The answer is A because it is 22 below zero.
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)
Answer:
63
Step-by-step explanation:
Set up the proportion
(6x+3)/17 = (8x - 1)/21 Cross Multiply
21 * (6x + 3) = 17 (8x - 1) Remove the brackets
126x + 63 = 136x - 17 Subtract 126x from both sides
63 = 136x - 126x - 17 Combine
63 = 10x - 17 Add 17 to both sides
63 + 17 = 10x Combine the left
80 = 10 x Divide by 10
80/10 = x
x = 8
Now you want 8x - 1
8*8 - 1 = 63
