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My name is Ann [436]

3 years ago

9

Find the 9th term of the geometric sequence whose common ratio is ⅔ and whose 1st term is 2. BRANLIEST ASAP TY!!!

1 answer:

Wittaler [7]3 years ago

4 0

Answer:

512/6561

Step-by-step explanation:

nth term = a×[r^(n-1)]

9th term: n = 9

2×[⅔^(9-1)]

2(⅔^8)

512/6561

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If JKLM is a rhombus, MK = 30, NL = 13, and mZMKL = 41°, find each measure.

oksian1 [2.3K]

Answer:

$NK = 15$

$JL = 26$

$KL = 19.85$

$\angle JKM =49$

$\angle JML =41$

$\angle MLK = 90$

$\angle MNL =90$

$\angle KJL =41$

Step-by-step explanation:

Given

$MK = 30$

$NL = 13$

$\angle MKL = 41$

Solving (a): NK

MK is a diagonal and NK is half of the diagonal. So:

$NK = \frac{1}{2} * MK$

$NK = \frac{1}{2} * 30$

$NK = 15$

Solving (b): JL

JL is a diagonal, and it is twice of NL.

$JL = 2 * NL$

$JL = 2 * 13$

$JL = 26$

Solving (c): KL

To solve for KL, we consider triangle KNL where:

$\angle KNL = 90$

and

$KL^2 = NL^2 + NK^2$

$KL^2 = 13^2 + 15^2$

$KL^2 = 394$

$KL = \sqrt{394$

$KL = 19.85$

Solving (d - h):

To do this, we consider triangle JKN

$\angle KNL = \angle LNM = \angle MNJ = \angle JNK = 90$ -- diagonals bisect one another at right angle

Alternate interior angles are equal. So:

$\angle MKL = \angle KMJ = \angle KJL = \angle JLM = 41$

Similarly:

$\angle MKJ = \angle KML = \angle MJL = \angle JLK = 90 - 41$

$\angle MKJ = \angle KML = \angle MJL = \angle JLK = 49$

So:

$\angle JKM =49$

$\angle JML =41$

$\angle MLK = \angle MLJ + \angle JLK$

$\angle MLK = 49 + 41$

$\angle MLK = 90$

$\angle MNL =90$

$\angle KJL =41$

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