Question

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

Answer 1

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 \(x = -2\sqrt{5}i\)