Hello,
Vertices are on a line parallele at ox (y=-3)
The hyperbola is horizontal.
Equation is (x-h)²/a²- (y-k)²/b²=1
Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and 2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)
Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6
c²=a²+b²==>b²=36-4²=20
Equation is:
3(x+2)+11>20
minus 11 both sides
3(x+2)>9
divide both sides by 3
x+2>3
minus 2 both sides
x>1
It’s E because if your multiply 5 x 2x u get 10x and 10 x5 = 50 so our answer is E 10x +50
Answer:
p(x) = (5x - 1) (x + 4) (x - 2)
Answer:
- sin(x) = 1
- cos(x) = 0
- cot(x) = 0
- csc(x) = 1
- sec(x) = undefined
Step-by-step explanation:
The tangent function can be considered to be the ratio of the sine and cosine functions:
tan(x) = sin(x)/cos(x)
It will be undefined where cos(x) = 0. The values of x where that occurs are odd multiples of π. The smallest such multiple is x=π/2. The value of the sine function there is positive: sin(π/2) = 1.
The corresponding trig function values are ...
tan(x) = undefined (where sin(x) >0)
sin(x) = 1
cos(x) = 0
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And the reciprocal function values at x=π/2 are ...
cot(x) = 0 . . . . . . 1/tan(x)
csc(x) = 1 . . . . . . .1/sin(x)
sec(x) = undefined . . . . . 1/cos(x)