How to find excluded values ?
1 answer:
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Answer:
1. -10
2.
3. -1+2i
4. -3-7i
5. 13
6. rectangular coordinates are (-4.3,-2.5)
7. rectangular coordinates are (-2.5,4.3)
8. x^2 + y^2 = 8y
9. Polar coordinates of point (-3,0) are (3,180°)
10. Polar coordinates of point (1,1) are (√2,45°)
Step-by-step explanation:
1) Simplify (2+3i)^2 + (2-3i)^2
Using formula (a+b)^2 = a^2+2ab+b^2
=((2)^2+2(2)(3i)+(3i)^2)+((2)^2-2(2)(3i)+(3i)^2)
=(4+12i+9i^2)+(4-12i+9i^2)
We know that i^2=-1
=(4+12i+9(-1))+(4-12i+9(-1))
=(4+12i-9)+(4-12i-9)
=(-5+12i)+(-5-12i)
=5+12i-5-12i
=-10
2. Solve x^2-2x+3 = 0
Using quadratic formula to find value of x
a=1, b=-2 and c=3
3. If u =1+3i and v =-2-i what is u+v
u+v = (1+3i)+(-2-i)
u+v = 1+3i-2-i
u+v = 1-2+3i-i
u+v = -1+2i
4. if u = 3-4i and v = 3i+6 what is u-v
u-v = (3-4i)-(3i+6)
u-v = 3-4i-3i-6
u-v = 3-6-4i-3i
u-v = -3-7i
5. if u=(3+2i) and v=(3-2i) what is uv?
uv = (3+2i)(3-2i)
uv = 3(3-2i)+2i(3-2i)
uv = 9-6i+6i-4i^2
uv = 9-4i^2
i^2=-1
uv = 9-4(-1)
uv = 9+4
uv = 13
6. Convert (5, 7π/6) to rectangular form
To convert polar coordinate into rectangular coordinate we use formula:
x = r cos Ф
y = r sin Ф
r = 5, Ф= 7π/6
x = r cos Ф
x = 5 cos (7π/6)
x = -4.3
y = r sin Ф
y = 5 sin (7π/6)
y = -2.5
So rectangular coordinates are (-4.3,-2.5)
7. Convert (5, 2π/3) to rectangular form
To convert polar coordinate into rectangular coordinate we use formula:
x = r cos Ф
y = r sin Ф
r = 5, Ф= 2π/3
x = r cos Ф
x = 5 cos (2π/3)
x = -2.5
y = r sin Ф
y = 5 sin (2π/3)
y = 4.33
So rectangular coordinates are (-2.5,4.33)
8. Convert r=8cosФ to rectangular form
r.r = (8 cos Ф)r
r^2 = 8 (cosФ)(r)
Let (cosФ)(r) = y and we know that r^2 = x^2+y^2
x^2 + y^2 = 8y
9. Convert(-3,0) to polar form
We need to find (r,Ф)
r = √x^2+y^2
r = √(-3)^2+(0)^2
r =√9
r = 3
and tan Ф = y/x
tan Ф = 0/-3
tan Ф = 0
Ф = tan^-1(0)
Ф = 0°
As Coordinates are in 2nd quadrant, so add 180° in the given angle
0+180 = 180°
So,Polar coordinates of point (-3,0) are (3,180°)
10) Convert (1,1) to polar form
We need to find (r,Ф)
r = √x^2+y^2
r = √(1)^2+(1)^2
r =√2
and tan Ф = y/x
tan Ф = 1/1
tan Ф = 1
Ф = tan^-1(1)
Ф = 45°
As Coordinates are in 1st quadrant, so Ф will be as found
So,Polar coordinates of point (1,1) are (√2,45°)