so i guess we could say M=KX
#help-0
magic anchor
north adder
@magic anchor yeah so
okay
im gonna use the variables that are defined in the question
L^3 = km as given
L^3/m = k
so this would mean that:
25^3/870 = k
k = 17.95977011
from the first equation:
L^3 = 17.95977011 x 1300
L = 28.5812603.... cm
@magic anchor Makes sense?
magic anchor
Oh my god you genius 🤣 tysm
magic anchor
<@603848537423478796> Makes sense?
yes
see I made a mistake at the beginning and did 870/25^3, no wonder I had such a hard time
thanks a ton Denton have a nice day/night
.close
lone heartBOT
north adder
thanks a ton Denton have a nice day/night
glad to help :)
lone heartBOT
Available help channel!
rapid kettle
|2x+3|>x-1
lone heartBOT
|2x+3|>x-1
rapid kettle
can someone help
rapid kettle
ummmm we just started with mod
so will it be
4x^+9+12x ?
and x^2+1-2x on the other side ?
north adder
mhm
rapid kettle
and then gimme a sec
north adder
remember if you get imaginary solutions
then it should be all real numbers
assuming youre still working with inequalities
jagged crater
I just can’t figure out how to do this apparently you need to use the Trapezoidal rule but you can’t really use that with what’s given
north adder
,rotate
ocean sealBOT
north adder
I just can’t figure out how to do this apparently you need to use the Trapezoida...
you can split them into 2 trapezoids
jagged crater
How would I calculate the side bits, would I need to rearrange the formula?
jagged crater
you can split them into 2 trapezoids
Also thank you for the recommendation
north adder
How would I calculate the side bits, would I need to rearrange the formula?
no youd just use the trapezoid rule
jagged crater
Really?
lament root
lone heartBOT
normal ingot
what did you get?
lament root
I got -1/(x+a)^2
normal ingot
yes
normal ingot
you mean -1?
normal ingot
yup
upper echo
why is there (35/2)^2 at the end and why is it squared?
lone heartBOT
why is there (35/2)^2 at the end and why is it squared?
normal ingot
expand the second equation fully, and you'll see why
upper echo
Oh so you need to cancel it out?
ocean sealBOT
lirmirit
kindred anchor
In how many ways can five distinct books be arranged in two bookshelves?
tight locust
How many books can a book shelf hold
kindred anchor
Maximum, it's 5
tawny schooner
The order of the books arranged in one bookshelf also matters?
kindred anchor
It's not given
tawny schooner
I'm gonna assume it does
kindred anchor
Ok
tawny schooner
In that case the answer will be 6!
kindred anchor
How it's 6! ?
languid bolt
it's 5!
kindred anchor
no
tawny schooner
Consider a stick between the books as a partition for the bookshelf
kindred anchor
For a single bookshelf, it's 5!
tawny schooner
Consider a stick between the books as a partition for the bookshelf
Then the problem becomes permuting these among themselves as each placement of the stick will give a different grouping of the books
The permutations of 5 books and 1 stick, so 6!
tawny schooner
For a single bookshelf, it's 5!
Is 6! Given in the options then?
kindred anchor
It's Numerical answer type question.
tawny schooner
I see
tawny schooner
Yes
kindred anchor
What's the significance of two bookshelves here?
tawny schooner
Two bookshelves is basically a partition placed between the books
One partition-> two groupings-> two bookshelves
tawny schooner
Sure, you too!
kindred anchor
.close
rancid rivet
lone heartBOT
rancid rivet
how do i solve this?
lone heartBOT
@rancid rivet Has your question been resolved?
nimble holly
i think you should solve how many times cindy+ sara must be chosen first
then bill+timm
then rest without them
then you add the total ways
not sure tho
lone heartBOT
@rancid rivet Has your question been resolved?
nimble holly
nice
lone heartBOT
@rancid rivet Has your question been resolved?
thick lynx
It can be proven that when factoring $x^2 + bx + c$ and $b$ and $c$ are integers, then $u$ and $v$ in the form $(x + u)(x + v)$ can never be fractions, right?
thick lynx
Is that a counterexample?
novel night
The rational root theorem says, in this case, that if the polynomial has a rational solution $x$, then that solution will in fact be an integer. It doesn't rule out irrational solutions -- and the solutions for $x^2 + x - 1$ are indeed irrational.
ocean sealBOT
daveamayombo
thick lynx
The rational root theorem says, in this case, that if the polynomial has a *rati...
