#

Divide both sides by 2

fading cedar
#

yes

willow bear
#

would have preferred if you had let shazz2005 figure this out for themselves, tenletters.

fading cedar
#

my apologies

weak goblet
#

W=18-2L

willow bear
#

are you sure?

weak goblet
#

Or W=18-L

willow bear
#

are you sure dividing 36-L by 2 gives you 18 - 2L? or 18 - L, for that matter?

#

why would the L miraculously stay intact (or, god forbid, get multiplied by 2?)

weak goblet
#

18-L/2

willow bear
#

yes, so now you have W = 18 - L/2.

willow bear
weak goblet
#

Im not too sure how

fading cedar
#

well, now you have an expression for W in terms of L

weak goblet
#

Yeah

willow bear
#

Area = LW

#

W = 18 - L/2

fading cedar
#

^

weak goblet
#

Ummm what am I supposed to do now @willow bear

willow bear
#

this will let us isolate one of the variables [we did this already!], and plug the result into the expression for area.

weak goblet
#

Im still not to sure I understand

willow bear
#

plug W = 18 - L/2 into Area = LW

weak goblet
#

W=18-L/2

willow bear
#

?????

fading cedar
#

not quite

willow bear
#

well now you just didn't do what i asked you

#

i did not ask you to just repeat the equation W = 18 - L/2.

weak goblet
#

Yea what do I plug in

willow bear
#

i TOLD you!

#

i told you EXACTLY what to plug in, and what to plug it into!

#

i asked you to replace W with (18 - L/2) in Area = LW !!!

#

i can't understand what part of this is remotely unclear!

weak goblet
#

Oh okay so L*(18-L/2)

willow bear
#

Area = L(18 - L/2)

fading cedar
#

yes

willow bear
#

yes

#

finally

#

great

#

this took way more effort than it should've

#

but ok

#

so now you have that

weak goblet
willow bear
#

you have A = L(18 - L/2)

weak goblet
#

Yup

willow bear
#

you can now change it to standard form if you want

#

this will make it more visible that it is in fact a quadratic in L, which is exactly what you were instructed to do

#

even though you did not need to know that, so the fact they said it explicitly should be treated only as a sanity check

weak goblet
#

So I change it to standard form?

#

How would I?

fading cedar
#

what do you normally do with brackets?

weak goblet
#

Remove them

fading cedar
#

and how do you do that?

willow bear
#

tenletters, do you want to take over?

#

you seem very eager to help

fading cedar
#

i suppose

willow bear
#

okay then i'm out

weak goblet
#

No @willow bear

#

Dont leave pleas

willow bear
#

tenletters just said they're going to take over.

#

i have better things to do.

weak goblet
#

Oh okay 😢

#

@fading cedar help me you made her leave

fading cedar
#

indeed i may have

#

so...

#

need to expand them brackets, no?

weak goblet
#

Yes

fading cedar
#

and what do you get from that?

weak goblet
#

Do I distribute the L

fading cedar
#

indeed you do

#

as a(b+c) = ab + ac

weak goblet
#

Okay so it would be

fading cedar
#

Yes

#

now

weak goblet
#

Yes

fading cedar
#

do you know what do to get a maximum?

weak goblet
#

I do not

fading cedar
#

ah

weak goblet
#

Please help me out with that

fading cedar
#

you're going to need to get the derivative

weak goblet
#

Whats that and how

fading cedar
#

in a general sense, it is the instantaneous rate of change

#

as in, for a curve, it will give you the exact slope at any given point

weak goblet
#

Oh okay

#

How do I find that

fading cedar
#

you take the power of the variable, and multiply it to the front

#

and then take away one from the power

#

(in this case anyway)

weak goblet
#

So 18^2

fading cedar
#

not like that

#

but 18(2)

