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Please help asap, ty!

 Apr 24, 2018
 #1
avatar+98044 
+2

71.  Arc length  =  2 * pi (8) (120/360)  = (16/3)pi mm ≈ 16.76 mm

 

72. 120 * pi  / 180  =   (120/ 180) *pi  =  (2/3) pi  rads

 

73.  Sector area  = (1/2) *r^2* (theta in rads)  =  (1/2) * (8)^2 * (2/3 pi)  =

 

(64/ 3) pi  mm^2  ≈  67.02 mm^2 

 

74.   (x - 3)^2  + (y + 4)^2  = 4 

 

75. x^2 - 6x + y^2 + 4y - 3  = 0

 

x^2 - 6x + y^2 + 4y  =  3            complete the square  on x and y

 

x^2 - 6x + 9  + y^2 + 4y + 4  =  3 + 9 + 4

 

(x - 3)^2 + (y + 2)^2  = 16

 

center ( 3,  -2)    radius  = 4 

 

 

cool cool cool

 Apr 24, 2018
 #2
avatar+164 
+2

The arc length of QR is 120/360 of 16pi, which is the circumference. That is 16pi/3

 

The central angle in radians is 2/3pi. 180 degrees in radiansis pi and 360 is 2pi.

 

The area of the sector is 64pi/3. Like 71, we just set up a proportion again. it is just a third of the total area

 

The equation is (x-3)^2+(y+4)^2=4

Because the basic form is (x-h)^2+(y-k)^2=r^2 and (h,k) is the center and r is the radius.

 

We can do the last one by completing the square.

We take the square of half of the coefficients.

(x-3)^2+(y+2)^2=-10

 

Sorry if these were not detailed, I didn't have a calculator and was in a hurry.

 Apr 24, 2018

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