Homework Help

[Pentium100 2,056]

31 minutes ago, Emerald Heart said:

Does anyone understand any of this?

That's a weird way to solve a quadratic equation.

Normally I would do it like this:

So, for this equation:

Anyway the idea is that:

When you multiply two sums, you multiply each part of one sum with each part of the other sum.

, squaring a sum looks like this:

So the example has you trying various combinations to find the solution instead of doing the way I normally do.

Edited April 29, 2020 by Pentium100
Correct a mistake

[Emerald Heart 24,571]

3 minutes ago, Pentium100 said:

That's a weird way to solve a quadratic equation.

Normally I would do it like this:

So, for this equation:

Anyway the idea is that:

When you multiply two sums, you multiply each part of one sum with each part of the other sum.

, squaring a sum looks like this:

So the example has you trying various combinations to find the solution instead of doing the way I normally do.

Awesome! I'll try that out.

Thank you very much! I was getting irritated...

[Pentium100 2,056]

12 minutes ago, Emerald Heart said:

Awesome! I'll try that out.

Thank you very much! I was getting irritated...

One small correction, it does not matter for this, but it would matter for the general thing:

[SasQ 1,355]

On 4/29/2020 at 9:10 PM, Emerald Heart said:

Does anyone understand any of this?

I've tried and tried but I can't get it right.

It is a method of factoring quadratics (and other polynomials as well) that doesn't require going through that messy quadratic formula. (If you practice it for a while, you might even be able to factor them in your head! :> )
It is based on the fact that if a polynomial has any rational solution (a whole number, like 7, or a fraction, like 2/3, but no radicals, like √2), then it must be a divisor of its constant term. (I can explain you why is this true in my next post if you wish )
So you start by looking at the constant term (49 in this case), and writing down its divisors.
What numbers divide into 49?
Well, it is a perfect square (49=7·7), and it also certainly can be divided by 1 and itself (as every number), so the divisors are (positive as well as negative):
       1,   7,    49
      -1,  -7,  -49
You can check if any of them is a solution by substituting it for x into your equation and checking whether it will produce 0 for the whole thing. (Notice that we add 49 as the last step, so the part z² – 14·z must somehow produce the opposite of it, -14, if we want them to cancel each other out and produce 0 as the final result.) Let's see:
    (1)² – 14·(1)    certainly won't work, because 1² = 1, and we subtract 14, so we get –13, not –49.  Same with
   (-1)² – 14·(-1)   because (-1)² = 1, but (-14)·(-1) = 14, so we get 14+1 = 15, not –49. We need something bigger as a factor. Maybe 7 will work?
    (7)² – 14·(7)    7² is 49, and (-14)·7 = -98, which is twice as big, so it looks promising:  49 – 98 = -49 indeed, so after adding the last 49 it indeed cancels out to 0.
That way we've just found or first factor (and first solution to the quadratic, for that matter) That factor will be (x–7), so you can either divide out by it to get the other factor, or use the table as the one you posted above. The table uses the fact that whatever the two factors of 49 are, they must add up to the middle term's coefficient (the one next to z in the first power; or for general polynomials: the coefficient of the almost-greatest power). The middle term has 14 as its coefficient, and we know one part of it: 7, so the other part must also be 7. That's what is shown in the table above. Therefore:
    z² – 14·z + 49  =  (z–7)·(z–7) = (z–7)²

Now why does that all work? (And when it doesn't?) Where doest that table come from?

Let's start with the factored result, in the most general form, in which instead of the 7's, we put some symbols for the solutions, let's say X and Y:
     (z–X)·(z–Y)
Now multiply it out (by the infamous FOIL dictum):
     z·z + (-X)·z + z·(-Y) + (-X)·(-Y)
     z² – X·z – Y·z + X·Y
Collect the z's together:
     z² – (X+Y)·z + X·Y
and now compare it with the general form of the quadratic (I used the same colors to help you match them):
     z² + b·z + a
Do you see it?
The product of our solutions, X·Y, matches the constant term a.  This is our 7·7=49.
The negative sum of our solutions, –(X+Y), matches b, the coefficient of z (the almost-greatest power). This is our -(7+7) = -14.
We can summarize it as two formulas, known as Viète's formulas (because he was a famous mathematician using this trick):
    a =     X·Y
    b = –(X+Y)
The table is a short-hand for that.

In case you asked what if there's some coefficient c with the greatest power of z?
Well, just divide everything by it so that you got just a single z² alone, and remember to scale your solutions accordingly:
    a/c =     X·Y
    b/c = –(X+Y)
but the overall process is exactly the same for fractions as for whole numbers.
Although the formulas are still true for radicals, it's harder for us to "factor" them with this method, so this formula is usually applied in schools only when the solutions are whole numbers or fractions. So let me spare you that rAdiCaL mAtHs for now

You can check that this trick works also for higher-degree polynomials, such as cubics, quartics, quintics etc. Just there's more factors to try out.
For cubics, there's always three solutions, so the constant term will be made of three factors:  X·Y·Z, while the coefficient of the almost-greatest power (z² in this case) will be X+Y+Z.
(Notice that it's not negative this time: the signs alternate; for quartics it will be negative again, –(X+Y+Z+W), for quintics positive, and so on...)

