B maybe or A
Because 15 percent of 212 is 31.8 so it’s closer to 40 maybe
The first is C. r = 4(4/9)
the second one is B. h = 130
the second one is A. k = 7
i’m so sorry, they’re all wrong
Answer: 4
Step-by-step explanation:
We know that a plane is 2 dimensional surface that extends infinitely far.
The number of points required to define a plane = 3
Here , we have 4 points A, B, C, and D.
So, the number of possible combinations of 3 points to make a plane from 4 points = 
[ 
]
Hence, the greatest number of planes determined using any three of the points A, B, C, and D if no three points are collinear = 4.
The answer is -1975. Hope I was able to help
Answer:
Step-by-step explanation:
Given the expression cosec (x) = 4 and tan(x)< 0
since cosec x = 1/sinx
1/sinx = 4
sinx = 1/4
From SOH, CAH TOA
sinθ = opposite/hypotenuse
from sinx = 1/4
opposite = 1 and hypotenuse = 4
to get the adjacent, we will use the Pythagoras theorem
adj² = 4²-1²
adj² = 16-1
adj ²= 15
adj = √15
cosx = adj/hyp = √15/4
tanx = opposite/adjacent = 1/√15
since tan < 0, then tanx = -1/√15
From double angle formula;
sin2x = 2sinxcosx
sin2x = 2(1/4)(√15/4)
sin2x = 2√15/16
sin2x = √15/8
for cos2x;
cos2x = 1-2sin²x
cos2x = 1-2(1/4)²
cos2x = 1-2(1/16)
cos2x= 1-1/8
cos2x = 7/8
for tan2x;
tan2x = tanx + tanx/1-tan²x
tan2x = 2tanx/1-tan²x
tan2x = 2(-1/√15)/1-(-1/√15)²
tan2x = (-2/√15)/(1-1/15)
tan2x = (-2/√15)/(14/15)
tan2x = -2/√15 * 15/14
tan2x = -30/14√15
tan2x = -30/7√15
rationalize
tan2x = -30/7√15 * √15/√15
tan2x = -30√15/7*15
tan2x = -2√15/7
