The answer is 45.10 you just need it to add a zero to it it's simple :) anything else I can help?
The answer:
the full question is as follow:
if A+B-C=3pi, then find sinA+sinB-sinC
first, the main formula of sine and cosine are:
sinC = 2sin(C/2)cos(C/2)
sinA +sinB = 2sin[(A+B)/2]cos[(A-B)/2]
therefore:
sinA+sinB-sinC = 2sin[(A+B)/2]cos[(A-B)/2] - 2sin(C/2)cos(C/2)
sin[(A+B)/2] = cos(C/2)
2sin(C/2)cos(C/2) = cos[(A+B)/2
and
A+B-C=3 pi implies A+B =3 pi + C, so
cos[(A+B)/2] = cos [3 pi/2 + C/2]
and with the equivalence cos (3Pi/2 + X) = sinX
sinA+sinB-sinC = cos(C/2)+ sin(C/2)
3/18=1/6
9/27=1/3
no one of these fractions is equivalent to 3/18
Answer:
A. C (3,5)
Step-by-step explanation:
This makes the most sense because of the position point C is at on the coordinate plane. It is above point D, which has a x-axis of 3. Option A is the only one that has an x-axis of 3.
Hope it helps!
8x^2 + 9 - 6x^2 - 2x - 2
= 2x^2 - 2x + 7
answer is B
2x^2 - 2x + 7
