Hello from MrBillDoesMath!
Answer:
See Discussion below
Discussion:
(sinq + cosq)^2 = => (a +b)^2 = a^2 + 2ab + b^2
(sinq)^2 + (cosq)^2 + 2 sinq* cosq => as (sinx)^2 + (cosx)^2 = 1
1 + 2 sinq*cosq (*)
Setting a = b = q in the trig identity:
sin(a+b) = sina*cosb + cosa*sinb
sin(2q) = (**)
sinq*cosq + cosq*sinq => as both terms are identical
2 sinq*cosq
Combining (*) and (**)
(sinq + cosq)^2 = 1 + 2sinq*cosq => (**) 2sinq*cosq = sqin(2q)
= 1 + sin(2q)
Hence
(sinq + cosq)^2 = 1 + sin(2q) => subtracting 1 from both sides
(sinq + cosq)^2 - 1 = sin(2q)
The last statement is what we are trying to prove.
Thank you,
MrB
Answer:
12.50 - 3.75 = 8.75
$8.75 / $1.25 * (1/2) (per mile) = 3.5 miles
1. Mile .5 (fee and 1st mile)
$3.75 + $1.25 = $5.00
2. Miles .5 - 3.5 (additional 3 miles)
$1.25 * (3*2) = $7.50
Total: $5.00 + 7.50 = $12.50
Step-by-step explanation:
Answer:
Hello,
P=(30,6)
Step-by-step explanation:
