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Recent problems solved by 'lwsshak3'

lwsshak3 answered: 6520 problems
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Trigonometry-basics/752182: sin3A / sinA + cos3A / cosA = 4cos2A Please to do this using as least steps as possible please?\ Thanks!! 1 solutions Answer 457725 by lwsshak3(6522)   on 2013-05-25 04:25:13 (Show Source): You can put this solution on YOUR website! A=acute angle sin3A / sinA + cos3A / cosA = 4cos2A I assume you meant to write given expression as follows: 4cos^2(A)=1 cos^2(A)=1/4 cos(A)=1/2 A=60º or π/3

Trigonometry-basics/752229: if A lies in the second quadrant and 3tan A +4=0 then the value of 2 cot A-5 cos A+sin A is equal to? 1 solutions Answer 457723 by lwsshak3(6522)   on 2013-05-25 04:10:10 (Show Source): You can put this solution on YOUR website! if A lies in the second quadrant and 3tan A +4=0 then the value of 2 cot A-5 cos A+sin A is equal to? *** 3tanA+4=0 tanA=-4/3 Working with a 3-4-5 reference right triangle in Q2 cotA=-3/4 cosA=-3/5 sinA=4/5 .. 2 cot A-5 cos A+sin A =2(-3/4)-5(-3/5)+4/5 =-6/4+3+4/5 =-3/2+3+4/5 =-15/10+30/10+8/10 =23/10

Trigonometry-basics/752296: verify this identity tanx(cotx + tanx) = secx^2 1 solutions Answer 457722 by lwsshak3(6522)   on 2013-05-25 03:50:06 (Show Source): You can put this solution on YOUR website! verify this identity tanx(cotx + tanx) = secx^2 Start with LHS tanx(cotx + tanx)=tanx((1/tanx)+tanx))=1+tan^2x=sec^2x verified: LHS=RHS

Equations/752312: Solve a) x^3+8x^2+17x+10=0 Can you please help me out? Thanks so much in advance:) Can you please show the steps it would really help me understand:) 1 solutions Answer 457690 by lwsshak3(6522)   on 2013-05-24 20:50:53 (Show Source): You can put this solution on YOUR website! Solve a) f(x)=x^3+8x^2+17x+10=0 Use rational roots theorem to solve: ....0...|......1.......8........17.......10........... ....1...|......1......9.........26.......36 upper boundary (all numbers positive) ..-1...|......1......7.........10.....-10 (-1 is a root) f(x)=(x+1)(x^2+7x+10=0 f(x)=(x+1)(x+2)(x+5)=0 roots: x=-5, -2,-1

Trigonometry-basics/752310: How can I use an addition or subtraction formula to simplify the following equation. In order to find all the solutions in the interval [0,2pi). sin 4 x *cos 3 x - cos 4 x *sin 3 x = 0 1 solutions Answer 457671 by lwsshak3(6522)   on 2013-05-24 17:29:35 (Show Source): You can put this solution on YOUR website! find all the solutions in the interval [0,2pi). sin 4 x *cos 3 x - cos 4 x *sin 3 x = 0 Identity: sin(x-y)=sinx*cosy-cosx*siny For given problem: sin 4 x *cos 3 x - cos 4 x *sin 3 x=sin(4x-3x)=0 sinx=0 x=0, π

logarithm/750607: how do you solve log3[4]+log3[(4x^+9)]=log3 60 1 solutions Answer 457670 by lwsshak3(6522)   on 2013-05-24 17:20:45 (Show Source): You can put this solution on YOUR website! how do you solve log3[4]+log3[(4x^2+9)]=log3 60 4(4x^2+9)=60 4x^2+9=15 4x^2=6 x^2=6/4=3/2 x=±√(3/2)

logarithm/751410: Log subscript3 (X)+logsubscript3 (x^2-8)= log38X 1 solutions Answer 457669 by lwsshak3(6522)   on 2013-05-24 16:50:20 (Show Source): You can put this solution on YOUR website! Log subscript3 (X)+logsubscript3 (x^2-8)= log38X (x)(x^2-8)=8x x^2-8=8 x^2=16 x=±√16 x=-4 (reject, 8x>0) or x=4

