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3 years ago

Find the length of LM if L is the midpoint of NM. NL = 3x + 1, LM = 8x - 24.

Mathematics

1 answer:

SashulF [63]3 years ago

8 0

Answer:

LM = 16

Step-by-step explanation:

From the question given:

L is the midpoint of NM

NL = 3x + 1

LM = 8x - 24

LM =?

Next, we shall determine the value of x.

This can be obtained as follow:

Since L is the midpoint of NM, it means that NL and LM are equal i.e

NL = LM

Thus, we can obtain the value of x as follow:

NL = 3x + 1

LM = 8x - 24

NL = LM

3x + 1 = 8x - 24

Collect like terms

3x - 8x = - 24 - 1

-5x = - 25

Divide both side by - 5

x = -25/-5

x = 5

Finally, we shall determine the length of LM as follow:

LM = 8x - 24

x = 5

LM = 8(5) - 24

LM = 40 - 24

LM = 16

Therefore, the length of LM is 16

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Answer:

$\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}$

Step-by-step explanation:

Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .

Given :-

• $\sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}$

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• $\sf\implies cos2A =\dfrac{\sqrt3}{2}$

Proof :-

We know that ,

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Therefore , here substituting the value of sinA , we have ,

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$\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8}$

Add the numbers in numerator ,

$\sf\implies cos2A = 1-2\times \dfrac{4-2\sqrt3}{8}$

Multiply it by 2 ,

$\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4}$

Take out 2 common from the numerator ,

$\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4}$

Simplify ,

$\sf\implies cos2A = 1 -\dfrac{ 2-\sqrt3}{2}$

Subtract the numbers ,

$\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2}$

Simplify,

$\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} }$

Hence Proved !

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