[Derpsieh 231]

I could use some help with general HTML/CSS/JS ^^'

There's nothing specifically I need help with right this second, but if anyone is proficient with those, I'd love to be able to talk to them and get their insights/views on certain things :3

[Muffinnz 3,219]

Im afraid to use this thread because people will see my lack of knowledge on a subject

[Splashee 28,537]

4 hours ago, Muffinnz said:

Im afraid to use this thread because people will see my lack of knowledge on a subject

Better to have us see your lack of knowledge, then your teachers/school/grade, etc. Get help when you can! Also, I would never think lesser of you for asking help with homework or anything else

[ExplosionMare 18,007]

On 10/28/2020 at 1:18 AM, Muffinnz said:

Im afraid to use this thread because people will see my lack of knowledge on a subject

If it makes you feel better, I had no idea for the longest time that Spain was a part of Europe 

The point here is that we’re helping you with stuff so it would be weird if you knew everything, you know? We don’t care if you don’t know stuff, we just wanna help.

[ExplosionMare 18,007]

All I need to do is graph these points but I’m not sure if I marked them right or got them all down. Could anyone take a look at this and let me know how I can improve this before I go ahead and connect the points?

[Pentium100 2,056]

55 minutes ago, ExplosionMare said:

All I need to do is graph these points but I’m not sure if I marked them right or got them all down. Could anyone take a look at this and let me know how I can improve this before I go ahead and connect the points?

It looks like you have marked the four points correctly. I cannot really comment on the rest of the task, because it looks like the question is missing here.

[ExplosionMare 18,007]

8 minutes ago, Pentium100 said:

It looks like you have marked the four points correctly. I cannot really comment on the rest of the task, because it looks like the question is missing here.

All I was told to do is plot what was given to me. Here’s the list of what I have to plot if this helps at all:

[Pentium100 2,056]

9 minutes ago, ExplosionMare said:

All I was told to do is plot what was given to me. Here’s the list of what I have to plot if this helps at all:

The task is a bit confusing. The way I understand it - 

Function graph starts at (-3,0) then goes down to (-2,-3), then up to (0,-2), then down to (1,-3), continuing to (3,-5), then up to (0,5). Something like this:

 

 

 

[ExplosionMare 18,007]

16 minutes ago, Pentium100 said:

The task is a bit confusing. The way I understand it - 

Function graph starts at (-3,0) then goes down to (-2,-3), then up to (0,-2), then down to (1,-3), continuing to (3,-5), then up to (0,5). Something like this:

 

 

 

I assume that’s how they want it. Thanks!

[SasQ 1,355]

On 2020-10-28 at 6:18 AM, Muffinnz said:

Im afraid to use this thread because people will see my lack of knowledge on a subject

Lack of knowledge is not something to be ashamed of – there is no person on this planet who would know everything; there will always be things that you don't have knowledge of. And the only way to gain knowledge is to ask people who have it and learn from them. It doesn't hurt And learning stuff this way is usually much more fun than in schools. The more damaging is not asking, because then you remain ignorant and it piles up. The most dangerous are those people who don't want to learn even if you give them knowledge on a silver plate, because they think that they already know everything. Because they usually don't, and their stubbornness makes them incapable of learning. That's how people get stupid. And this is a reason to be ashamed.

On 2021-09-12 at 8:12 PM, ExplosionMare said:

All I need to do is graph these points but I’m not sure if I marked them right or got them all down.

Yup, you pretty much got them right, except one: the (-3 , 5) one, because the problem says that [-3, 5] is the DOMAIN of this function, so it's not supposed to be a point. It is a RANGE (along the X axis) in which this function has values. If you pick any point within this range on the X axis, and move straight up or down, you should be guaranteed to find a point of the graph there, representing the value of that function for that particular argument X. But if you pick some point outside of that range and move straight up or down, you shouldn't be able to find such a point, because outside of that domain, function is supposed to have no values.

But other than that, your points seem to be ok. Now you just need to connect them, and @Pentium100 showed you one possible way of doing it.

Keep in mind though that these lines don't necessarily have to be straight (in fact, they don't even have to be continuous curves). As long as the problem doesn't specify what kind of function you are supposed to draw, you can pretty much draw any squiggle that goes through those points, provided that it meets all the required conditions. E.g. from x = -3 to x = -2 they require it to decrease, so you can move down (either slow or fast), but no up.