"When factoring the quadratic x² + bx + c, where b and c are integers, why do we not consider the case where u and v in the factored form (x + u)(x + v) are fractions?"
Is this the rational root theorem?
last ether
No
The statement is only true for polynomials with only rational root solutions
But the statement is not the theorem itself
novel night
The rational root theorem is true for any polynomial with integer coefficients. But factoring a polynomial into a form like (x+u)(x+v) is the same as finding roots -- in this case -u and -v would be roots of the original polynomial.
thick lynx
I see, the proof uses a fraction for u and v, which means that the solutions have to be rational
thx
.close
alpine sable
help
clear stump
no
alpine sable
what
clear stump
Ask your math question in a clear, concise manner.
alpine sable
find value of 2 beta
options
pi/4-alpha
3pi/4-alpha
pi/8-alpha/2
3pi/8-alpha/2
ok
.close
lone heartBOT
lone heartBOT
Available help channel!
daring salmon
5 + 10 =?
lone heartBOT
@daring salmon Has your question been resolved?
soft ridge
I need some help with a epsilon-delta proof
lone heartBOT
I need some help with a epsilon-delta proof
soft ridge
dont know how to do the fancy typing so bear with me
the limit as x approaches 2 of (x^2 - 4x + 5) is equal to 1
whenever i mess around with the epsilon inequality I get mod(x-2) times mod(x-2) is less than epsilon, I don't really know where I am meant to go after this to write up the proof/
merry depot
Doesn’t delta put a bound on |x-2|?
Or are you asking how to get delta from where you’re at?
soft ridge
how do i get delta?
i said that delta is equal to min{1, sqrt(epsilon)} but I dont think its right, at least I can't seem to finish to the proof
novel night
One cute trick in this case is write $x = 2 + a$. Plug this into the formula for $x^2 - 4x + 5$ and simplify. You'll get something that depends on $a$. The question is, how small must $a$ be in order for the expression above to be within $\epsilon$ of 1. Try that, and see if the answer pops out.
ocean sealBOT
daveamayombo
novel night
(And $\min(1, \sqrt{\epsilon})$ will turn out to be a valid choice for $\delta$, though actually $\sqrt{\epsilon}$ works just as well.)
ocean sealBOT
daveamayombo
soft ridge
how the hell did you see that?
novel night
The cute trick part? The idea is that $x$ is getting close to 2. So think of $x$ as being "2 plus a little bit", and make that explicit by saying $x = 2 + a$.
ocean sealBOT
daveamayombo
lone heartBOT
@soft ridge Has your question been resolved?
neon girder
why is 1 rejected
lone heartBOT
why is 1 rejected
neon girder
and minus 1 of course, why ?
alpine sable
because you are writing the absolute value as $$y=\begin{cases}x\quad \text{if $x\geq0$}\ -x\quad\text{if $x<0$}\end{cases}$$, then $x=-1$ is not in the domain of the $y=x$ section when $x\geq0$, and similarly for the other rejection
ocean sealBOT
Toby
lime bobcat
It is because of the absolute value. The number you get is in the region you're not considering at that moment. When y=x (x positive) then you get a -1 (negative)
when y=-x (x negative) you get a 1 (positive)
BTW beautiful calligraphy
neon girder
because you are writing the absolute value as $$y=\begin{cases}x\quad \text{if $...
god why is it so hard to read this
alpine sable
BTW beautiful calligraphy
took four tries to get rid of the errors lol
alpine sable
god why is it so hard to read this
look at the message from TeXit for the compiled version
lime bobcat
Oh, I meant @neon girder's calligraphy
alpine sable
ah lmao
lime bobcat
But your LaTeX is cool too. I'm used to use array enviroment with \left\{ and \right.
