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3 years ago

Q#3. if cos x = sqrt(2)/2, and 3pi/2 < x < 2pi, what is sin(x + pi/3)

1 answer:

5 0

#3 sin(x) = -sqrt(2)/2 so x = 315 degrees. So we have sin(x+pi/3) = sin(375) = sin(15) which in exact form would be (sqrt(3) -1)/(2sqrt(2))

#4 to find x we take the arctan(20/21) which is approximately 43.6 degrees and add 180 to it since it's in the 3rd quadrant which is 223.6 degrees so we do sin(223.6 + 90) = sin(313.6) which equals -21/29. This can also be done with the pythagorean theorem. -21/29.

#5 13^2 - 5^2 = 12^2. so sin is -12/-13 = 12/13. we take arcsin of that and add 180 and then divide by 2 and take the sin of that and its approximately 3/sqrt(13).

#6 sin is 4/5 so we take arcsin of that and get 143.13 which -60 is 83.13 and we take the sin of that to get 3/10 + 2sqrt(3)/5

Hope this helped :L

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Alenkasestr [34]

Answer:

$(x,y)=(0,\frac{-2}{3})$

Step-by-step explanation:

Given : Vertices of Triangle $P(x_1,y_1)=(-4,-1)$ , $Q(x_2,y_2)=(2, 2)$ and $R(x_3,y_3)=(2,-3)$

To find : Centroid of a triangle

The centroid of a triangle is the point of intersection of the three medians of the triangle.

The formula to calculate the centroid of a triangle is:

$(x,y)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

Where, $x_1,x_2,x_3,y_1,y_2,y_3$ are the coordinates of the centroid

Substituting the values in the formula gives us:

$(x,y)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

$(x,y)=(\frac{-4+2+2}{3},\frac{-1+2-3}{3})$

$(x,y)=(\frac{0}{3},\frac{-2}{3})$

Therefore, $(x,y)=(0,\frac{-2}{3})$

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3 years ago

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Rewrite this standard form quadratic equation into vertex form by completing the square. Upload your work here to show your thin

Alex787 [66]

Solution :

Given the equation :

$y = x^2 - 6x+3$

$y=1(x^2 - 6x +n)+3$

$n = \left(\frac{b}{2}\right)^2$

$n = \left(\frac{6}{2}\right)^2$

n = 9

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