Answer:
Step-by-step explanation:
8. B is the midpoint of AD
A (x,y)
D( -1,3) B ( (x-1)/2, (y+3)/2) and B ( 1,1)
then (x-1)/2=1, x-1=2 ( +1) , x= 2+1, x=3
(y+3)/2=1, y+3=2 ( -3) y=2-3, y= -1
then A (3, -1)
9. (-8,3)
(s,3) the midpoint (0,u)
then (s-8)/2=0, s=8
(3+3)/2=u, u=6/2, u= 3
10. a=1 ( the y coordinate is the same as the midpoint
A(3,1)
B(x,y) the midpoint (5,1)
(3+x)/2=5, 3+x=10 , x=10-3, x=7
(y+1)/2=1, y+1=2,y=2-1, y=1
B(7,1)
to calculate pi you can use the Nilakantha Series.
we know that
if angle B is equal to 90 degrees
so
we have a right triangle
but
the opposite side to the right angle is the hypotenuse
and
the hypotenuse
is the long side of the triangle
in this problem, the hypotenuse is b=12
and
12 is not greater than 14
therefore
with these dimensions you can not form a triangle
there is no solution
Answer:
y = 2sin(x/2 - pi/4) + 1
Step-by-step explanation:
The normal sine has a range of [-1, 1]. This one has a range of [-1, 3], so the amplitude is 2 times higher (4 vs 2).
After that, 2sin(x) has a range of [-2, 2], but this one has a range of [-1, 3], so we need to shift it upwards by 1.
We now know that the answer is 2sin(____)+1.
A normal sine has it's middle point while rising at x = 0, but this one has it at x = pi/2 - so we need to shift a sine to the right by pi/2.
Lastly, a sine has a cycle of length 2pi, but this one has a cycle of length 4pi. We can achieve that by dividing the argument 2.
We now know that the answer is 2sin(x/2 - pi/4) + 1. We can plot it to verify.