I should consider the cases enviroment
tidal trout
$$\left(a^{\frac{1}{x-y}}\right)^{\frac{1}{x-z}}:\cdot \left(a^{\frac{1}{y-z}}\right)^{\frac{1}{y-x}}\cdot \left(a^{\frac{1}{z-x}}\right)^{\frac{1}{z-y}}$$
ocean sealBOT
Nayan RBLX
tidal trout
i got confused
tidal trout
I did
but in top I got 1-1-1/the denominator
0-1 is -1
$$=a^{\frac{1}{\left(x-y\right)\left(x-z\right)}+\frac{1}{\left(y-z\right)\left(y-x\right)}+\frac{1}{\left(z-x\right)\left(z-y\right)}}$$
ocean sealBOT
Nayan RBLX
tidal trout
got this now what
high rapids
You made some mistakes multiplying the fractions
high rapids
Let's do the first one for example
Oh my bad
You are right
The common denominator here is all them denominators multiplied right?
lone heartBOT
@tidal trout Has your question been resolved?
high rapids
. reopen
tidal trout
ok so it will be 1/the denominator right?
high rapids
Let's see
tidal trout
1+1-1/ the denominator, 2-1 = 1
high rapids
What is the common denominator you got?
high rapids
So it's zero?
lone heartBOT
Available help channel!
warm gull
I don't really understand which step I messed up on. I assume something with 0r, but I thought the variable must stay for it to have a proper answer, instead of just having "-1=11"? What steps am I missing to get the answer of 1? Thank you
harsh girder
why 6r+6r =0r?
final olive
.close
lone heartBOT
final olive
is this set of forumlas consistent {(P → Q),(Q → R),¬(P → R)}
lime bobcat
No
P -> Q and Q->R implies P->R
But the set has also not(P->R), so the set derives a formula and its negation which is inconsistent.
lone heartBOT
Available help channel!
wide swift
Where can I be pointed to begin solving this problem? A link/explanation as to the steps of solving work be best
harsh girder
you can simplify $\log_4(9)$ first
ocean sealBOT
秋水
harsh girder
\log_{2^2}(3^2)
wide swift
$\log_{2^2}(3^2)$
ocean sealBOT
aaPattles
ocean sealBOT
aaPattles
harsh girder
Then I get rid of the bottom two so I get $x-5 = 2^{(3^2)}$ right
no
wide swift
I'm lost 
harsh girder
do you know $\log_{2^2}(3^2) = \log_2 (3)$
ocean sealBOT
秋水
harsh girder
do you know change of base formula?
harsh girder
$$\log_{4}{9} = \frac{\log_{2}(9)}{\log_2 (4)}$$
ocean sealBOT
秋水
wide swift
Then I apply that to the main formula?
harsh girder
$$\log_{4}{9} = \frac{\log_{2}(9)}{\log_2 (4)} = \frac{2 \log_{2}(3)}{2} = \log_2(3)$$
ocean sealBOT
秋水
wide swift
This means $\log{4}{9} is the same as \log{2}{3}$?
ocean sealBOT
aaPattles
wide swift
I give up on this bot
harsh girder
\log_{4}(9)
wide swift
\log_{4}(9) = \log{2}{3}
harsh girder
yes, they are equal
wide swift
And this is always be the case
And his applies to similar logs too?
For example, $\log{5}{50} = \log{25}{2500}$?
harsh girder
no
ocean sealBOT
aaPattles
wide swift
I need a documentation for this bot
harsh girder
the formula is
$$\log_{a^{m}} (b^n) = \frac{n}{m} \log_a(b)$$
ocean sealBOT
秋水
harsh girder
wide swift
So the reason behind why the bottom fraction in the first fraction = 2 is because it turns to be
harsh girder
$\log_2(2^2) =2$ because of $2^2$, it's not $1$
ocean sealBOT
秋水
wide swift
Gotcha
harsh girder
or you can write
$$\log_2(2^2) = 2 \log_2(2)$$
ocean sealBOT
秋水
wide swift
Do we ever get rid of the $\log_2$ similar to how we get rid of the $2^x$ in this equation?
