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IgorC [24]
4 years ago
9

If LMC ~ BJK, then _____ ~ JB a- LC b-ML c-CM d-CL

Mathematics
2 answers:
bixtya [17]4 years ago
6 0

Answer:   \overlien{ML}

Step-by-step explanation:

We know that the corresponding parts of two congruent triangles are congruent.

If \triangle{LMC}\cong\triangle{BJK}

Then , \angle{L}\cong\angle{B}

\angle{M}\cong\angle{J}                            (1)

\angle{C}\cong\angle{K}                           (2)

Then from (1) and (2) the side congruent to \overlien{JB} is \overlien{ML}.

11Alexandr11 [23.1K]4 years ago
4 0

The answer would be b- ML.

to do this you should look at the similarity statement and match up the letters in the correct order... if that makes sense?

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Distribution is NOT binomial.

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What is the modulus and argument after (StartRoot 3 EndRoot) (cosine (StartFraction pi Over 18 EndFraction) + I sine (StartFract
fgiga [73]

Answer:

|z| = 27 -- Modulus

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Step-by-step explanation:

Given

((\sqrt 3)(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18}))^6

Required

Determine the modulus and the argument

We have that:

z = ((\sqrt 3)(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18}))^6

Expand:

z = (\sqrt 3)^6(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18})^6

z = 27(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18})^6

A complex equation can be expressed as:

z = |z| e^{i\theta}

Where

|z| = modulus

\theta = argument

Where

e^{i\theta} = (cos\frac{\pi}{18} + i\ sin\frac{\pi}{18})

So:

z = 27(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18})^6 becomes

z = 27(e^{i\frac{\pi}{18}})^6

By comparison:

e^{i\theta} = (e^{i\frac{\pi}{18}})^6

This gives:

{i\theta} = i\frac{\pi}{18}}*6

{i\theta} = i\frac{6\pi}{18}}

{i\theta} = i\frac{\pi}{3}}

Divide through by i

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Hence, the modulus, z is:

|z| = 27

And the argument \theta is

\theta = \frac{\pi}{3}

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