Answer:
An Arc
Step-by-step explanation:
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
Answer:
i think the answer would be 53 CM
Step-by-step explanation:
Step-by-step explanation:
here's the answer to your question
Answer:
Step-by-step explanation:
The wording on this is not the best. It sounds like the 1 zero has even multiplicity (that's because of where the modifier is). On top of that it has an odd power. You could try this. y =x*(x^2+1)^2
The problem is not with the power. It gives x^5. The problem is with the multiplicity of the one place where it crosses. (X^2 + 1) does factor, but it gives a complex root. I'm not sure that's allowed. However, it is the best I can do.