ocean sealBOT
aaPattles
harsh girder
$\log_2(x-5)=\log_2(3)$,
then you should use $x-5=3$
ocean sealBOT
秋水
harsh girder
yes
wide swift
Do we ever have to bring them back? I remember in school they still play a part of the equation
harsh girder
you can have a check, notice the domain of log(x) is x>0
harsh girder
yes
queen lodge
What is the graph of √1+sin2x
lone heartBOT
What is the graph of √1+sin2x
modest dragon
What is the graph of √1+sin2x
use the inequation 1+sin(2x)=>0
and then get and idea
queen lodge
This will always be >=0
alpine sable
,w plot sqrt(1+sin(2x))
ocean sealBOT
tribal oxide
Do you know how to draw the right side graph?
queen lodge
Right side?
tribal oxide
Rhs
queen lodge
Yes
tribal oxide
Epic
lone heartBOT
queen lodge
You cannot cancel an unknown variable
last ether
You forgot to set sin(x) = 0
I mean you can cancel it but it ends up removing a solution
And you generally shouldn't
Factor sin(x) out
Just factor it out
sin(x)(3-5cos(x)) = 0
Plus you forgot -53.1
queen lodge
You even cancelled cos x
Which we cannot
You cannot muntiply an variable with zero and say it as zero
queen lodge
Let me solve and explain
last ether
They didn't cancel cos(x) out
I mean they kinda did but it wouldn't matter since they multiplied by cos/cos
I got -53.1, 0, 53.1
Yeah
Also
The cosine in the denominator
queen lodge
ocean sealBOT
Umbraleviathan
last ether
You also added an extra cos(x)
You can take an unknown value and multiply it by 0 to make it 0, but it has the possibility of removing a solution
queen lodge
Yes I got it if x =π/2 it will be infinite
last ether
What axi0m did was fine
Which is what I would've done, apart from just straight out removing a solution
Use triangles
Or the fact that cos(x) = cos(-x)
That's a shit drawing
They share the same adjacent and the hypoentuse is always positive (and equivalent)
The what
British moment
Useless
Too much to memorize
Just draw triangles
lone heartBOT
topaz palm
ayy
lone heartBOT
Available help channel!
alpine sable
lone heartBOT
alpine sable
what am I doing wrong
balmy warren
yes
alpine sable
lone heartBOT
tacit arch
• Show your work, and if possible, explain where you are stuck.
alpine sable
My bad
I don't know how to go about this problem
I was originally gonna square root 1+3^2
tacit arch
do you know what $\arctan$ or $\tan^{-1}$ is?
ocean sealBOT
riemann
alpine sable
Yea45 degrees?
tacit arch
Yea45 degrees?
no. arctan is a function
tacit arch
do you know what it gives you in a triangle?
alpine sable
Theta?
tacit arch
right. now all you need to do is draw the associated triangle with your given vectors
tacit arch
$\tan(\theta) = \frac{\text{opp}}{\text{adj}}$ so just be careful which right triangle you draw so you get the right angle
ocean sealBOT
riemann
alpine sable
lone heartBOT
alpine sable
what did I do wrong
placid zinc
prime badge
we don't see what you did
alpine sable
Oh it is 125
so then I do 2/root 125
and then inverse cos
and change it to degrees
alpine sable
if you mean 5 by 2
wait 5/2?
pliant orchid
where does this 2 come from? cheers
lone heartBOT
where does this 2 come from? cheers
@pliant orchid Has your question been resolved?
pliant orchid
<@&286206848099549185>
hollow shale
Yep
ocean sealBOT
lone heartBOT
lone heartBOT
Available help channel!
alpine sable
What does it mean to add two equations
lone heartBOT
What does it mean to add two equations
lone heartBOT
green wigeon
how do i do part b and c?
rustic jungle
so whats your question
why this matrix is good?
you can just put down each beta into a linear combination of alfas
like b 1= 1a1+1a2+0a3+0a4
you should put them in row
not column i think
cuz they form a system of linear equations
lone heartBOT
@spare sierra Has your question been resolved?
lone heartBOT
@spare sierra Has your question been resolved?
lone heartBOT
@spare sierra Has your question been resolved?