#

giving you 36

#

and when you derive a function, you usually write it as dy/dx

weak goblet
#

Oh okay

fading cedar
#

with y being the function, and x being the variable

#

so, in this case, dA/dL

weak goblet
#

I would now have A=36L-L/2

fading cedar
#

you forgot something

weak goblet
#

Oh A=36L- L^2/2

fading cedar
#

you're forgetting to reduce the power by 1

#

(of the variable)

weak goblet
#

L^1

fading cedar
#

not quite

#

36l^0

#

and what is anything to the zero power

#

1

#

therefore you get 36 for the first term

weak goblet
#

36-L^2/2

fading cedar
#

still got the second term to fix

#

remember:bring the power to the front and multiply it, and reduce the power by 1

weak goblet
#

36L^0

fading cedar
#

2-1=0?

weak goblet
#

No

fading cedar
#

are you sure

#

so?

weak goblet
#

2-1=1

fading cedar
#

yes

weak goblet
#

36-L^1/2

fading cedar
#

still forgot to multiply by the old power

#

should be (-2L^1)/2

weak goblet
#

Ohhhh okay so 36-(2L^1)/2

fading cedar
#

yes

weak goblet
#

Okay

fading cedar
#

but what is something over itself?

#

2/2 = ?

weak goblet
#

1

fading cedar
#

yes

#

so you can simplify to 36-L

#

now

weak goblet
#

36-L

#

Got it

fading cedar
#

in order to find a maximum or a minimum, you equal the resulting derivative/slope function to zero

#

as when this occurs, the gradient/slope at the point is zero

weak goblet
#

Yea it is

fading cedar
#

so, knowing that

#

what to do next

weak goblet
#

Make one side 0

fading cedar
#

yes

#

and then?

weak goblet
#

I have A=36-L

fading cedar
#

not a = 36-l

weak goblet
fading cedar
#

rather A'(L) = 36 - L

weak goblet
#

Okay

fading cedar
#

to indicate it's the derived function

#

well, just solve for your unknown

#

and why involve the quadratic formula?

weak goblet
#

Idk

fading cedar
#

well, 36 - L = 0

#

what is L

weak goblet
#

36

fading cedar
#

Yes

weak goblet
#

Then what next

fading cedar
#

well, we had two variables to begin with

#

or, you can check if a function is maximum or minimum

#

by taking it's second derivative/ or drawing the graph

#

(of the original function)

#

the question is about max area is it not?

weak goblet
#

Yea we have to find the max area

fading cedar
#

well, in this case, since the highest power of the original function is 2

#

(therefore making it a quadratic)

#

there is only 1, turning point

#

and therefore only one value we can attain

weak goblet
#

Yea

fading cedar
#

if you take the second derivative, the resulting function will be simply 1

#

as all constants go away

weak goblet
#

@willow bear told me to write my answer like this. How would I
The area of the rectangle is maximized when its length is ___ and its width is ___; this gives it an area of ___.

fading cedar
#

yes that's right

#

just wondering about this question

weak goblet
#

So whats the length

#

36?

fading cedar
#

as if you do the min/max test using the second derivative

#

it is greater than 1

#

which means it's technically a minimum

#

but lesser than zero is a maximum

#

regardless

#

you now have your value of L

weak goblet
#

36??

fading cedar
#

yes

weak goblet
#

Then what is w

fading cedar
#

but the original expression was 2W + L = 36

weak goblet
#

Yea

fading cedar
#

L = 36

#

2W + L = 36

#

then?

weak goblet
#

2*0+36=36

fading cedar
#

oh no

#

i just realized what went wrong

#

my apologies

#

we were working with 36 this whole time

#

it should've been 18

weak goblet
#

So my length is 18

fading cedar
#

A'(L) = 18 - L = 0

#

from this yes

#

yes so solve for W

weak goblet
#

Im getting 9 as my answers

#

Answer*

fading cedar
#

and if we do take the second derivative (properly this time) we get -1

#

which is lesser than zero

#

and therefore a maximum

#

okok this is better

weak goblet
#

So what is my length?

fading cedar
#

18, as you stated

#

but solve for W

weak goblet
#

Correct than width is 9?

