I believe the answer is F(x)= -4x so the last option.
Yes, because I think two could work
Answer:
In ∆ABC AND ∆DEF
ABC=DEF...........each 90°
SIDE AB =SIDE ED...........given
SIDE BC =SIDE EF............B-F-C and E-C-F
∆ABC =∆DEF....................by SAS test
Answer:
He could make 16 piles.
Step-by-step explanation:
Divide 48 by 3: 48 ÷ 3 = 16
4 cos² x - 3 = 0
4 cos² x = 3
cos² x = 3/4
cos x = ±(√3)/2
Fixing the squared cosine doesn't discriminate among quadrants. There's one in every quadrant
cos x = ± cos(π/6)
Let's do plus first. In general, cos x = cos a has solutions x = ±a + 2πk integer k
cos x = cos(π/6)
x = ±π/6 + 2πk
Minus next.
cos x = -cos(π/6)
cos x = cos(π - π/6)
cos x = cos(5π/6)
x = ±5π/6 + 2πk
We'll write all our solutions as
x = { -5π/6, -π/6, π/6, 5π/6 } + 2πk integer k