fierce prairie
lone heartBOT
fierce prairie
Is it assumed to be a right triangle?
void niche
lone heartBOT
@fierce prairie Has your question been resolved?
drowsy willow
no, but using the information, you can get that the triangle is right because of median to hypotenuse theorem.
and the answer is 3√3 btw
fierce prairie
no, but using the information, you can get that the triangle is right because of...
ahh okay that’s what I thought, just was unsure if a theorum exists or something
fierce prairie
and the answer is 3√3 btw
yea that’s what I got using law of sines
thank you
.close
gloomy wolf
lone heartBOT
gloomy wolf
my teacher said i have to use the concept of special right triangle
but im kinda confused
.close
fierce prairie
lone heartBOT
fierce prairie
is there some rule that says the 3 angles add up to 360?
wary stream
No, but you can apply something very similar
Drawing a straight line, you created 2 right angles
And that's a 5 sided shape, determine the sum of interior angles of a 5 sided shape
wary stream
Yes
crimson fractal
Hi um I hope I don't have to be in university to get help. I am just trying to figure out how many pounds it takes at 100 mph to create 75000 psi. Its not homework or anything but something I've been trying to figure out how many pounds at 100 mph is needed to punch through steel at 75000 psi.
knotty spire
-
That is not nearly enough information to give an answer. Your question is vague in many aspects.
-
This is not mathematics, this is a problem for physics and engineering.
crimson fractal
Whats that?
It popped up and thought you sent
How do I calculate psi into pounds? Is it a pounds per second thing? If a punch is at 100 mph?
abstract fractal
psi is pounds per square inch. You need to know the area in which the force is acting upon
crimson fractal
Oh okay.
abstract fractal
Moreover, you need more info than just speed to calculate force
crimson fractal
Well I know the psi is 75000 so does that mean if the speed is 100 mph the force to reach 75000 is 750 lbs?
@abstract fractal
abstract fractal
No
You need to know the area the force is acting upon to know the force required
crimson fractal
Like if its a steel door or plate the size of the fist that would do the punching?
abstract fractal
Moreover, you need to know information like the acceleration of the steel as a result of the punch, which is going to depend heavily on the steel itself, how thick is the steel, and things like that. It's gonna be almost impossible to predict to any accuracy from just the thought alone
crimson fractal
Oh okay.
abstract fractal
Like if its a steel door or plate the size of the fist that would do the punchin...
You're going to need to know the surface area making contact with the steel
crimson fractal
So lets see the fist radius is 6.8 inches the speed is still 100 mph with it being low carbon steel of 2 inches thick? Is there anything else needed for the calculation? @abstract fractal
abstract fractal
I'm not an engineer
You'll find there're very few engineers here
I do not know how to calculate the force from just the speed and type of steel
I doubt many do
The best way would be to do manual experimentation, but I doubt anyone here has access to the required machinery to do so
crimson fractal
Oh ok
lone heartBOT
@crimson fractal Has your question been resolved?
crimson fractal
Its fine though. Decided to do something else.
lone heartBOT
Available help channel!
gusty quiver
lone heartBOT
gusty quiver
If I want to treat a problem like this rigorously, then it's enough to show that all the variables in a system of equations are equal to each other (or each one is just multiplied by a constant compared to each of the other variables) and one of the variables is equal to zero, right?
lone heartBOT
@gusty quiver Has your question been resolved?
normal ingot
Let $(a_1,a_2,\ldots,a_{24})$ be a sequence consisting of the integers 1 to 24, in some random order. Show that there must be three consecutive $a_i$ with sum at least 38.
ocean sealBOT
lirmirit
normal ingot
I've been trying to prove it with a contradiction so far, but didn't have much luck
Any ideas on how to approach this?
lone heartBOT
@normal ingot Has your question been resolved?
normal ingot
<@&286206848099549185>
mortal trellis
What is the sum of all a_i? What happens if a1,a2,a3 and a4,a5,a6 and a7,a8,a9,... all sum to less than 38?
normal ingot
What is the sum of all a_i? What happens if a1,a2,a3 and a4,a5,a6 and a7,a8,a9,....
