you can also plot two curves, 1st x*y=1 and x+y=0.5. where they intersect is the solution. since they font intersect anywhere, theres no solution to this

Ask them what is 1/00000000000000.1 ?

Teacher went to Trump academy

This is how we end up with Terrance Howard math

This is overthinking on a whole other level. 1 divided by zero is 1 due to the fact that in this case zero literally stands for nothing. So 1 divided by nothing equals 1.

If I divide an orange 0 times then I still have a whole orange

25 +/- sqrt(525) ===> 47.9129 or 2.0871.

X=a/0=infinity

Area is a square law, perimter is linear with scale... the ratio of area to perimeter has to increase with larger, and reduce for smaller... This ratio will follow k x^2 curve... But k depends on the units in use.

Long division is still the best methodology in my world (old guy). But what I'm most impressed by in this video is your mad marker skills. Love it!!

So can a rectangles area be equal or bigger than its perimeter ??? Yes every rectangle can. just change the units used to nanometres or something ??

Before watching the video: the 60 degree 1.50$ one is definitely the better deal I’d say

16 is that number

It is easy -1/12

It is not possible. The largest area a rectangle can have is a square. A square ¼ on a side has an area of 1/16 and a perimeter of 1. A rectangle with a perimeter of 1 can, therefore, have any area between 0 and 1/16, but never greater

infinity is not a number !!! Take care!!!

Yes. A square of 1x1. A square is a rectangle of a=b

ax=b if a is non zero , x=b/a Else if a=0 if b is non zero no solution Else if b=0 any number is a solution . That's the whole story. Stop disturbing the mind of the people.

Are you building a rocket? This is not 3rd grade math😂

The rectangle with the smallest perimeter to its area is a square. So if we have a square of 1 cm2 the perimeter is 4. It cannot be smaller. Hence it is not possible to have a rectangle with an area 1 cm2 less than 4. Now with Area 100 cm2, the smallest perimeter is a square with side 10 cm. The smallest possible perimeter is 40 cm. So what happen if we decrease the width and increase the length to say 1x100 = 100 cm2? Now we have a perimeter of 202 cm. Between those two rectangles P=40 and P=202, there will be one with the P=100 cm. Because all values of width 1-10 are valid and all length from 10-100 are valid. I do not need algebra to find it. Just logic with examples of both sides and a continuum between, to know the rectangle must be there somewhere.

25±5√21

L < 1/2 W < 1/2 L×W < 1/4 < 1

We need a tutorial on how to switch between two markers so perfectly. 1:23

I JUST MET SIMILAR SITUATION. Two mothers of state school students who are my friend insist that 1:0=0 I argue with them but then i go silent. Is this what they teach in school now?

Arithmetic mean is always either greater than or equal to the Geometric mean

Does this mean 12inches is better than 14"? I'm dumb and won't even try this calculation. I'll subscribe juat for this😂😂😂😂

You do nothing. The teacher is correct. Any number divided by zero is zero. I learned this 40 years ago in advanced mathematics.

0:03 0=-4🔥🔥

You can just substitute a for 1/b and you get the quadratic b²-½b+1 =0 and the discriminant is basically √¼-4 which means the value of b is not real and by extension, a is also not real so no, such a rectangle doesn't exist

short: p = 2l + 2w = 1; l + w = 0.5, but if both are positive l < 1 and w < 1 so l*w < 1.

I wonder if the teacher has an ENTJ personality, because my manager does too and they can respond to errors by doubling down. I hope I don't sound prejudiced by saying this. I think I wrote clearly enough. And FYI it's 'doubled down' rather than 'doubled it down'. ❤

AM - GM: x + y ≥ 2 * sqrt(xy) = 2

I thought 1 divided by zero = ∞

I would have guessed 1 before I guessed 0. If you take something, and divide it by nothing, what are you left with? The same thing you started with. 🤷🏽‍♀️ but I guess that's wrong too so.....😅

In the fifth dimension, anything is possible.

Easy to prove. Divide by 0.1 to show that it becomes 10. Then by 0.01 to show it becomes 100. Then by 0.0001 to show it becomes 10,000. Then ask them if you see that getting closer to 0 any time soon.

For it to work the Area has to be more or equal than 16. A>=16 Where: l = (A±sqrt((A^2)-16A)) / 4 w = (A∓sqrt((A^2)-16A)) / 4

lol I thought a / 0 = ∞, actually I remember my mat A formulas I had " a / 0 = ∞ " , anyways been undefined makes math broken let's go Terrance Howard!

I’m glad he wasn’t my teacher! He made that way more complicated then it had to be

0 is nothing and undefined means can't be done or no answer or nothing so 0 can still be acceptable right

So yeah, let's see... 2a+2b = 1 ab = 1 a+b = 0.5 ab = 1 b = 0.5-a a(0.5-a)=1 -a² + 0.5a - 1 = 0 0.5² - 4(-1*-1) = -3.75 < 0, therefore we got no solutions, assuming that the sides have positive real number lengths Although you could try making imaginary shapes... Reminds me of the i,-1,0 triangle

This question is poorly asked, or I'm dumb. How can you have f(sin2x) = f(x)? The x on the left is not equivalent to the one on the right.

I could learn from this guy all day.

As of what I know, we can only split roots when there's at least one +ve value in a root, so we just can't do √1 = (√-1)(√-1)

Given a Euclidean space, you're absolutely correct. If you allow curved space, though, postulates 4 and 5 don't necessarily apply and you can do many weird things.

Isn't i strange that 1x0=0 , 0x0=0 and 17x0=0? If we agree that this is not the same outcome, than it makes completely sense to say that 0/0=1

I don’t understand how this is so hard. A zero is nothing. There is nothing there. If you have a number, any number, and you divide it by nothing then it remains the same number. Why does it take a long math equation to prove nothing is nothing?

If X÷Y=Z, then X=YxZ. So, if 1÷0=0, then 1= 0x0. Zero times zero does not equal one. So, one divided by zero cannot equal zero.

Maybe if the school had a student newspaper this should have been put on the front page with an article talking about the slipping standards of teachers

Sue the teacher and principal and school board. That level of incompetence cannot be tolerated.