Answer:
Step-by-step explanation:
A complex number is defined as z = a + bi. Since the complex number also represents right triangle whenever forms a vector at (a,b). Hence, a = rcosθ and b = rsinθ where r is radius (sometimes is written as <em>|z|).</em>
Substitute a = rcosθ and b = rsinθ in which the equation be z = rcosθ + irsinθ.
Factor r-term and we finally have z = r(cosθ + isinθ). How fortunately, the polar coordinate is defined as (r, θ) coordinate and therefore we can say that r = 4 and θ = -π/4. Substitute the values in the equation.
![\displaystyle \large{z=4[\cos (-\frac{\pi}{4}) + i\sin (-\frac{\pi}{4})]}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7Bz%3D4%5B%5Ccos%20%28-%5Cfrac%7B%5Cpi%7D%7B4%7D%29%20%2B%20i%5Csin%20%28-%5Cfrac%7B%5Cpi%7D%7B4%7D%29%5D%7D)
Evaluate the values. Keep in mind that both cos(-π/4) is cos(-45°) which is √2/2 and sin(-π/4) is sin(-45°) which is -√2/2 as accorded to unit circle.
Hence, the complex number that has polar coordinate of (4,-45°) is
The width of the rectangle is twice the length of the rectangle
<h3>How to compare the dimensions of the rectangle?</h3>
The given parameters are:
Width = 4 cm
Length = 2 cm
Express 4 cm as 2 * 2 cm
Width = 2 * 2 cm
Substitute Length = 2 cm in Width = 2 * 2 cm
Width = 2 * Length
This means that the width of the rectangle is twice the length of the rectangle
Hence, the true statement is that the width of the rectangle is twice the length of the rectangle
Read more about rectangles at:
brainly.com/question/25292087
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Answer:
x^2 - 8xy + 3y^2 - 2
Step-by-step explanation:
(-8xy + 2x^2 + 3y^2) - unknown = x^2 + 2
- unknown = x^2 + 2 + 8xy - 2x^2 - 3y^2
- unknown = -x^2 + 8xy - 3y^2 + 2
Unknown = x^2 - 8xy + 3y^2 - 2
Check:
(-8xy + 2x^2 + 3y^2) - (x^2 - 8xy + 3y^2 - 2)
= -8xy + 2x^2 + 3y^2 - x^2 + 8xy - 3y^2 + 2
= -8xy + 8xy + 2x^2 - x^2 + 3y^2 - 3y^2 + 2
= x^2 + 2
Answer:
is outside the circle of radius of 
centered at 
.
Step-by-step explanation:
Let 
and 
denote the center and the radius of this circle, respectively. Let 
be a point in the plane.
Let 
denote the Euclidean distance between point and point .
In other words, if is at 
while is at 
, then 
.
Point would be inside this circle if 
. (In other words, the distance between 
and the center of this circle is smaller than the radius of this circle.)
Point would be on this circle if 
. (In other words, the distance between and the center of this circle is exactly equal to the radius of this circle.)
Point would be outside this circle if 
. (In other words, the distance between and the center of this circle exceeds the radius of this circle.)
Calculate the actual distance between and :
.
On the other hand, notice that the radius of this circle, 
, is smaller than . Therefore, point would be outside this circle.