I got (sum of all a_i)=1+2+...+24=300 and if a1,a2,a3 and a4,a5,a6,... are all less than 38, the sum of all a_i<38*8=304
mortal trellis
If it's smaller than 38 it's at most 37
normal ingot
oh, that's right
so the sum of all a_i would be less than or equal to 37*8=296 which is a contradiction
mortal trellis
Yes
upbeat sleet
lone heartBOT
upbeat sleet
someone tell me the formula for quad with 90 degree angle
north adder
someone tell me the formula for quad with 90 degree angle
depends on which quadrilateral
north adder
but i think its 1/2 (A + B) or something
uhhhh whats A and B
north adder
I see
okay can you tell me the side of the rhombus
if the perimeter is 28
and all the sides are equal
upbeat sleet
okay can you tell me the side of the rhombus
that makes no sense
north adder
that makes no sense
the magnitude of the side
youre given that the perimeter is 28
and you know that the sides are all equal
right?
which part are you struggling to understand
north adder
im gonna guide you there....
upbeat sleet
bc the is a 90 degree angle
north adder
the triangle is 90 degrees
doesnt mean that the sides are perpendicular t oeach other
upbeat sleet
they dont intersect?
north adder
no
okay wait
what EXACTLY are you confused about
the formula is 1/2 ab where a and b are the diagonals of the rhombus's magnitude
does that make sense?
upbeat sleet
no
so u do one side
then the other then add them together as usual
so u add 28 x 28?
north adder
what are you saying
north adder
so u do one side
what does this even mean
upbeat sleet
one triangle
upbeat sleet
what doesnt make sense
do u do 28 + 28 then half it double the answer
north adder
do u do 28 + 28 then half it double the answer
no why would that work
north adder
okat
good
so as you can see
the graph here we have diagonals of the rhombus
does this diagram make sense?
in the righthand figure AC is d1 and BD is d2
upbeat sleet
does this diagram make sense?
yes
north adder
good
tell me what you know in youre question
lets start with that
its good to state the obvious
because then you know what you are given
upbeat sleet
everyone line is 28 cm
upbeat sleet
oh i didnt read
north adder
perimeter is the magnitude of the enclosed shape
yeah
so what does each side equal to?
upbeat sleet
2cm apex
north adder
@upbeat sleet you know the pythagorean theorem right?
upbeat sleet
yes
north adder
gib so the diagonal is what we need in this case
north adder
nono it still is
you need to find the base of the triangle
because you want to know how much the diagonal is
upbeat sleet
base is 7 cm?
north adder
a side is defined for a shape as a line segment that connects 2 vertices
upbeat sleet
doing home school
upbeat sleet
i cant even think
north adder
anyways can you find the base side of the triangle?
upbeat sleet
8?
upbeat sleet
i estamated
north adder
remember that the pythag theorem is
a^2 + b^2 = c^2
so what would these be
you know c^2 is the hypotenuse
so what would c be?
upbeat sleet
c^2 base?
north adder
you know c^2 is the hypotenuse
.
north adder
no
your concepts are mixed up
the right angle
is inbetween the 2 sides
and the hypotenuse is opposite of the right angle
does it make sense?
can you redo your equation?
north adder
how do i know the length of say b
you have to find that out
show me what you have so far
youre doing great!
north adder
you still havent found the base?
upbeat sleet
its 7 cm
upbeat sleet
so this is how i see it
north adder
remember this?
upbeat sleet
wait
lemme talk
the 2cm mark is the height
is isnt were the length ends
all the way around is 7 cm
upbeat sleet
huh
hidden pier
$Area = base \cdot height$
ocean sealBOT
🙛𝕍ѳrtєx🙙
upbeat sleet
yes ik
north adder
finding area is just bh is it not
ight, take over if you want.
upbeat sleet
am i trippin
thats what i said
upbeat sleet
ight, take over if you want.
i appreciate your help tho
hidden pier
i mean yeah you can split it up into two triangles and a thin rectangle in the middle as well
north adder
i mean yeah you can split it up into two triangles and a thin rectangle in the m...
