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2 years ago

Multiply the complex numbers: (1∕2 + 4i)2

2 answers:

8 0

[I'm supposing that you have to find the value of (1/ 2 + 4i)²]

Answer:

$= \frac{ - 63}{4} + 4i$

Step-by-step explanation:

we must know that

(a + b)² = a² + b² + 2ab

${( \frac{1}{2} + 4i) }^{2} = { (\frac{1}{2} })^{2} + {(4i)}^{2} + 2 . \frac{1}{2} . 4i$

$= \frac{1}{4} + 16 {i}^{2} + 4i$

since i²= -1

$= \frac{1}{4} - 16 + 4i$

$= \frac{1 - 64}{4} + 4i$

$= \underline{ \frac{ - 63}{4} + 4i}$

[I'm also assuming that instead of -153/4 + 4i option A says -63/ 4 + 4i.]

so the answer is option A.

andrezito [222]2 years ago

3 0

Step-by-step explanation:

153∕4 + 8i is the correct answer

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The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

Norma-Jean [14]

The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$

So,

- Derivating to get the velocity function, we have:

$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

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