Angle BAC is 75 because you subtract 27 from 102.
Answer:
No they don't.
LINK TO PICTURE OF WORK - My uploads aren't working my apologies.
https://tex.z-dn.net/?f=A%3A%5C%5C(2*2*2)%2B(3*2*2)%2B(3*2*2)%3D%5C%5C(4*2)%2B(6*2)%2B(6*2)%3D%5C%5C8%2B12%2B12%3D%5C%5C20%2B12%3D%5C%5C32%20cm%5E2%5C%5C%5C%5CB%3A%5C%5C(1*3*2)%2B(1*4*2)%2B(3*4*2)%3D%5C%5C(3*2)%2B(4*2)%2B(12*2)%3D%5C%5C6%2B8%2B24%3D%5C%5C14%2B24%3D%5C%5C38%20cm%5E2
Welp. I sure hope you like the Pythagorean theorem...
Top line:
One point is (-2,-2) while the other is (3,-3)
Thus the distance in between is sqrt((3-(-2))^2+(-3-(-2))^2)=sqrt(5^2+(-1)^2)=sqrt(26)
Most right line:
One point is (4,-6) while the other is (3,-3)
Thus the distance in between is sqrt((3-4)^2+(-3-(-6))^2)=sqrt((-1)^2+3^2)=sqrt(10)
Most bottom line:
One point is (1,-6) while the other is (4,-6)
Thus the distance in between is sqrt(4-1)^2+(-6-(-6))^2)=sqrt(3^2+0^2)=sqrt(9)=3
Most bottom left line:
One point is (1,-6) while the other is (-2,-4)
Thus the distance in between is sqrt((1-(-2))^2+(-6-(-4))^2)=sqrt(3^2+(-2)^2)=sqrt(13)
Lastly the most left line:
One point is (-2,-2) while the other is (-2,-4)
Thus the distance in between is sqrt((-2-(-2))^2+(-2-(-4))^2)=sqrt(0^2+(2)^2)=sqrt(4)=2
Thus to find the perimeter, we add up all the sides to get
sqrt(26)+sqrt(10)+3+sqrt(13)+2=16.8668 or B
-(-3)power (n )
n is 1;2;3;4;....
This should help just try to use this
