How do you find (2+4i)^3 in rectangular form?

Mathematics

2 answers:

marshall27 [118]3 years ago

8 0

$(2+4i)^3=8+48i-96-64i=-88-16i$

Paha777 [63]3 years ago

4 0

note that (a + b)³ = a³ + 3a²b + 3ab² + b³

(2 + 4i)³ ← with a = 2 and b = 4i noting that i² = (√-1)² = -1

= 2³ + (3 × 4 × 4i ) + ( 3 × 2 × 16i² ) + ( 4i )³

= 8 + 48i - 96 - 64i

= - 88 - 16i

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alexandr1967 [171]

Answer:

3).

$- 4y = 8x \\ \frac{y}{x} = \frac{8}{ - 4} \\ \frac{y}{x} = - 2 \\ y = - 2x \\ hence \: constant \: is \: - 2$

4).

$y \alpha x \\ y = kx \\ where \: k \: is \: the \: constant \: of \: proportionality \\ when \: y = 24 \: and \: x = 8 \\ substitute \: in \: y = kx \\ 24 = 8k \\ k = \frac{24}{8} \\ k = 3 \\ value \: of \: constant = 3$

If it was by 10, k = 3 by 10

k = 30

5).

$general \: equation \: of \: line \\ y = mx + c \\ but \: m = \frac{2}{3} \: and \: c = 9 \\ substitute \: for \: m \: and \: c \: in \: y = mx + c \\ y = \frac{2}{3} x + 9 \\ alternatively \\ 3y = 2x + 27$