am i balding or
hidden pier
but i just see a parallelogram with height 2cm and base 7cm
north adder
bruh
im sorry @upbeat sleet i wasted your time
youre both right
i overcomplicated the solution
upbeat sleet
i overcomplicated the solution
its fine
upbeat sleet
i mean yeah you can split it up into two triangles and a thin rectangle in the m...
so i was correct with 7cm times 2 cm then half the answer to get one triangle?
hidden pier
yes it would im dumb
also why get one triangle? the question asks for the whole area
upbeat sleet
you can just double one triangle can u not?
hidden pier
so i was correct with 7cm times 2 cm then half the answer to get one triangle?
sorry yeah you can do this
upbeat sleet
**yes it would im dumb**
im getting trolled
hidden pier
but the question asks for the entire area
hidden pier
you mean 7cm?
upbeat sleet
i mean to get both
hidden pier
both whats?
upbeat sleet
so the rombus basically
hidden pier
the area of this whole parallelogram is just $b\cdot h = 7\cdot2=14$
ocean sealBOT
🙛𝕍ѳrtєx🙙
hidden pier
because if you picture it as just a parallelogram then the area of a parallelogram is $A = bh$
ocean sealBOT
🙛𝕍ѳrtєx🙙
upbeat sleet
because if you picture it as just a parallelogram then the area of a parallelogr...
my answer any good?
hidden pier
no
it's 14
the base is 7 and the height is 2
so the area of the rhombus is the base times the height = 7*2
=14
ocean sealBOT
🙛𝕍ѳrtєx🙙
upbeat sleet
do you understand what $Area = Base \times Height$ is
yea
hidden pier
then what part don't you understand
upbeat sleet
there is a right angle in it tho
hidden pier
the base is 7 and the height is 2 right?
hidden pier
this is the formula
base times height is the formula where the height is perpendicular to the base
upbeat sleet
base times height ***is*** the formula where the height is perpendicular to the ...
ive never heard of that before
hidden pier
it's a simple "calculate the area of a parallelogram" question
base times height = area
hidden pier
@upbeat sleet does this picture make sense to you
hidden pier
so what about 7*2=14 do you not get
upbeat sleet
but i was looking more towards its different if there is a right angle in a rombus
its a diffeernt equation
hidden pier
um no
OH
if you take the two diagonals and multiply them, then halve the result, that is also the area (of a rhombus)
however this problem you have here is just base times height
it's just like a parallelogram
upbeat sleet
i see
hidden pier
and the second image even has a right angle (because the height is always perpendicular to the base; if it wasn't then it wouldnt be. height)
upbeat sleet
so im thinking of something different
hidden pier
maybe you're thinking of this
which was also what the previous helper was suggesting
but in this problem it's just much simpler to just do base times height instead of going through the trouble of finding the diagonals that we don't even know
alright it's 1:00 am over here so ima sleep
lone heartBOT
@upbeat sleet Has your question been resolved?
karmic falcon
what is the scientific form of 2^11*5^13
lone heartBOT
what is the scientific form of 2^11*5^13
karmic falcon
.close
dry drum
Is this move legal?
lone heartBOT
Is this move legal?
lone heartBOT
@dry drum Has your question been resolved?
vocal tapir
lone heartBOT
vocal tapir
I'm failing to understand how the denominator becomes x^2(x-1) and not x^3(x-1)
isn't the LCD [1st denominator] * [2nd denominator] after factored?
median oar
The first fraction needs to multiply by (x-1) the second by x and they are the same
Pretend x=3, 9 and 6, the LCM is 18, which is 3 * 3 * 2 = x * x * (x-1)
@vocal tapir
vocal tapir
ugh
vocal tapir
The first fraction needs to multiply by (x-1) the second by x and they are the s...