If the problem required it to be a polynomial curve of the lowest degree that goes through these points, this question would get much more interesting And in that case we would need to ask Mr. Lagrange for help

Edited September 17, 2021 by SasQ
typo

[SasQ 1,355]

@ExplosionMare in case you wanted to know how the simplest polynomial curve going through your points looks like: just for fun I found it (with the help of Mr. Lagrange)
Here it is, along with its formula:

You can play with it yourself here: https://www.desmos.com/calculator/ehgi7ngaor

Try moving the red points to some other coordinates by clicking and dragging them The curve should then rearrange itself to still fit through all those points.
The original curve you started with will still be visible as a dashed line, for comparison.

Notice however that in my curve, points B and E aren't local minima, and point C is not a local maximum. Your problem had these requirements, so I can try come up with a curve that satisfies these requirements as well, but something tells me that its formula might end up even more complicated

Interestingly, if it wasn't limited to the [-3, 5] domain, there would be one more point at which it crosses the X axis at roughly 6.559 (if you're curious, I can tell you the exact formula for that point's X coordinate too ).

Edited September 18, 2021 by SasQ
Fun fact

[ExplosionMare 18,007]

13 hours ago, SasQ said:

@ExplosionMare in case you wanted to know how the simplest polynomial curve going through your points looks like: just for fun I found it (with the help of Mr. Lagrange)
Here it is, along with its formula:

You can play with it yourself here: https://www.desmos.com/calculator/ehgi7ngaor

Try moving the red points to some other coordinates by clicking and dragging them The curve should then rearrange itself to still fit through all those points.
The original curve you started with will still be visible as a dashed line, for comparison.

Notice however that in my curve, points B and E aren't local minima, and point C is not a local maximum. Your problem had these requirements, so I can try come up with a curve that satisfies these requirements as well, but something tells me that its formula might end up even more complicated

Interestingly, if it wasn't limited to the [-3, 5] domain, there would be one more point at which it crosses the X axis at roughly 6.559 (if you're curious, I can tell you the exact formula for that point's X coordinate too ).

Thanks, I’ll keep this bookmarked so I can use the graph in the future! That looks about like the graph I got previously. You and @Pentium100 had a point with the domain being a bit limiting. It’s too bad there weren’t many directions given besides “graph these points”.

[ExplosionMare 18,007]

So I have to complete this step on the Word Doc I’m editing:

This is what I have so far:

Does anyone know where I can find the button to change the style? I am on the Apple version of Word so some of the button placements may be different.

[Bluish 22]

I’m good with Algebra and Geometry, but im a complete boss with equations such as slope intercept form and stuff like that. I’d be happy to help ʕ•́ᴥ•̀ʔっ✏️ 

[ExplosionMare 18,007]

13 hours ago, Dominic_The_Pony said:

I’m good with Algebra and Geometry, but im a complete boss with equations such as slope intercept form and stuff like that. I’d be happy to help ʕ•́ᴥ•̀ʔっ✏️ 

Are you any good at College Algebra? Might need some help with that later if you’re interested 

[Silly Druid 6,138]

I should be able to help as well, I think I've got College Algebra covered.

[Bluish 22]

10 hours ago, ExplosionMare said:

Are you any good at College Algebra? Might need some help with that later if you’re interested 

IMMA FRESHMAN 🤓

[DemeowHiya 250]

On 2021-11-23 at 12:05 AM, Dominic_The_Pony said:

IMMA FRESHMAN 🤓

haha, I feel that. Some of my friends ask me for help in school and I'm like, "dude I'm only a Sophomore!"

I'm wishing you luck though, ExplosionMare! I hope it all goes well 

Edited December 4, 2021 by DemeowHiya

[ExplosionMare 18,007]

I'm so close to finishing this little soda can here (it isn't meant to be super complex) and all it needs is this dip/hole here:

I've been able to make something similar on a cube but I'm struggling to make it on my soda can (usually involves pressing i to make the circle then e to extrude it downwards). Here is what I have so far:

[Splashee 28,537]

 

@ExplosionMare

If you are using Blender:

 

In edit mode (TAB to switch from object mode to edit mode), press K for the knife tool (or look up online what the key is for the knife tool, or use Blender's search to find it). When you are hovering over a vertex, it turns highlighted (green  color on my old old Blender version). Click on it, and then click on another vertex. Once you made the cut you want, press Enter to apply.

 

Then, you can select a face, and delete it, to make a hole:

 

Then you can play around with the vertices as much as you want (select a vertex, press G twice to have it slide, so the face stays the same, etc, etc):

 

 

(I did this cylinder really quickly, and just started cutting to remember how the knife tool worked)

[Splashee 28,537]

Loop Cut is the thing you are looking for:

Create Mesh->Cylinder

Go to Ortho View (because modelling in perspective is not good, ever)

Go to Edit Mode (object mode can really mess up your object if you make any changes to it)

Select->Deselect All

And Loop Cut. Do as many as needed!

Quote

Mode:	Edit Mode
Panel:	Toolbar ‣ Tools ‣ Mesh Tools ‣ Add: Loop Cut and Slide
Hotkey:	Ctrl-R

Mesh->Transform->Scale to model the edges as you want them.

Video from super old Blender on PC: