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Method 1) (- x= 3) equation is given Multiplying both sides by (-1) -1*-x=-1*3 Then x=-3 or Method 2) Let the equation be (- x= 3) If we multiply both sides with "MINUS" sign -(- x)= -(3) Then x= -3. Which one is correct or both methods are correct .

Thank you 🎉

Can you reduce 4/8?

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110

20° if the maximum is 90° and 110° if the maximum is 180

110?? Idk

So true bro

Bro? Bro? Are you serious? If this shit will shoot I will copy your content

Bruh

Wtf

Since you know that multiplying 3 with four gives you 12, then also multiply 2 by four and get 8. So why the extra steps.

or you could just subtract the nearest round number and add/remove .50 dependinh if the number is higher or lower. For example 10-6.5 is roughly 6-10 which is 4 and the cus its lower you subtract the .5 and get 3.5

So, Normal math skills, got it 👍

Fourth grade math

Aussie formula

36 + 54 = 90 90 + 18 = 108…

40 and 50... 90... 90 and 20... 110.. minus 2, 108 Much easier

Isn't it like the normal method or am I twitching?

What.. what is the alternative to this?

What if the area was 66 cm2 ? :) What mental process would you try?

So you "solve" a problem by taking 3 guesses! Wow, great solving skills!

There's a much simpler solution to this question. 1. A square is a 4 sided shape with equal side lengths which are perpendicular to each other let's put the square in the cartesian plane with its sides parallel to the x and y axis, let's call the side lengths S. 2. We can determine the area of any square by "infinitely summing" the inner "corner shells" of the square to do that we get the semi-perimeter of a square with side lengths x, which is 2x, because each side is of length x and there are 4 sides/2 which is x+x which is 2x. 3. Now we can integrate. Our bounds of integration are from 0 to S, and integration works because we can multiply the perimeters by their width, dx. Now, this may not be accurate if we have a real width, but as it is infinitesimal (dx) we can move along to the integration. ∫2x dx is x^2, which from 0 to S is S^2-0^2 which is S^2. 4. Now we set S^2 equal to 64 cm^2. We move the 64cm^2 to the other side to get S^2-64cm^2=0 which we recognize as the auxiliary equation of the differential equation S''-64cm^2*S=0 5. To solve this differential equation, we can find the Laplace transforms of both sides (I'll call the Laplace transform of S, F(s)), which nets us s^2*F(s)-s*S(0)-S'(0)-64cm^2*F(s)=0. Let's call S(0) and S'(0) A and B respectively and get rid fo the cm^2 because it's a constant for now for simplicity. We move everything to the other side except for s^2*F(s)-64F(s). This results in s^2*F(s)-64F(s)=As+B and F(s)=As/(s^2-64)+B/(s^2-64). Now we find the inverse Laplace transform of both sides. As the inverse Laplace transform is a linear, constant coefficient function, we can split the terms and we get Asin(8ix)-Bi/8*cos(8ix) 6.We recognize the imaginary number's places to be hyperbolic sin and cos and get iAsinh(8x)+B/8*cosh(8x) which we can get in terms of the e^x definitions of hyperbolic trig to get (8A-B)/16*e^(-8x)+(8A+B)/16*e^(8x). (8A+B)/16 and (8A-B)/16 can be represented by separate constants, and we take the coefficients of x in e^(-8x) and e^(8x), -8 and 8. To finally get our answer we flip the signs on -8 and 8 to get the solutions 8 and -8. We slap on the cm from before and eliminate the extraneous solution of -8cm (since sides can't be negative) to get 8cm.

I loved this video, it teaches something simple for people to understand geometry.

Easy *draws a square with 8 cms*

Awesome!

Very helpful!

Ummmmmm. This is a joke, right? a square with an area of 64cm SQUARED will have sides of 64cm. A square with 8cm sides has an area of 8cm^2 (squared), or an area of 64cm (not squared).

idk if this is for elementary schoolers, but honestly probably just easier to tell them how to do square root, since guess and check will take way too long

When are you going to use logarithms to solve this.

stop the yap this took me a second to solve and I'm in middle school you just memorize squares 1 > 1 2 > 4 3 > 9 ... 7 > 49 *8 > 64*

My son is loving this! … he just passed his latest math test for the first time! … we can’t thank you enough!🎉