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3 years ago

8

19. Given that the lines are parallel, what angle is congruent to angle 6?

1 answer:

nikklg [1K]3 years ago

6 0

It is C bc it is on the opposite side and on the outside

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Pi/6 is the reference angle for:<br><br> 3pi/6<br><br> 5pi/6<br><br> 13pi/6<br><br> 8pi/6

Elanso [62]

Answer:

$\frac{5\pi}{6}, \frac{13\pi}{6}$

Step by step

Remember, a reference angle is a positive acute angle that represents an angle θ of any measurement.

a)

$\frac{3\pi}{6}=90^\circ$ , then its reference angle is $\frac{\pi}{2}$

b)

$\frac{5\pi}{6}=150^ \circ$ is in the second cuadrant, then its reference angle is $180^ \circ -150^{\circ}=30^{\circ}=\frac{\pi}{6}$

c)

$\frac{13\pi}{6}=390^\circ=360^\circ+30^\circ$ , then it is in the first cuadrant and the reference angle is $30^\circ =\frac{\pi}{6}$ .

d)

$\frac{8\pi}{6}=240^\circ$ and it is in the third cuadrant, then its reference angle is $240^\circ -180^\circ=60^\circ=\frac{\pi}{3}$

5 0

3 years ago

Hi whats the answer<br> √548

harkovskaia [24]

23.409399821439250218

6 0

2 years ago

5.<br> {wawawawawawawawawawawa}

Harrizon [31]

Answer:

35mc

Step-by-step explanation:

7 0

2 years ago

WILL GIVE BRAINLIEST, THANKS, AND 5 STARS!<br> PLEASE HELP ME!

Bogdan [553]

Answer:

-4 and 4

Step-by-step explanation:

<u>Method 1</u>

Apply Difference of Two Squares Formula: $a^2-b^2=(a+b)(a-b)$

Given $x^2-16=0$

Rewrite 16 as 4²

Therefore, $a^2=x^2$ and $b^2=4^2$

$\implies a = \sqrt{x^2}=x$

$\implies b=\sqrt{4^2} =4$

$\implies x^2-16^2=(x+4)(x-4)$

$\implies (x+4)(x-4)=0$

$\implies (x+4)=0 \implies x=-4$

$\implies (x-4)=0 \implies x=4$

<u>Method 2</u>

Given equation:

$x^2-16=0$

Add 16 to both sides:

$\implies x^2-16+16=0 + 16$

$\implies x^2=16$

Square root both sides:

$\implies \sqrt{x^2}=\sqrt{16}$

$\implies x=\pm4$

Therefore, x = -4, x = 4

6 0

2 years ago

Have sum points ;) ;) ;) ;) ;) ;) ;)

OleMash [197]

Ahh thank you so much !! <3

5 0

2 years ago

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