<span>
since y = –1.5x² its domain is R=] -infinity, infinity[
the inverse of y
</span>–1/1.5 y = x² , and x = sqrt ((–1/1.5) y) , and f^-1 (x) = sqrt ((–1/1.5) x)
its domain is –1/1.5) x>=0, so x<=0,
from where, the range of y = –1.5x² is the interval ]-infinity, 0]
Answer:
√(2 + √3)/4
Step-by-step explanation:
Sine 5π/12 = Sine (5π/6)/2
Recall
π = 180°
Thus,
Sine (5π/6)/2 = Sine (5×180 /6)/2
= Sine 150/2
Recall
Sine θ/2 = √(1 – Cos θ)/2
Thus,
Sine 150/2 = √(1 – Cos 150)/2
But, Cosine is negative in the 2nd quadrant. Thus,
Cos 150 = – Cos 30 = –√3/2
Thus,
√(1 – Cos 150)/2 = √(1 – –√3/2 )/2
= √(1 + √3/2 )/2
= √[(2 + √3)/2 ÷ 2]
= √[(2 + √3)/2 × 1/2]
= √(2 + √3)/4
Therefore,
Sine 5π/12 = √(2 + √3)/4
Answer:
what table
Step-by-step explanation:
and how can you tell if some one is trying to help if you don't give them a chance ?
Answer: b. 70%
Step-by-step explanation:
Given : Nik logs the following hours meeting with clients (c) or doing other work (o) :
Mon: 6 c, 4 o
Tue: 8 c, 2 o
Wed: 9 c, 1 o
Thu: 7 c, 3 o
Fri: Off
The number of hours Nik spend with clients on Thursday = 7 [Number of corresponding to c is 7 in the table]
Total hours he spend in work on Thursday = 7+3 = 10
The percent of time Nik spent with clients on Thursday :
Hence, the Nik spent 70% of his time with clients on Thursday.
Thus , the correct option is b. 70%.
