Question

Does anyone has Algebra ll and if so can you please help me with this problem 5i/-2-6i And also I can't pay you but I hope God can give more good luck

Answer 1

Answer:

$s = \frac{5i}{-2-6i} = -\frac{30}{40}-\frac{10}{40}i$

Step-by-step explanation:

We have the following complex number $s = \frac{5i}{-2-6i}$, we proceed to simplify the expression as follows:

1) $\frac{5i}{-2-6i}$ Given.

2) $(5i)\cdot (-2-6i)^{-1}$ Definition of division.

3) $[(5i)\cdot (-2-6i)^{-1}]\cdot [(-2+6i)\cdot (-2+6i)^{-1}]$ Modulative and associative properties/Existence of the additive inverser

4) $[(5i)\cdot (-2+6i)]\cdot [(-2-6i)^{-1}\cdot (-2+6i)^{-1}]$ Commutative and associative properties.

5) $[(5i)\cdot (-2+6i)]\cdot [(-2-6i)\cdot (-2+6i)]^{-1}$ $a^{c}\cdot b^{c} = (a\cdot b)^{c}$

6) $[(5i)\cdot (-2+6i)]\cdot [4+36]^{-1}$    $(a+b)\cdot (a-b) = a^{2}-b^{2}$/Definition of complex number/  $a\cdot (-b) = -a\cdot b$

7) $[(5i)\cdot (-2)+(5i)\cdot (6i)]\cdot 40^{-1}$ Definition of sum.

8) $(-10\cdot i+30\cdot i^{2})\cdot 40^{-1}$    $a\cdot (-b) = -a\cdot b$/Associative and commutative properties.

9) $(-30-10i)\cdot 40^{-1}$ Commutative properties/Definition of complex number/ $a\cdot (-b) = -a\cdot b$

10) $-30\cdot 40^{-1}-(10i)\cdot 40^{-1}$ Distributive property.

11) $-\frac{30}{40}-\frac{10}{40}i$ Definition of division/Result.

$s = \frac{5i}{-2-6i} = -\frac{30}{40}-\frac{10}{40}i$