Let set C = {1, 2, 3, 4, 5, 6, 7, 8} and set D = {2, 4, 6, 8}.
g100num [7]
<span>1. If it is intersection then it SHOULD be included in both the sets right?
Now we know that odd numbers from 1-100 but the second set are multiples of 5 from 50-150! So we mainly need to look for common numbers which are ODD and are a MULTIPLE OF 5 BETWEEN 50 - 100!!
So
A={51,53,57,59,61......99}
B={55,60,65,70.......95} [We stop till 100 because set A has no such element]
So what is A ∩ B here?
A ∩ B = {All odd numbers and multiples of 5 between 50 - 100}
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Separate the vectors into their <em>x</em>- and <em>y</em>-components. Let <em>u</em> be the vector on the right and <em>v</em> the vector on the left, so that
<em>u</em> = 4 cos(45°) <em>x</em> + 4 sin(45°) <em>y</em>
<em>v</em> = 2 cos(135°) <em>x</em> + 2 sin(135°) <em>y</em>
where <em>x</em> and <em>y</em> denote the unit vectors in the <em>x</em> and <em>y</em> directions.
Then the sum is
<em>u</em> + <em>v</em> = (4 cos(45°) + 2 cos(135°)) <em>x</em> + (4 sin(45°) + 2 sin(135°)) <em>y</em>
and its magnitude is
||<em>u</em> + <em>v</em>|| = √((4 cos(45°) + 2 cos(135°))² + (4 sin(45°) + 2 sin(135°))²)
… = √(16 cos²(45°) + 16 cos(45°) cos(135°) + 4 cos²(135°) + 16 sin²(45°) + 16 sin(45°) sin(135°) + 4 sin²(135°))
… = √(16 (cos²(45°) + sin²(45°)) + 16 (cos(45°) cos(135°) + sin(45°) sin(135°)) + 4 (cos²(135°) + sin²(135°)))
… = √(16 + 16 cos(135° - 45°) + 4)
… = √(20 + 16 cos(90°))
… = √20 = 2√5
Step-by-step explanation:
p = x / n
p = 550 / 1083
p = 0.5078
Answer:
5.29 units
Step-by-step explanation:
By Pythagoras theorem:
They both have a lab not eh twelfth day because 4,8,12 And then 3,9,12
