There would be 3 pigs in each group.
2 groups would get 1 extra pig each.
Please mark brainliest if helpful!!
<span>Use the formula: r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ] where k = 0,1,2,3,4
</span><span>First 5th root:
k = 0
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]
2*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]
2*[ cos( (280+0)/5 ) + i*sin( (280+0)/5 ) ]
2*[ cos( 280/5 ) + i*sin( 280/5 ) ]
2*[ cos( 56 ) + i*sin( 56 ) ]
-------------------------------------------------------------------
Second 5th root:
k = 1
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]
2*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]
2*[ cos( (280+360)/5 ) + i*sin( (280+360)/5 ) ]
2*[ cos( 640/5 ) + i*sin( 640/5 ) ]
2*[ cos( 128 ) + i*sin( 128 ) ]
-------------------------------------------------------------------
Third 5th root:
k = 2
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]
2*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]
2*[ cos( (280+720)/5 ) + i*sin( (280+720)/5 ) ]
2*[ cos( 1000/5 ) + i*sin( 1000/5 ) ]
2*[ cos( 200 ) + i*sin( 200 ) ]
-------------------------------------------------------------------
Fourth 5th root:
k = 3
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]
2*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]
2*[ cos( (280+1080)/5 ) + i*sin( (280+1080)/5 ) ]
2*[ cos( 1360/5 ) + i*sin( 1360/5 ) ]
2*[ cos( 272 ) + i*sin( 272 ) ]
-------------------------------------------------------------------
Fifth 5th root:
k = 4
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]
2*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]
2*[ cos( (280+1440)/5 ) + i*sin( (280+1440)/5 ) ]
2*[ cos( 1720/5 ) + i*sin( 1720/5 ) ]
2*[ cos( 344 ) + i*sin( 344 ) ]</span>
Answer:
Step-by-step explanation:
use a math app
Area is 2-dimensional like a carpet or an area rug. A parallelogram is a 4-sided shape formed by two pairs of parallel lines. Opposite sides are equal in length and opposite angles are equal in measure. To find the area of a parallelogram, multiply the base by the height.
