Answer:All coordinate pairs that are in Quadrant I will have a positive x value and a positive y value. All coordinate pairs that are in Quadrant II will have a negative x value and a positive y value. All coordinate pairs that are in Quadrant III will have a negative x value and a negative y value.
Step-by-step explanation:ebob
The answer is (((((A ))))))
Consider the exponential function 
By definition, the domain of a function is the set of input argument values for which the function is real and defined.
Let we take 
then 
then 
then 
If we chose larger values of x, we get larger function values.
For example, If we take 
Thus if we choose smaller and smaller values of x. the f unction values will be smaller and smaller functions.
Thus the domain of the function is the set of all real numbers.
Thus the range is limited to the set of positive real numbers. That is, 
If we choose larger values of x, we will get larger function values, as the function values will be larger powers of 2.
If we choose smaller and smaller x values, the function values will be smaller and smaller fractions.
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
