Rectangular form:
z = -2.1213203-2.1213203i
Angle notation (phasor):
z = 3 ∠ -135°
Polar form:
z = 3 × (cos (-135°) + i sin (-135°))
Exponential form:
z = 3 × ei (-0.75) = 3 × ei (-3π/4)
Polar coordinates:
r = |z| = 3 ... magnitude (modulus, absolute value)
θ = arg z = -2.3561945 rad = -135° = -0.75π = -3π/4 rad ... angle (argument or phase)
Cartesian coordinates:
Cartesian form of imaginary number: z = -2.1213203-2.1213203i
Real part: x = Re z = -2.121
Imaginary part: y = Im z = -2.12132034
Answer:
2^x
64 bacterium
10 hours
Step-by-step explanation:
doubles every hour
if x = hours
2^x
2^x --> 2⁶ = 64
2^x = 1024
1024 = 2^10 ---> x=10
Let's solve the equation 2k^2 = 9 + 3k
First, subtract each side by (9+3k) to get 0 on the right side of the equation
2k^2 = 9 + 3k
2k^2 - (9+3k) = 9+3k - (9+3k)
2k^2 - 9 - 3k = 9 + 3k - 9 - 3k
2k^2 - 3k - 9 = 0
As you see, we got a quadratic equation of general form ax^2 + bx + c, in which a = 2, b= -3, and c = -9.
Δ = b^2 - 4ac
Δ = (-3)^2 - 4 (2)(-9)
Δ<u /> = 9 + 72
Δ<u /> = 81
Δ<u />>0 so the equation got 2 real solutions:
k = (-b + √Δ)/2a = (-(-3) + √<u />81) / 2*2 = (3+9)/4 = 12/4 = 3
AND
k = (-b -√Δ)/2a = (-(-3) - √<u />81)/2*2 = (3-9)/4 = -6/4 = -3/2
So the solutions to 2k^2 = 9+3k are k=3 and k=-3/2
A rational number is either an integer number, or a decimal number that got a definitive number of digits after the decimal point.
3 is an integer number, so it's rational.
-3/2 = -1.5, and -1.5 got a definitive number of digit after the decimal point, so it's rational.
So 2k^2 = 9 + 3k have two rational solutions (Option B).
Hope this Helps! :)
Answer:
y=1/x-1/5
Step-by-step explanation:
tell me if u have more questions
y=-x+5
is the line so to make it perpendicular
y=1/x-1/5 i gtg
The next larger thousandth is 36.994 .
The next smaller thousandth is 36.992 .
Neither of those is any nearer to 36.993
than 36.993 already is.
The last '3' at the end of 36.993 is in the thousandths' place.
There is no more piece of another thousandth after it.
So 36.993 is already on a complete thousandth, and
there's no rounding required.