I'm still having a hard time grasping why it shouldn't be x^3(x-1)
isn't it x^2(x(x-1))
median oar
What’s the lowest common multiple of x² and x(x-1)
vocal tapir
What’s the lowest common multiple of x² and x(x-1)
I guess not x^2(x(x-1))
median oar
How do you find the lowest common multiple of 6 and 9
vocal tapir
How do you find the lowest common multiple of 6 and 9
find the nearest positive number both can divide into
median oar
Have you ever done it the way where you divide both by prime numbers until the results are no longer divisible
Like say 54 and 72
You divide it by 2, 27 36
Divide by 3, 9 12
Divide by 3, 3 4
Then you take everything you divided and the remaining bits and multiply them together
vocal tapir
Have you ever done it the way where you divide both by prime numbers until the r...
yes, but never solidified my skills in it
ocean sealBOT
median oar
,w LCM 54 and 72
ocean sealBOT
vocal tapir
Divide by 3, 3 4
hmm yeah that makes sense
median oar
yes, but never solidified my skills in it
Math is very foundation based it’s really hard to advance in math if you don’t have the previous stuff down pat
vocal tapir
Math is very foundation based it’s really hard to advance in math if you don’t h...
yeah, but I'm almost excellent to this point, but you can't just have everything in math at 100% skill level ig at least for me
let me think through what you've written
median oar
Ask if u got any problems I’ll be home soon and can write some stuff to make it a bit easier to understand if you need
vocal tapir
Like say 54 and 72
ok so the way I understand is that:
54 is divisable by 3, 9, and 12
3, 9, and 12 by
3, 3, 4
then 3,3,4
72 by 2, 27, and 36
so
2, 9, 6,
2, 3, 6
do I have it right up until this point?
meaning 9=3^2 and if it's squared, I just take the root
median oar
You’re thinking about it too complicatedly
vocal tapir
😭 It's not a problem with me, but am I right?
median oar
I honestly don’t know what you’re doing
median oar
Do you at least know that every integer can be represented as a product of primes?
vocal tapir
Do you at least know that every integer can be represented as a product of prime...
yes
1 = 11
2= 21 for e.g
median oar
I’m nearly home gimme like 5mins
ocean sealBOT
median oar
the number 2x3x3x3x4 is divisible by 2x3x3x3 (54) and is left with 4
the number 2x3x3x3x4 is divisible by 2x3x3x4 (72) and is left with 3
vocal tapir
ohhh that makes a lot of sense
median oar
This should be obvious at this point now
Btw if u want to find the GCF (greatest common factor) then you multiply everything on the side and ignore the bottom 2 remaining numbers
In 54 and 72’s case it’d be 2x3x3
vocal tapir
This should be obvious at this point now
that makes everything a whole lot easier
never got taught such a great way
of handling this
ty @median oar
median oar
:D
steady dagger
what is R? there are two such areas...
lone heartBOT
what is R? there are two such areas...
steady dagger
and how do we know the bounds of integration
because I certainly racked my brains in vain to attempt to come up with a justification
dense bloom
sum of first 6 members in arithmetic progression is ?
lone heartBOT
sum of first 6 members in arithmetic progression is ?
dense bloom
how to find first n members with this?
gray ingot
sum of first 6 members in arithmetic progression is ?
your handwriting is not legible
what does the second line say
gray ingot
thats the only information you have?
dense bloom
yes
gray ingot
what was the original question
dense bloom
moment
the sum of the third and fourth members of the arithmetic sequence is 7. the sum of the first six members of that sequence is?
dense bloom
**Swas.py**
what is this?
gray ingot
Sum of first 4 terms - Sum of first 2 terms
dense bloom
ok
gray ingot
$S_n = \frac{n}{2}[2a+(n-1)d]$
gray ingot
hmm but this just gives $3a_1 + 5d = 7$
gray ingot
$S_6 = \frac{6}{2}[2a_1+5d]$
gray ingot
I dont think this is solvable 
gray ingot
$S_6 = 3(7-a_1) \$
$S_6 = 21-3a_1$
dense bloom
i don't know
gray ingot
it depends on the first term and we don't know what that is
there can be infinite such series
wait
yeah
Wait
gray ingot
mb I made a dumb mistake
so we had
a3 + a4 = 7
a1 + 2d + a1 + 3d = 7
2a1 + 5d = 7
alright?
dense bloom
yes
dense bloom
oh shit
gray ingot
so $S_6 = 3 \times 7 = 21$
dense bloom
good job
gray ingot
and thats your answer
gray ingot
there can be infinite such series
btw this still holds