Let's simplify and solve each problem step by step.
1. Simplify \((4-12i) - (-8+4i)\)
To simplify this expression, we'll distribute the negative sign and then combine like terms:
\[ (4 - 12i) - (-8 + 4i) = (4 - 12i) + (8 - 4i) = (4 + 8) + (-12i - 4i) \] \[ = 12 - 16i \]
2. Simplify \((5 + 2t)(1 + 3t)\)
We'll use the distributive property (FOIL method):
\[ (5 + 2t)(1 + 3t) = 5 \cdot 1 + 5 \cdot 3t + 2t \cdot 1 + 2t \cdot 3t \] \[ = 5 + 15t + 2t + 6t^2 \] \[ = 5 + 17t + 6t^2 \]
3. Find the total amps in the circuit with currents \(3 - 2j\) and \(2 + 4y\)
To find the total current, we'll simply add the two currents:
\[ (3 - 2j) + (2 + 4y) = (3 + 2) + (-2j + 4y) \] \[ = 5 + (-2j + 4y) \] So the total current is: \[ 5 + (4y - 2j) \text{ amps} \]
4. Simplify \(\frac{4 - 2i}{7 + 3i}\)
To simplify this expression, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{4 - 2i}{7 + 3i} \cdot \frac{7 - 3i}{7 - 3i} = \frac{(4 - 2i)(7 - 3i)}{(7 + 3i)(7 - 3i)} \]
Calculating the denominator first:
\[ (7 + 3i)(7 - 3i) = 7^2 - (3i)^2 = 49 - 9(-1) = 49 + 9 = 58 \]
Now calculating the numerator:
\[ (4 - 2i)(7 - 3i) = 4 \cdot 7 + 4 \cdot (-3i) - 2i \cdot 7 - 2i \cdot (-3i) \] \[ = 28 - 12i - 14i + 6 = 34 - 26i \]
Putting it all together:
\[ \frac{34 - 26i}{58} = \frac{34}{58} - \frac{26i}{58} = \frac{17}{29} - \frac{13i}{29} \]
5. Solve the equation \(5x^2 + 100 = 0\)
First, isolate \(x^2\):
\[ 5x^2 = -100 \] \[ x^2 = -20 \]
Now taking the square root of both sides:
\[ x = \pm \sqrt{-20} \] \[ x = \pm \sqrt{20}i \] \[ x = \pm 2\sqrt{5}i \]
Summary of Results
- \((4 - 12i) - (-8 + 4i) = 12 - 16i\)
- \((5 + 2t)(1 + 3t) = 5 + 17t + 6t^2\)
- Total current: \(5 + (4y - 2j)\) amps
- \(\frac{4 - 2i}{7 + 3i} = \frac{17}{29} - \frac{13i}{29}\)
- \(x = \pm 2\sqrt{5}i\)