Examples of using This expression right here in English and their translations into Thai
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Colloquial
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Ecclesiastic
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Ecclesiastic
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Computer
I got this expression right here.
So it's times the cosine of this expression right here.
And this expression right here is our Q of xy.
Times ds, which is this expression right here.
This expression right here is a transformation of a.
When you just look at this expression right here, you're.
So this expression right here is going to be non-negative.
So we solved da dx, that is equal to this expression right here.
So this expression right here is x minus 5, times x minus 5.
But here we want to take the negative root because this expression right here is going to be negative.
So this expression right here, this right here is positive.
So you still could solve for y, but this expression right here, or this equation, defines y implicitly.
So this expression right here, it's actually going to be a surface in three dimensions.
Now to solve for dy dx, we just have to divide both sides of this equation by this expression right here.
And then this expression right here-- so let me just write the left again.
So in order for this left-hand side of the equation to be 0, this term, this expression right here, has to be 0.
And then this expression right here, I can actually separate out the exponents.
But they usually involve starting from this point right here and algebraically manipulating it until you get-- until this expression right here looks something.
This expression right here, this term right here, is going to be equal to 0.
And that's because if x was negative 3, this expression right here, you would be dividing by 0, it would be undefined.
So this expression right here is the same thing as negative 2 times-- what's negative 12f squared divided by negative 2?
And just like we saw in the last video, this, let me do it in orange, this expression right here, actually let me draw that dy a little further out so it doesn't get all congested.
And then this expression right here simplifies to minus minus, it's 6x minus x squared, and then you have a 4 minus 9 minus 5.
And then from here, you know that this expression right here is going to be equal to x plus b over 2a squared.
So this expression right here simplifies to a squared cosine squared of s plus this over here: a squared sine squared of s.
But this right here, what I just did, this expression right here, this is the derivative with respect to x of this. .
So this expression right here, which was just a dot b can be rewritten as-- I just told you that the length of vector a times cosine of theta is equal to this little magenta adjacent side.
And the only way that I can ensure that this expression right here is equal to the absolute value of x dot y, the only way I can assure this is that c is positive.
For this expression right here, you have to add the constraint that a cannot equal negative 2, negative 1, or 1, that a can be any real number except for these.
If x was 1, this expression right here would be dividing by 0 and it would be undefined.