1. Convert 148 pounds to kilograms Round your answer to the nearest tenth. Kg 2. In Z3i)

  1. Convert 148 pounds to kilograms Round your answer to the nearest tenth. Kg
  2. In Z3i), a. Write the additive inverse of 2 + i. b. Write the multiplicative inverse of (21) (1 + i)2 in the form a + bi.

the correct answer and explanation is:

1. Convert 148 pounds to kilograms:

To convert pounds to kilograms, use the conversion factor 1 pound=0.453592 kilograms1 \, \text{pound} = 0.453592 \, \text{kilograms}. 148 pounds×0.453592 kg/pound=67.131024 kg148 \, \text{pounds} \times 0.453592 \, \text{kg/pound} = 67.131024 \, \text{kg}

Rounded to the nearest tenth: 67.1 kg67.1 \, \text{kg}

Answer: 67.1 kg67.1 \, \text{kg}


2. Z3i)

a. Write the additive inverse of 2+i2 + i:

The additive inverse of a complex number a+bia + bi is −a−bi-a – bi. For 2+i2 + i: Additive inverse=−2−i\text{Additive inverse} = -2 – i

Answer: −2−i-2 – i

b. Write the multiplicative inverse of (2+i)2(2 + i)^2 in the form a+bia + bi:

  1. Expand (2+i)2(2 + i)^2:

(2+i)2=(2+i)(2+i)=4+4i+i2(2 + i)^2 = (2 + i)(2 + i) = 4 + 4i + i^2

Since i2=−1i^2 = -1: (2+i)2=4+4i−1=3+4i(2 + i)^2 = 4 + 4i – 1 = 3 + 4i

  1. Find the multiplicative inverse of 3+4i3 + 4i: The multiplicative inverse of a complex number z=a+biz = a + bi is given by:

1z=conjugate of z∣z∣2\frac{1}{z} = \frac{\text{conjugate of } z}{\lvert z \rvert^2}

For z=3+4iz = 3 + 4i, the conjugate is 3−4i3 – 4i, and the magnitude squared is: ∣z∣2=32+42=9+16=25\lvert z \rvert^2 = 3^2 + 4^2 = 9 + 16 = 25

Thus: 13+4i=3−4i25=325−425i\frac{1}{3 + 4i} = \frac{3 – 4i}{25} = \frac{3}{25} – \frac{4}{25}i Multiplicative inverse=0.12−0.16i\text{Multiplicative inverse} = 0.12 – 0.16i

Answer: 0.12−0.16i0.12 – 0.16i


Explanation:

Complex numbers combine a real part and an imaginary part in the form a+bia + bi. To solve the problems in Z3i, understanding key properties like the additive and multiplicative inverses is crucial.

The additive inverse of a complex number is the number that, when added to the original number, gives zero. For example, the additive inverse of 2+i2 + i is −2−i-2 – i because: (2+i)+(−2−i)=0(2 + i) + (-2 – i) = 0

The multiplicative inverse involves finding a number that, when multiplied with the original, gives 11. For z=3+4iz = 3 + 4i, its multiplicative inverse is: 1z=conjugate of z∣z∣2\frac{1}{z} = \frac{\text{conjugate of } z}{\lvert z \rvert^2}

Here, ∣z∣2=25\lvert z \rvert^2 = 25 is derived from the Pythagorean theorem applied to a=3a = 3 and b=4b = 4. The conjugate, 3−4i3 – 4i, ensures the imaginary part cancels when dividing. Simplifying yields: 325−425i\frac{3}{25} – \frac{4}{25}i

Understanding these inverses helps in fields like signal processing or solving differential equations, where complex numbers play a key role.

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