No se de que país seas y pues si lo traduces espero que tengas más éxito en todo y bueno no será mucho pero ojalá que te valla bien amiga
@ATMathAcademy8 ай бұрын
Thankyou so much!
@aprendiz48 ай бұрын
Let the result be seen for more than one millisecond please
@ATMathAcademy8 ай бұрын
Sure!
@josephwaters27928 ай бұрын
The other instructional videos aren't much better right side up. Things are more described rather than explained. Performing is not teaching. Rethink your approach and rework it. Good luck!
@ATMathAcademy8 ай бұрын
Thankyou for the suggestion. I will try best next time.
@FatimetouMedlemine-bp3gv8 ай бұрын
🫡👏👏👏
@ATMathAcademy8 ай бұрын
Thankyou!😊
@tet8768 ай бұрын
or just 2×36
@ATMathAcademy8 ай бұрын
Yes
@ChiefHitemBong8 ай бұрын
Wait...I thought any number divided by itself is 1..?
@ATMathAcademy8 ай бұрын
Yes,number divided by itself is 1. Here,in this problem for 1÷3(sqrt(10)) Sqrt(10) is irrational number. So, I am Multiplying with sqrt(10) for both numerator and denominator so that sqrt(10)×sqrt(10)=sqrt(10^2)=10. So, in this way, we made denominator as a whole number 10 from irrational number sqrt(10). So, by Multiplying both numerator and denominator with sqrt(10 ) we converted it from irrational to whole number. Thus, we rationalized the denominator.
@ГошаЛайв8 ай бұрын
Теорема виетта
@ATMathAcademy8 ай бұрын
Yes! Nice For standard quadratic equation :ax^2+bx+c=0 Vietta's theorem: Sum of roots=-b÷a Product of roots=c÷a Sum of Roots=-6+-3=-9=-b÷a=-9÷1=-9 Product of roots=-6×-3=18=c÷a=18÷1=18
@ГошаЛайв8 ай бұрын
Эээ, просто 400*2
@ATMathAcademy8 ай бұрын
👍
@srikanthraavi8 ай бұрын
Great short
@ATMathAcademy8 ай бұрын
Thankyou!
@Jeffreydafox8 ай бұрын
School: Please Excuse My Dear Aunt Sally Me: Please Eat My Dear Awesome Soup
@ATMathAcademy8 ай бұрын
Thankyou for the idea for PEMDAS!
@johnroache328 ай бұрын
And don't forget… We will all use this in every day life !
@ATMathAcademy8 ай бұрын
Nice!
@Onenonlyprincess8 ай бұрын
Damn, that is some beautiful handwriting
@ATMathAcademy8 ай бұрын
Thankyou!
@totalsoupsloud8 ай бұрын
Please connect your x's!!!
@ATMathAcademy8 ай бұрын
Yes, added correct value of x greater than 14 in the description and comments section. Sorry for the confusion and Thankyou for correcting it!
@kankri998 ай бұрын
At Math Academy: x > -14; On KZbin Comment: x > 14
@ATMathAcademy8 ай бұрын
Sorry,it's X>14.
@c4yourself3198 ай бұрын
Where did the 3 and 4 go that were before X?
@ATMathAcademy8 ай бұрын
In the inequality of: (X-5)÷3>(X-2)÷4 We do cross multiplication So, 4(X-5)>3(X-2) 4X-20>3X-6 Then, moving variables on one side and constants on the other side of the inequality. We get, 4X-3X>-6+20 X>14
@surajsahu88519 ай бұрын
Wait isnt iota square 1
@ATMathAcademy9 ай бұрын
I is an imaginary number. For example: To solve x^2+1=0 X^2=-1 x=squareroot of -1 So, for solving such equations the value of squaroot of -1 is named as an imaginary number called "i". So, as i=-1 Hence the value of i^2=(squaroot of -1)×(squaroot of -1) Which is i^2=-1. Thankyou!