Throw some math problems at me or someone else

TROLLDLLR · 2018-03-01 04:15:19 UTC

There are n integers, where n is the number of integers that satisfies that requirement

Find the roots of the graph y=x!

Ezth · 2018-03-02 15:01:30 UTC

Kappa

AKLDragon · 2018-03-02 17:51:15 UTC

.
1st is y = -1/3 x - 2, you can eliminate 4th option right off, and then you plot the point and go to the y-axis with that slope to find the y-intercept.

2nd problem:

y = 1/2 x - 5
y = 1/2 x + 3
y = -1/2 x - 2
y = -1/2 x

Tauntauned · 2018-03-02 18:09:25 UTC

A story regarding math,

A “colleague” of mine if you could even call them that was talking about getting his masters degree

and talking about how he needs to do 6 courses of Calculus, confused i asked why so much calculus and no trig

he responded “calculus is just advanced trigonometry”

I looked at him with a same look that you would give when aiming a gun at the person who had killed your entire family and merely walked away

dishonor upon his family dishonor upon him and dishonor upon his career

this is why i work from wherever i am living instead of going to a office

CandyShi · 2018-03-02 20:04:21 UTC

Does… he have google?

Tauntauned · 2018-03-02 23:57:25 UTC

that is a “professional” working environment for you. I would say 1/5 dont know what they are doing but are in the field that applies to most fields really

Mynamerr · 2018-03-03 00:17:31 UTC

As a mathematican I am certified to tell you that tartar and trigonometry are basically the same

Tauntauned · 2018-03-03 00:41:24 UTC

and tartar is refereeing to this of course, the tartars not the disgusting images i found of teeth

EpicNecros · 2018-03-17 05:43:49 UTC

$\int \sec x dx$

if you have already worked with this retarded integral and/or you already know the answer or you used a damn website to figure it out then at the very least don’t spoil it for people who want to subject themselves to u-sub hell

again, if you have the answers then don’t be a killjoy
already solved the first one which is pretty easy, working on the second

i hate differentials though >:[

Unselfish · 2018-03-17 06:07:10 UTC

Answer to the first one if anyone's interested

EpicNecros · 2018-03-17 08:55:22 UTC

fix details, add the /

EpicNecros · 2018-03-17 21:15:20 UTC

solving #3

original problem:
solve for y given
$e^y(2x^2+4x+1)\frac{dy}{dx}=(x+1)(e^y+3), y(0)=2$

separate x and y to get

$\frac{e^y}{e^y+3}\frac{dy}{dx}=\frac{x+1}{2x^2+4x+1}$

this next step is optional, but it made it way easier for me to do u-sub

$e^y*\frac{1}{e^y+3}\frac{dy}{dx}=(x+1)\frac{1}{2x^2+4x+1}$

integrate both sides with respect to x

$\int e^y*\frac{1}{e^y+3}\frac{dy}{dx}dx=\int (x+1)*\frac{1}{2x^2+4x+1}dx$

$\int e^y*\frac{1}{e^y+3}dy=\int (x+1)*\frac{1}{2x^2+4x+1}dx$

now the u-sub comes. the left side is pretty easy, and all you need to do with the right side is look at the numbers and apply the appropriate multiplications and such

$u=2x^2+4x+1$
$du=4x+4dx$

$v=e^y+3$
$dv=e^ydy$

now apply the substitutions

$\int e^y*\frac{1}{e^y+3}dy=\int \frac{4x+4}{4}*\frac{1}{2x^2+4x+1}dx\rightarrow \int \frac{1}{v}dv=\frac{1}{4}\int \frac{1}{u}du$

integrate to get

$ln|v|=\frac{1}{4}ln|u|+c$

and sub the values for v and u then apply the rule of multiplication of logarithms to get

$ln|e^y+3|=ln|2x^2+4x+1|^{\frac{1}{4}}+c$

now sub in the initial values for x and y

$ln|e^2+3|=ln|2(0)^2+4(0)+1|^{\frac{1}{4}}+c$

to get

$ln(e^2+3)=ln(1)^{\frac{1}{4}}+c$

and solve to obtain the value of c

$c=ln(e^2+3)$

then sub in c and solve to get the equation for y

$ln(e^y+3)=ln|2x^2+4x+1|^{\frac{1}{4}}+ln(e^2+3)$

$ln(e^y+3)=ln[|2x^2+4x+1|^{\frac{1}{4}}(e^2+3)]$

$e^y+3=\sqrt[4]{|2x^2+4x+1|}(e^2+3)$

$e^y=\sqrt[4]{|2x^2+4x+1|}(e^2+3)-3$

and the solution is finally obtained by applying natural log to both sides

$y=ln[\sqrt[4]{|2x^2+4x+1|}(e^2+3)-3]$

i messed up once after i forgot to include “+c” while integrating lmao

Ezth · 2018-03-19 13:40:37 UTC

Smh

AKLDragon · 2018-03-19 14:44:57 UTC

For problems 1-4, the slope of a parallel line has the same slope as the original line.
EX.-2/5 x + 4, the first problem, the slope of a parallel line is also -2/5.

Answers 1-4

m = -2/5 (m is slope)
m = 6/5
m = -1/5
m = 3

For 5-8, the perpendicular slope is the negative reciprocal of the line’s slope. This means that you take the slope of the line, then you switch the top and bottom of the fraction and turn it negative. (for a slope such as 3, imagine it as slope 3/1 (the same thing) and switch it to 1/3, then turn it negative.

Answers 5-8

m = 2/3
is a special problem, it is a vertical line. slope of perpendicular line is 0 in this case
m = 4/3
m = 1/2

ok, class gets out in 2 minutes (english). I’ll come back and do the rest in my next class (business).

AKLDragon · 2018-03-19 15:34:47 UTC

For 9-12, the slope is [the change in y / the change in x]. So the formula is
_{1st y-value - 2nd y-value}
^{1st x-value - 2nd x-value.}
Ex. For #9, 1st x-value = -9, 2nd x-value = 17, 1st y-value = -9, 2nd y-value is -10. Plug that in and get
_{(-9) - (-10)} => ₁
^{(-9) - (17)} =>^-28
So the answer is -1/28. (dont ask how fun all that was to format)

Answers 9-12

m = -1/28
m = 19/12
m = -2
m = -1/2

EpicNecros · 2018-03-19 18:16:55 UTC

i’d solve them but they’re sideways

Ezth · 2018-03-19 18:17:18 UTC

Smh

Awdreg · 2018-03-20 03:09:02 UTC

Lol this has literally become a homework help thread

BLUbeans · 2018-03-20 03:54:52 UTC

lmao yeah, there’s actually a real homework thread buried somewhere in this subforum

but no one uses it :’(

EpicNecros · 2018-03-20 05:15:04 UTC

hey man i just enjoy a good math problem
i also enjoy mental torture

other people have other things they enjoy, like having others do their hw