simplify (4-12i)-(-8+4i)

simplify (5+2i)(1+3i)

the current in one part ofg a series circuit is 3-2j amps. the current in another part of the circuit is 2+4y amps. find the total amps in the circuit.

simplify 4-2i over 7+3i

solve 5x^2 + 100 = 0

1 answer

Let's go through each problem step by step.

  1. Simplify \((4 - 12i) - (-8 + 4i)\)

    \[ = (4 - 12i) + (8 - 4i) \quad \text{(distributing the negative sign)} \] \[ = (4 + 8) + (-12i - 4i) \] \[ = 12 - 16i \]

    The simplified expression is \(12 - 16i\).

  2. Simplify \((5 + 2i)(1 + 3i)\)

    Using the distributive property (FOIL): \[ = 5 \cdot 1 + 5 \cdot 3i + 2i \cdot 1 + 2i \cdot 3i \] \[ = 5 + 15i + 2i + 6i^2 \] Since \(i^2 = -1\): \[ = 5 + 17i + 6(-1) \] \[ = 5 + 17i - 6 \] \[ = -1 + 17i \]

    The simplified expression is \(-1 + 17i\).

  3. Find the total current in the circuit given \(3 - 2j\) amps and \(2 + 4y\) amps.

    Add the two currents: \[ (3 - 2j) + (2 + 4y) = (3 + 2) + (-2j + 4y) \] \[ = 5 + (4y - 2j) \]

    The total current in the circuit is \(5 + (4y - 2j)\).

  4. Simplify \(\frac{4 - 2i}{7 + 3i}\)

    To simplify, multiply the numerator and the denominator by the conjugate of the denominator: \[ = \frac{(4 - 2i)(7 - 3i)}{(7 + 3i)(7 - 3i)} \] First, compute the denominator: \[ (7 + 3i)(7 - 3i) = 49 + 21i - 21i - 9i^2 = 49 + 9 = 58 \] Now compute the numerator: \[ (4 - 2i)(7 - 3i) = 28 - 12i - 14i + 6i^2 = 28 - 26i + 6(-1) \] \[ = 28 - 26i - 6 = 22 - 26i \] Thus, we have: \[ \frac{22 - 26i}{58} = \frac{22}{58} - \frac{26}{58}i \] Simplifying further: \[ = \frac{11}{29} - \frac{13}{29}i \]

    The simplified expression is \(\frac{11}{29} - \frac{13}{29}i\).

  5. Solve \(5x^2 + 100 = 0\)

    First, isolate \(x^2\): \[ 5x^2 = -100 \] \[ x^2 = -20 \] Now take the square root of both sides. This gives: \[ x = \pm \sqrt{-20} = \pm \sqrt{20}i = \pm 2\sqrt{5}i \]

    The solutions are \(x = 2\sqrt{5}i\) and \(x = -2\sqrt{5}i\).

In summary:

  1. \(12 - 16i\)
  2. \(-1 + 17i\)
  3. \(5 + (4y - 2j)\)
  4. \(\frac{11}{29} - \frac{13}{29}i\)
  5. \(x = 2\sqrt{5}i\) and \(x = -2\sqrt{5}i\)