what are the asymptotes of csc x?
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The vertical asymptotes for y=csc(x) y = csc ( x ) occur at 0 , 2π , and every πn , where n is an integer.
816483
The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.
435660
Ans. 400 pm 1.8 Crystalline CsBr has a bcc structure. Calculate the unit cell edge length if the density of CsBr crystal is 4.24 g cm . (Atomic masses : Cs =133; Br = 80)
1201709
For a 95% confidence interval, we have α=0.05, which gives 2.5% of the area at each end of the chi-square distribution. We find values of χ20.975=13.844 and χ20.025=41.923. Evaluating (n−1)s2χ2, we obtain 21.297 and 64.492.
728468
The GCF for the numerical part is 6 . The factors for x4 are x⋅x⋅x⋅x x ⋅ x ⋅ x ⋅ x .
1076899
In inverse variation, the graph is a hyperbola (y = x a ). An inverse variation hyperbola never passes through the origin. An inverse variation hyperbola never crosses the x or y axis; the x and y axes are called the asymptotes of the hyperbola.
823799
The bisector of ∠BAC and ∠CAD intersect the sides BC and CD at the points E and F respectively. Prove that EF||BD.
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ABCD is a parallelogram X and Y are the midpoints of BC and CD respectively. Prove that ar(AXY) = 3/8ar(parallelogram ABCD).
384822
The secant of x is 1 divided by the cosine of x: sec x = 1 cos x , and the cosecant of x is defined to be 1 divided by the sine of x: csc x = 1 sin x . = tan 5π 4 . ... The period of sec x is precisely the same as that of cos x, which means that the period of sec x is 2π.
724011
C is the point on circumference such that angle COB = x. The area of the minor segment cut off by AC is equal to twice the area of the sector BOC.
342120
The inverse cosecant function - arccsc For every trigonometry function such as csc, there is an inverse function that works in reverse. These inverse functions have the same name but with 'arc' in front. So the inverse of csc is arccsc etc. When we see "arccsc A", we interpret it as "the angle whose cosecant is A".
688049
['^0 – Black.', '^1 – Red.', '^2 – Green.', '^3 – Yellow.', '^4 – Blue.', '^5 – Light Blue.', '^6 – Purple.', '^7 – White.']
908556
["x(13âˆ'4x)=0,", '7(y+12)=0,', "(xâˆ'19)(x+3)=0,", "(yâˆ'6)(3zâˆ'4)=0."]
306409
The formula to use is a derivation of the better-known midpoint formula. Letting (m1, m2) represent the given midpoint, (x1, y1) represent the given endpoint, and (x2, y2) represent the unknown endpoint, the formula is: (x2, y2) = (2_m1 – x1, 2_m2 – y1).
1020675
The concavity changes at the asymptote. ... There is no point of the graph of f(x)=1x at which the concavity changes, so the graph has no inflection point. As Alan P. said in his answer, a graph can have a point of inflection that lies on its asymptote.
846535
Here are the rules: If the graph has a gap at the x value c, then the two-sided limit at that point will not exist. If the graph has a vertical asymptote and one side of the asymptote goes toward infinity and the other goes toward negative infinity, then the limit does not exist.
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The secant of x is 1 divided by the cosine of x: sec x = 1 cos x , and the cosecant of x is defined to be 1 divided by the sine of x: csc x = 1 sin x .