1. cos^-1((-sqrt3)/2) 5pi/6 2. arccos(cos7pi/2)

1. cos^-1((-sqrt3)/2)5pi/6 2. arccos(cos7pi/2) -pi/2 3. sin^-1(sin7pi/4) -pi/4 4. csc(arctan11/60) 61/11
1. 2 answers
2. asked by Anonymous
3. 486 views
Find the exact solution(s) of the system: x^2/4-y^2=1 and x=y^2+1A)(4,sqrt3),(4,-sqrt3),(-4,sqrt3),(-4,-sqrt3)
1. 3 answers
2. asked by Jon
3. 717 views
Find the exact solution(s) of the system: x^2/4-y^2=1 and x=y^2+1A)(4,sqrt3),(4,-sqrt3),(-4,sqrt3),(-4,-sqrt3)
1. 0 answers
2. asked by Jon
3. 617 views
(sqrt3 + 5)(sqrt3 -3) ?multiply that out sqrt3*sqrt3 -3sqrt3 +5sqrt3 -15 now combine and simplify terms. yes the problems is:
1. 0 answers
2. asked by Jasmine20
3. 797 views
State the period and phase shift of the function y=-4tan(1/2x + 3pi/8)a) 2pi, -3pi/4 b) pi, 3pi/8 c) 2pi, 3pi/8 d) pi, -3pi/8
1. 4 answers
2. asked by Lucy
3. 1,473 views
Calculate the limit: lim(((arccos x)^(n+1))/((arccos x)^n) with x tends to 1-, for every n that belongs to N, and then prove
1. 0 answers
2. asked by Zisis
3. 511 views
Find a polynomial function of degree 3 with the given numbers as zeros.-5, sqrt3, -sqrt3 Would this work as a function for this
1. 2 answers
2. asked by Alyssa
3. 727 views
Assume that no denominator equals 0.sqrt12 - sqrt18 + 3sqrt50 + sqrt75 = (sqrt2^2*3) - (sqrt2*3^2) + (3sqrt2*5^2) + (sqrt3*5^2)
1. 1 answer
2. asked by Jon
3. 712 views
Rationalize denominator, all variables positive real numbers and denominator is not zero.sqrt3/3 sqrt2 - sqrt3
1. 1 answer
2. asked by Lori
3. 673 views
Problem:cos[arccos(-sqrt3/2)-arcsin(-1/2)] This is what I have: cos [(5pi/6)-(11pi/6)] cos (-6pi/6) or -pi giving an answer of
1. 2 answers
2. asked by Dennis
3. 653 views