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Alja [10]
2 years ago
5

Given: LP=NP ML=MN Prove: LQ=QN

Mathematics
1 answer:
julsineya [31]2 years ago
4 0

Answer:

The Proof is below.

Step-by-step explanation:

Given:

\overline {LP} \cong \overline{NP}

\overline {ML} \cong \overline{MN}

To Prove:

\overline {LQ} \cong \overline{QN}

Proof:

In  ΔLPM  and ΔNPM  

\overline {LP} \cong \overline{NP}  ……….{Given}

\overline {ML} \cong \overline{MN}  ……….{Given}

\overline {LP} \cong \overline{NP}  ……….{Reflexive Property}

ΔLPM ≅ ΔNPM      ….{ By Side-Side-Side congruence test}

∴ ∠LMP ≅ ∠NMP  ...{Corresponding parts of congruent triangles (c.p.c.t).}.....( 1 )

Now In ΔLMQ  and ΔNMQ  

\overline {ML} \cong \overline{MN}  ……….{Given}

∠LMQ ≅ ∠NMQ                                 ..........{From 1 above}

\overline {MQ} \cong \overline{MQ}  ……….{Reflexive Property}

ΔLMQ ≅ ΔNMQ         ....{ By Side-Angle-Side Congruence test}

∴ \overline {LQ} \cong \overline{QN} ...{Corresponding parts of congruent triangles (c.p.c.t).}.....Proved

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musickatia [10]

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D. 4

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Ulleksa [173]

Answer:

I) X+Y

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(Iv)

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5 0
1 year ago
PLEASE HELP WILL GIVE BRAINLIEST
Svetach [21]

Answer:

Check explanation

Step-by-step explanation:

I'll give you my way of solving it and you can figure it out from there.

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5 0
3 years ago
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Step-by-step explanation:

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3 years ago
Read 2 more answers
Determine which regions contain cube roots of −1. Check all that apply.
pashok25 [27]

cube roots of -1 lies in the region given in the best options are  A. on real axis, B. quadrant 1, F. quadrant 4.

<h3>What is cube roots of -1?</h3>

Cube roots of -1 mean "finding solution of the equation z³ = -1 since the equation is of order 3, the equation has 3 roots in complex field."

According to the question,

Cube roots of -1 can written as (-1)^\frac{1}{3}

First, write the given number in polar form

x = ( cos 0 - i sin 0)^\frac{1}{3}

Add 2kπ to the argument

x = (cos2k\pi - i sin 2k\pi )^\frac{1}{3}

Apply De Moivre's  theorem

(cos\alpha  - i sin\alpha) = cosn\alpha + i sin\alpha

x = \frac{2k\pi }{3} - i sin\frac{2k\pi }{3}

Put k = 0,1,2..., (n-1)                   [sin (-θ) = -sin θ, cos(-θ) = cos θ]

The three roots are

cos 0 -  sin 0, cos \frac{2\pi }{3}, i sin\frac{2\pi }{3} , cos \frac{4\pi }{3},  isin\frac{4\pi }{3}

-1, -\frac{1}{2} ,- i \frac{\sqrt{3} }{2}, -\frac{1}{2} , +\frac{\sqrt{3} }{2} ,  are the three roots of cube roots of -1

It can be seen that the three roots lies in the regions of real axis, quadrant 1 and quadrant 4.

Hence, cube roots of -1 lies in the region given in the best options are  A. on real axis, B. quadrant 1, F. quadrant 4.

Learn more about cube roots here:

brainly.com/question/310302

#SPJ1

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