Answer:
the answer is 2 + i.
Step-by-step explanation:
Let the square root of 3 + 4i be x + iy.
So (x + iy) (x +iy) = 3 +4i
=> x^2 + xyi + xyi + i^2*y^2 = 3 + 4i
=> x^2 – y^2 + 2xyi = 3 + 4i
Equate the real and complex terms
=> x^2 - y^2 = 3 and 2xy = 4
2xy = 4
=> xy = 2
=> x = 2/y
Substitute in
=> x^2 - y^2 = 3
=> 4/y^2 - y^2 = 3
=> 4 – y^4 = 3y^2
=> y^4 + 3y^2 – 4 = 0
=> y^4 + 4y^2 – y^2 – 4 =0
=> y^2(y^2 + 4) – 1(y^2 + 4) =0
=> (y^2 – 1) (y^2 + 4) =0
Therefore y^2 = 1, we ignore y^2 = -4 as it gives complex values of y.
Therefore y = 1 and x = 2/1 = 2
The required square root of 3 + 4i is 2 + i.
Answer:
Step-by-step explanation:
The standard form for this function is
y = Asin(Bx - C) + D where
A is the amplitude,
B is used to find the period in the formula 
,
C is used to find the phase shift in the formula 
,
and D is the midline which either moves the function up or down from its natural midline of 0.
We have no given phase shift, so there is no C value. The amplitude is easy; we just fill in A as 4. The period is another story:
and we solve for B by cross multiplying:
3πB = 4π and
so
so the function is
Answer:
You need to use steps to find the answer.
Step-by-step explanation:
So you would solve -3/8 by dividing -3 by 8. Then you would do the same for -1/4, divide -1 by 4. When your done you add both answers together and bam!
It's false. It's a product so...
Derivative of the first TIMES the second PLUS derivative of second TIMES the first.
Derivative of the first (x^3) = 3x^2
Times the second = 3x^2 * e^x
Derivative of the second = e^x (remains unchanged)
Times the first = e^x * x^3
So the answer would be (3x^2)(e^x) + (e^x)(x^3)
which can be factorised to form x^2·e^x(3 + x)