logarithm/752315: Solve for x a) log(base 6)4 +2log(base 6)v = 2 Can you please help me out? Thanks so much in advance:) Can you please show the steps it would really help me understand:) 1 solutions Answer 457668 by lwsshak3(6522)   on 2013-05-24 16:35:41 (Show Source): You can put this solution on YOUR website! solve for v a) log(base 6)4 +2log(base 6)v = 2 convert to exponential form: base(6) raised to log of number(2)=number(4v^2) 6^2=4v^2 v^2=36/4=9 v=±√9=±3

logarithm/748069: 1 solutions Answer 457660 by lwsshak3(6522)   on 2013-05-24 16:08:58 (Show Source): You can put this solution on YOUR website! u^2-u-3=0 solve by quadratic formula: u≈-1.303 (reject, log>0) u≈2.303= x=3^2.303≈12.51 .. Check: =(2.30)^2-2.30≈3

logarithm/752092: Rewrite the expression f(x) = ln (x^2-16 / x^5 )  using properties of logarithms. 1 solutions Answer 457588 by lwsshak3(6522)   on 2013-05-24 02:12:13 (Show Source): You can put this solution on YOUR website! Rewrite the expression

logarithm/752135: How to find log value for square root of 2*pi*n 1 solutions Answer 457582 by lwsshak3(6522)   on 2013-05-24 01:54:48 (Show Source): You can put this solution on YOUR website! How to find log value for square root of 2*pi*n

logarithm/752143: 2 log10x+1=log10 250 1 solutions Answer 457576 by lwsshak3(6522)   on 2013-05-24 01:31:30 (Show Source): You can put this solution on YOUR website! (x+1)^2=250 x+1=√250 x=-1±√250≈1±15.81 x≈-16.81(reject, (x+1)>0) or x≈14.81

Linear_Algebra/752150: Use the given conditions to write an equation for the line in point-slope form. Passing through (-5, -7) and (-8, -6) 1 solutions Answer 457574 by lwsshak3(6522)   on 2013-05-24 01:11:54 (Show Source): You can put this solution on YOUR website! Use the given conditions to write an equation for the line in point-slope form. Passing through (-5, -7) and (-8, -6) ** m=∆y/∆x=(-6-(-7))/(-8-(-5))=1/-3=-1/3 using coordinates of any one of the points:(-5,-7) 3y+x=26

Trigonometry-basics/751757: Given cosA = -1/4 and 90 degrees < A < 180 degrees, find tanA in surd form. 1 solutions Answer 457541 by lwsshak3(6522)   on 2013-05-23 19:36:40 (Show Source): You can put this solution on YOUR website! Given cosA = -1/4 and 90 degrees < A < 180 degrees, find tanA in surd form sec(A)=1/cos(A)=-4 (in Q2 where tan<0) .. Check with calculator: cos(A)=-1/4 A≈104.4775º tan(A)=tan(104.4775º)≈-3.873.. Exact value=-√15≈-3.873..

Trigonometry-basics/752005: find the exact value of cos 2A and sin 2A given sin A = 12/13, A is in quadrant I 1 solutions Answer 457535 by lwsshak3(6522)   on 2013-05-23 19:01:56 (Show Source): You can put this solution on YOUR website! find the exact value of cos 2A and sin 2A given sin A = 12/13, A is in quadrant I ( working with 5-12-13 right triangle) .. .. Check with calculator: sin(A)=12/13 (in Q1) A=67.38º 2A=134.76 sin(2A)=sin(134.76º)≈0.71 exact value=120/169≈0.71 cos(2A)=cos(134.76º)≈-0.704 exact value=119/169≈-0.704

Trigonometry-basics/752009: given sin theta= -root 7/4 and theta is in quadrant IV, find the exact value of cos theta/2 1 solutions Answer 457532 by lwsshak3(6522)   on 2013-05-23 18:44:53 (Show Source): You can put this solution on YOUR website! given sin theta= -root 7/4 and theta is in quadrant IV, find the exact value of cos theta/2 *** ( in Q3 where cos<0) .. Check with calculator: sinx=-√7/4 (in Q4) x=318.49º x/2=159.30º (in Q3) cos(x/2)=cos(159.30º)≈-0.9354.. Exact value=-√(7/8)≈-0.9354..

