We have been given two functions 
and 
. We are asked to find 
.
We will use composite function property 
to solve our given problem.
Now we will combine like terms as:
Therefore, the value of would be 
.
Y = 1/3x + b
Plug in the point
3 = 1/3(2) + b
3 = 2/3 + b, b = 7/3
Solution: y = 1/3x + 7/3
Which of the following relations represents a function? A. (2,3), (1,3), (3,3) B. (1,3), (2,3), (2,4) C. (1,3), (2,3), (1,4) D.
insens350 [35]
A function will not have ANY repeating x values....it can have repeating y values...just not the x ones
so ur function is : (2,3) ,(1,3), (3,3) ....u see how u have no repeating x values
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: B. 4
Step-by-step explanation: Divide 20 by 5 and get 4.
