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Graph y = sec ( x )
Solution: As with tangent we will have to avoid x's for which cosine is zero (recall that sec x =1/ cos x)
Secant will not present at
x = ........, - 5 ∏/2 , - 3 ∏/2 , - ∏/2 , ∏/2, 3 ∏/2 , 5 ∏/2 ,........
and the graph will have asymptotes at these points. Following is the graph of secant on the range - 5 ∏/2 Notice as well that the graph is always greater than 1 & less than -1. It should not be terribly surprising. Remember that -1 ≤ cos ( x ) ≤ 1 . Hence, one divided by something less than one will be greater than 1. Also, 1 / ±1 = ±1 and hence we get the following ranges for secant. sec (? x) ≥ 1 and sec (? x) ≤ -1
Notice as well that the graph is always greater than 1 & less than -1. It should not be terribly surprising. Remember that -1 ≤ cos ( x ) ≤ 1 . Hence, one divided by something less than one will be greater than 1. Also, 1 / ±1 = ±1 and hence we get the following ranges for secant.
sec (? x) ≥ 1 and sec (? x) ≤ -1
how do you graph y+3=-x+3x on a TI-83 graphing calculator?
1. Consider the code of size 4 (4 codewords) and of length 10 with codewords listed below. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1
Integrate following. ∫ -2 2 4x 4 - x 2 + 1dx Solution In this case the integrand is even & the interval is accurate so, ∫ -2 2 4x 4 - x 2 + 1dx = 2∫ o
1. In an in finite horizon capital/consumption model, if kt and ct are the capital stock and consumption at time t, we have f(kt) = ct+kt+1 for t ≥ 0 where f is a given production
case 2:when center is not known proof
solve the in-homogenous problem where A and b are constants on 0 ut=uxx+A exp(-bx) u(x,0)=A/b^2(1-exp(-bx)) u(0,t)=0 u(1,t)=-A/b^2 exp(-b)
Diffrent type of rectillinar figure..
area of r=asin3x
compare: 643,251: 633,512: 633,893. The answer is 633,512.
Area with Parametric Equations In this section we will find out a formula for ascertaining the area under a parametric curve specified by the parametric equations, x = f (t)
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