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Help with my trig homework?

Help with my trig homework? Topic: trig homework
June 17, 2019 / By Algernon
Question: Only 2 problems im stuck on. I just started learning trig. 1. Convert an angle 76 degrees to radians. Express your answer as a multiple of pi. 2. Convert an angle 7pi/6 to degrees. Steps & explanation would help a lot. Thanks!
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Best Answers: Help with my trig homework?

Tania Tania | 8 days ago
to change to a radian you mutiply by pi/180 so then you have (76pi)/180 so that simplifies to 19pi/45 and then you pretty much do the opposite to change radians to degrees. You will multiply by 180/pi. So 7pi/6 times 180/pi, your pis will cancel and you have 7/6 times 180/1 which equals 210 degrees.
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Tania Originally Answered: Help with trig homework! (solving inequalities)?
1: |5 - 2x| < 1 Split into two equations: 5 - 2x < 1 and -(5 - 2x) < 1 Solve each: 5 < 2x + 1 4 < 2x 2 < x x > 2 -5 + 2x < 1 2x < 6 x < 3 Solution: 2 < x < 3 Check: x = 4 |5 - 2(4)| < 1 |5 - 8| < 1 |-3| < 1 3 < 1 False x = 5/2 |5 - 2(5/2)| < 1 |5 - 5| < 1 |0| < 1 0 < 1 TRUE x = 0 |5 - 2(0)| < 1 |5| < 1 5 < 1 False So, the solution is: 2 < x < 3 2: (1 - x)/4 >= (-1 - 2x)/2 Multiply everything by 4: 1 - x >= 2(-1 - 2x) 1 - x >= -2 - 4x -x >= -3 - 4x 3x >= -3 x >= -1 Check: x = -2 (1 - (-2))/4 >= (-1 - 2(-2))/2 3/4 >= 3/2 FALSE x = 0 (1 - (0))/4 >= (-1 - 2(0))/2 1/4 >= -1/2 TRUE 3: (x - 3)/4 - (1 - x)/5 >= (2x + 3)/10 Multiply everything by 20: 5(x - 3) - 4(1 - x) >= 2(2x + 3) 5x - 15 - 4 + 4x >= 4x + 8 9x - 19 >= 4x + 8 5x - 19 >= 8 5x >= 27 x >= 27/5 Check: x = 0 (0 - 3)/4 - (1 - 0)/5 >= (2(0) + 3)/10 -3/4 - 1/5 >= 3/10 -15/20 - 4/20 >= 3/10 (-19/20) >= 3/10 FALSE x = 6 (6 - 3)/4 - (1 - 6)/5 >= (2(6) + 3)/10 3/4 + 1 >= (12 + 3)/10 7/4 >= 15/10 7/4 >= 3/2 7/4 >= 6/4 TRUE Check 4: ((x - 2)(x + 1)^2)/(x - 1) <= 0 x cannot = 1 (x - 2)(x + 1)^2 <= 0 x - 2 <= 0 x <= 2 (x + 1)^2 <=0 x + 1 <= 0 x <= -1 You have 3 critical points: -1, 1, 2 You need to test points left of, between and right of these critical points to determine the solution. You will use 4 values for x to perform your tests. x = -2 (-2 - 2)(-2 + 1)^2/(-2 - 1) <= 0 (-4)(-1)^2/(-3) <= 0 -4(1)/(-3) <= 0 4/3 <= 0 FALSE x = 0 (0 - 2)(0 + 1)^2/(0 - 1) <= 0 (-2)(1)^2/(-1) <= 0 -2/-1 <= 0 2 <= 0 FALSE x = 3/2 (3/2 - 2)(3/2 + 1)^2/(3/2 - 1) <= 0 (-1/2)(5/2)^2/(1/2) <= 0 (-1/2)(25/4)/(1/2) <= 0 -25/4 <= 0 TRUE x = 3 (3 - 2)(3 + 1)^2/(3 - 1) <= 0 (1)(4)^2/2 <= 0 16/2 <= 0 8 <= 0 FALSE The only region where the solutions are true: 1 < x <= 2 5: x^3 + 2x^2 - 8x >= 0 x(x^2 + 2x - 8) >= 0 x(x + 4)(x - 2) >= 0 x >= 0 x + 4 >= 0 x >= -4 x - 2 >= 0 x >= 2 Again, 3 critical points: -4, 0, 2 Choose values to check: x = -5 (-5)^3 + 2(-5)^2 - 8(-5) >= 0 -125 + 2(25) + 40 >= 0 -125 + 50 + 40 >= 0 -35 >= 0 FALSE x = -1 (-1)^3 + 2(-1)^2 - 8(-1) >= 0 -1 + 2 + 8 >= 0 9 >= 0 TRUE x = 1 (1)^3 + 2(1)^2 - 8(1) >= 0 1 + 2 - 8 >= 0 -5 >= 0 FALSE x = 3 (3)^3 + 2(3)^2 - 8(3) >= 0 27 + 2(81) - 24 >= 0 27 + 162 - 24 >= 0 165 >= 0 TRUE Two areas with true solutions: x >= 2 and -4 <= x <= 0
Tania Originally Answered: Help with trig homework! (solving inequalities)?
1) split into two 5 - 2x < 1 and 5 - 2x > -1 2x > 4 x > 2 and 2nd eq 2x < 6 x < 3 so 2 < x < 3 fulfills it. From there its just addition/multiplication, remembering if you multiply by a negative # you reverse the sign.
Tania Originally Answered: Help with trig homework! (solving inequalities)?
1. two answers since in absolute value brackets, -(5-2x)<1 -5+2x<1 2x<6 x<3....ans 1 5-2x<1 -2x<1 -2x<-4 (signs change because dividing by negative number) x>2...ans 2

Rickena Rickena
just MEMORIZE this conversion: 180 degrees=pi radians It's like converting pounds to kilogram (1kilogram=2.2 lbs). You multiply it to the ratio such that its unit will be cancelled: 1. 76 deg * (pi rad/180 deg) = 0.422 pi rad 2. (7pi/6) * (180 deg/pi rad) = 210 deg goodluck :)
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Rickena Originally Answered: I need help with my calculus homework (trig functions)?
u have to use the unit circle for this. So u take the arc sin of both sides to make the x by itself so it looks like: x = sin^(-1) root 2/2 and this is asking "where is the sin of an angle root 2/2. and that is at pie/4 and 5pie/4 if the interval is from 0 to 2pie. hope i helped
Rickena Originally Answered: I need help with my calculus homework (trig functions)?
Like with any quadratic equation, you solve by factoring (if feasible); otherwise fall back to the quadratic formula. Your 2cos^2x=√3cosx becomes 2(cos² x) -(√3) cos x =0 which factors: (cos x) [2 cos x -√3] = 0 Either factor can be zero; so if cos x = 0 then x =90° If the other factor is zero, cos x =(√3)/2 so x = 60° ========== BTW you should use grouping symbols judiciously so there's no ambiguity (abt what the radicand is; and what the exponent is);

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