How do you simplify i²⁹? (If you can't tell, that's the imaginary number i^29).

1 answer:

6 0

The imaginary number ' i ' is the square root of -1.

i = √-1
i² = -1
i³ = -√-1
i to the 4th power = +1
i to the 5th power = √-1
etc.

and all the higher powers keep going in the same 4-step cycle.

What's 29 divided by 4 ?
It's 7 with a remainder of 1.

So 29 with all the 4s thrown away is 1, and ' i ' to the 29th power is
the same thing as ' i ' to the 1st power = √-1 or just plain ' i '.

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In ΔJKL, k = 9.6 cm, l = 2.7 cm and ∠J=43°. Find ∠L, to the nearest 10th of a degree.

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Answer:

L = 10.64°

Step-by-step explanation:

From the given information:

In triangle JKL;

line k = 9.6 cm

line l = 2.7 cm; &

angle J = 43°

we are to find angle L = ???

We can use the sine rule to determine angle L:

i.e

$\dfrac{j}{SIn \ J} = \dfrac{l}{ SIn \ L}$

Using Pythagoras rule to find j

i,e

j² = k² + l²

j² = 9.6²+ 2.7²

j² = 92.16 + 7.29

j² = 99.45

$j = \sqrt{99.45}$

j = 9.97

∴

$\dfrac{9.97}{Sin \ 43} = \dfrac{2.7}{ Sin \ L}$

${9.97 \times Sin (L ) = (2.7 \times Sin \ 43)$

$= Sin \ L = \dfrac{ (2.7 \times Sin \ 43)}{9.97 } \\ \\ = Sin \ L = \dfrac{ (2.7 \times 0.6819)}{9.97 } \\ \\ = Sin \ L = 0.18466 \\ \\ L = Sin^{-1} (0.18466) \\ \\ L = 10.64 ^0$