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Answer:
The area of the parallelogram is
.
Step-by-step explanation:
Let's rewrite these two vectors:

Let's recall that the area of the parallelogram is the magnitude of the cross product between these vectors.
We can use the Determinant method to find it.
![u \times v=\left[\begin{array}{ccc}i&j&k\\1&-2&2\\0&3&-1\end{array}\right] = i((-2)*(-1)-2*3)-j(1*(-1)-2*0)+k(1*3-(-2)*0)=i(2-6)-j(-1)+k(3)=-4i+j+3k](https://tex.z-dn.net/?f=u%20%5Ctimes%20v%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C1%26-2%262%5C%5C0%263%26-1%5Cend%7Barray%7D%5Cright%5D%20%3D%20i%28%28-2%29%2A%28-1%29-2%2A3%29-j%281%2A%28-1%29-2%2A0%29%2Bk%281%2A3-%28-2%29%2A0%29%3Di%282-6%29-j%28-1%29%2Bk%283%29%3D-4i%2Bj%2B3k)
Now, the magnitude is the square root of each component squared. It will be:

Therefore the
.
I hope it helps you!
Lets say... 7 divided by 21. ok? So... Just take 7, (usually the smaller number, but not in all cases.) and then multiply it by 1. 2. 3. So lets see. 7x1=7 so thats not it. 7x2 is 14 so thats not it. And you want to keep doing that until you get the other number. 7x3 is 21!!! That is the correct number so if you were doing long division like this
___3______ <--- the answer would be there.
7|21
Step 4: 3>x. (The sign flips because you divided a negative)
I hope this helps love! :)