What is the answer to r(t)=(2t^2-5t)i+(4t-1) ?

The answer to r(t)=(2t^2-5t)i+(4t-1) is

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r(t)=(2t^2-5t)i+(4t-1)

r (t) = (2 t^{2} - 5 t) i + (4 t - 1)

(r = \frac{18}{5}, t = \frac{5}{2})

Solution steps

r (t) = (2 t^{2} - 5 t) i + (4 t - 1)

Rewrite

r (t)

in standard complex form:

r (t) + 0 i

r (t) + 0 i = (2 t^{2} - 5 t) i + (4 t - 1)

Rewrite

(2 t^{2} - 5 t) i + (4 t - 1)

in standard complex form:

(4 t - 1) + (2 t^{2} - 5 t) i

r (t) + 0 i = (4 t - 1) + (2 t^{2} - 5 t) i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[r (t) = 4 t - 1 0 = 2 t^{2} - 5 t]

[r (t) = 4 t - 1 0 = 2 t^{2} - 5 t] : (r = \frac{18}{5}, t = \frac{5}{2})

(r = \frac{18}{5}, t = \frac{5}{2})

z^{2} = \frac{- 2}{3 i}

z^{2} + 3 i z + 4 = 0

z^{2} - z^{1} = (4 + 3 i) - (3 - 2 i)

5 x + 7 y i = - x - 7 i

z^{2} - (1 - 2 i) z + (2 + 14 i) = 0

What is the answer to r(t)=(2t^2-5t)i+(4t-1) ?
The answer to r(t)=(2t^2-5t)i+(4t-1) is