Voorbeelden van het gebruik van Multiply both sides in het Engels en hun vertalingen in het Nederlands
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Multiply both sides by this.
So I'm going to multiply both sides by 12 over 5.
Multiply both sides by 2.
What happens if we multiply both sides by 1 over 21?
So multiply both sides of the equation times dx.
If we want to solve this, we can multiply both sides by 4.
Let's multiply both sides times 1.5 minus x.
I just don't like that 12 sitting there, so I'm going to multiply both sides by 12.
So let's multiply both sides by change in time.
Let me just get rid of the 12 first. Let me multiply both sides of this equation by 12.
Let's multiply both sides of this equation by 4.
Or, just the differential du, if we just multiply both sides by dx is equal to 1 over x dx.
Now let's multiply both sides of this equation by the square root of y.
So the first way I'm going to do it, is I'm going to multiply both sides of this equation by the inverse of 5/12.
Multiply both sides times the inverse of this, so times nine fifths on both sides. .
I can multiply both sides by the inverse of this coefficient.
which means multiply both sides by 10.
So we can multiply both sides of this by 2/3.
is the square root of x over the square root of y, and I multiply both sides of that times dx.
What if we were to multiply both sides of this equation by 12 over 5?
2 another way to think about it, if you multiply both sides by change in time you get velocity times change in time.
And then we just multiply both sides times the reciprocal of seven/ nine.
Let's multiply both sides of this equation by dx until you get dy is equal to x over y dx.
I'm dividing it by 3 right now So if I were to multiply both sides of this equation by 3 that would isolate the x.
Now I can multiply both sides of this equation by 1/5,
we can multiply both sides by the reciprocal of this right here.
And let's multiply both sides of this equation by 3/2, so these two will become 1.
So what I'm going to do is I'm going to multiply both sides of this equation by this expression right over here.
We can multiply both sides of this by the diameter and we could say that the circumference is equal to pi times the diameter.
Now it's a level one problem. We just have to multiply both sides times the reciprocal of the coefficient on the left-hand side.