fading cedar
#

2W + L (L is 18) = 36

#

yes

#

(36-18)/2 = 9

weak goblet
#

Okay

fading cedar
#

i suppose

weak goblet
#

The area of the rectangle is maximized when its length is 18 and its width is 9 this gives it an area of

fading cedar
#

18*9

weak goblet
#

18cm 9cm*

#

Okay

#

162cm

fading cedar
#

as for justifying it is a maximum area

#

you could say if the graph of the original function is drawn

#

it is a local maximum due to being a negative parabola

weak goblet
#

👍🏼

fading cedar
#

which the turning point of the graph (where the gradient of the graph is zero) is the highest point

weak goblet
#

Alright

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Is that it?

fading cedar
#

or you could say you took the second derivative, which simply gave you -1

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and since -1 is lesser than zero

weak goblet
#

But I cant have a negative width or length

fading cedar
#

the point is a maximum

#

no this is the second derivative

weak goblet
#

Ahhh okay

fading cedar
#

if it's greater over zero, its a minimum

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if it's lesser than zero, it's a maximum

weak goblet
#

Oh okay thanks

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Are we done?

fading cedar
#

yea that's about it

weak goblet
#

Thank you so much for your help today 😊😊 I have one more question but if you cant do it now tomorrow is fine as it is late for me

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Ill leave it here and if you get a chance please look at it and tell me the steps I will do it tomorrow morning thanks for your help

magic magnet
#

why the answer is only A, why not A and E?

echo wagon
#

For E, f(2) = 28 but f(-2) = -36

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@magic magnet

willow bear
#

,rccw

obsidian monolithBOT
magic magnet
#

oh okay , i got it, i need to place value for both sides.
i was just plugging negative value on left side.

plush birch
#

mhm

amber marsh
#

given that this is true:
if log(u) = log(v), then u = v
does it follow that:
if log(u) = -(log(v)), then u = -v
is also true?

obsidian monolithBOT
sick steppe
#

$-\log(v)=\log(\frac{1}{v})$

obsidian monolithBOT
amber marsh
#

ooookay, thank you

viscid thistle
#

hello one textbook I was reading stated that 6a9 is not a term, why is this?

weak goblet
#

Help pleasee

viscid thistle
#

You're more likely to get help if you put effort into presenting the problem

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Instead of sending a crummy photo

viscid thistle
#

😳

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but fr I can't read that I have wack eye sight

solid stirrup
#

sussy

sharp lagoon
#

how can i approach this question?

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Most of the examples i can find have denom,

pliant gyro
#

@sharp lagoon shouldn't it be set of all real no.s

sick steppe
#

and all polynomials have a domain of R

sharp lagoon
sharp lagoon
sick steppe
#

$(-\infty ,\infty)$

obsidian monolithBOT
viscid thistle
sharp lagoon
viscid thistle
#

@sharp lagoon im pretty sure the domain is X element of R

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r as in real numbers

sharp lagoon
#

okok!

sharp lagoon
#

Ok, this one I know the radical (x-3) would be x>_to 3 but what do i do with the 6 outside

trim hemlock
#

the 6 outside does not affect the domain at all

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for x>=3, f(x) is always defined

sharp lagoon
#

I see,,,

sharp lagoon
#

what is wrong with my answer

willow bear
#

well for one the domain is just R (or (-∞, +∞) if you so desire and/or are required to write)

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but the range is not (-∞, +∞)

willow bear
sharp lagoon
#

OH i shouldve used desmo to graph it

willow bear
#

i mean you didn't need to use desmos specifically but you would've done well to graph it yes

tribal coral
#

f(x) = ln(10x)e^-x

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How do I prove ln(10x) < e^x for all x

steel venture
#

start by graphing it @tribal coral

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then think about manipulating the expression ln(10x)e^(-x) into the inequality ln(10x) < e^(x)

lusty veldt
#

Hello, can I ask where did the 4 and 1 came from?

steel venture
#

it's completing the square

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they're on the other side of the equation as well

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if you're wondering why it's 4 and 1 specifically,
it's because x^2 - 4x can be turned into x^2 - 4x + (4 - 4) = (x^2 - 4x +4) - 4 = (x-2)^2 - 4
and similarly
y^2 + 2y = y^2 + 2y + (1 - 1) = (y^2 +2y +1) - 1 = (y + 1)^2 - 1

lusty veldt
#

Thankyouuu^^

hollow fog
sick steppe
mystic cloak
#

Please prove