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evablogger [386]
3 years ago
5

Which inequality represents the range of the relation graphed below?

Mathematics
1 answer:
Zarrin [17]3 years ago
5 0

Answer: I think it’s d

Step-by-step explanation:

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△klm, lm=20 sqrt 3 m∠k=105°, m∠m=30° find: kl and km
Anarel [89]

Answer:

KL =  \frac{20\sqrt{6}}{1+\sqrt{3}} = 17.93

MK =  \frac{40\sqrt{3}}{1+\sqrt{3}} = 25.36


Explanation:

According to the Law of Sines:

\frac{a}{sinA}=\frac{b}{sinB}= \frac{c}{sinC}

where:

A, B, and C are angles

a, b, and c are the sides opposite to the angles


First of all, let's find m∠L: the sum of the angles of a triangle is 180°, therefore

m∠K + m∠L + m∠M = 180°

m∠L = 180° - m∠K - m∠M

m∠L = 180° - 105° - 30°

m∠L = 45°


Now, we can apply the Law of Sines to our case (see picture attached):

\frac{LM}{sinK}=\frac{MK}{sinL}=\frac{KL}{sinM}


Let's solve one side at the time:

\frac{LM}{sinK}=\frac{MK}{sinL}

\frac{20\sqrt{3}}{sin(105)}=\frac{MK}{sin(45)}

MK = \frac{20\sqrt{3} }{sin(105)} \cdot sin(45)

MK = \frac{40\sqrt{3} }{1+\sqrt{3} } = 25.36


Similarily:

\frac{LM}{sinK}=\frac{KL}{sinM}

\frac{20\sqrt{3}}{sin(105)}=\frac{KL}{sin(30)}

KL = \frac{20\sqrt{3} }{sin(105)} \cdot sin(30)

KL = \frac{20\sqrt{6}}{1+\sqrt{3}} = 17.93

8 0
3 years ago
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