Trigonometry-basics/752001: verify the identity sin^2 theta (1+ cot^2 theta) = 1 1 solutions Answer 457522 by lwsshak3(6522)   on 2013-05-23 18:07:31 (Show Source): You can put this solution on YOUR website! verify the identity sin^2 theta (1+ cot^2 theta) = 1 start with LHS Verified: LHS=RHS

Trigonometry-basics/750407: Find all solutions in the interval o and 2pi 1.) Cos^2x+ sinx +1= 0 1 solutions Answer 457376 by lwsshak3(6522)   on 2013-05-22 19:59:13 (Show Source): You can put this solution on YOUR website! Find all solutions in the interval o and 2pi 1.) Cos^2x+ sinx +1= 0 1-sin^2(x)+sin(x)+1=0 sin^2(x)-sin(x)-2=0 (sin(x)-2)(sin(x)+1)=0 .. sin(x)=2 (reject, -1 ≤ sinx ≤ 1) sin(x)=-1 x=3π/2

Trigonometry-basics/751287: Simplify: 1-sin^2((pi/2)-x) 1 solutions Answer 457373 by lwsshak3(6522)   on 2013-05-22 19:36:22 (Show Source): You can put this solution on YOUR website! 1-sin^2((pi/2)-x) sin(π/2-x)=cos(x) 1-sin^2((pi/2)-x)=1-cos^2(x)=sin^2(x)

Trigonometry-basics/751257: Can you help me solve the equations for x, if 0is less than equal to x is less than equal to 2pi. a) 8cos^2x=4 1 solutions Answer 457370 by lwsshak3(6522)   on 2013-05-22 19:30:25 (Show Source): You can put this solution on YOUR website! solve the equations for x, if 0is less than equal to x is less than equal to 2pi. a) 8cos^2x=4 cos^2(x)=4/8=1/2 cos(x)=1/√2=√2/2 x=π/4, 3π/4 (in Q1 and Q2 where sin>0)

Trigonometry-basics/751467: Complete the identity. 1). cos θ - cos θ sin^2 θ = ? 2). (1 / cos^2 θ) - (1 / cot^2 θ) = ? 1 solutions Answer 457369 by lwsshak3(6522)   on 2013-05-22 19:24:49 (Show Source): You can put this solution on YOUR website! Complete the identity. 1). cos θ - cos θ sin^2 θ = ? 2). (1 / cos^2 θ) - (1 / cot^2 θ) = ?

Trigonometry-basics/750962: how to find the value of sin 1380 cos 225 and tan 420 1 solutions Answer 457366 by lwsshak3(6522)   on 2013-05-22 19:05:06 (Show Source): You can put this solution on YOUR website! how to find the value of sin 1380 cos 225 and tan 420 *** 1380-3*360=1380-1080=300 sin(1380)=sin(300)=sin(60)=-√3/2 (in Q4 where sin<0) .. cos(225)=cos(45)=-√2/2 (in Q3 where cos<0) .. 420-360=60 tan(420)=tan(60)=√3 (in Q1 where tan>0)

Trigonometry-basics/751013: Find the exact value of csc . Some quick help would be greatly appreciated. :) 1 solutions Answer 457363 by lwsshak3(6522)   on 2013-05-22 18:40:36 (Show Source): You can put this solution on YOUR website! Find the exact value of csc sin(5π/4)=-√2/2 (in Q3 where sin<0) csc(5π/4)=1/sin(5π/4)=-2/√2=-√2

Trigonometry-basics/751014: Find the exact value of cot . Some quick help would be greatly appreciated. :) 1 solutions Answer 457361 by lwsshak3(6522)   on 2013-05-22 18:35:16 (Show Source): You can put this solution on YOUR website! Find the exact value of cot . tan(2π/3)=-√3 (in Q2 where tan<0) cot 2π/3=1/tan 2π/3=-1/√3=-√3/3

Trigonometry-basics/751521: sec^-1 (-1) 1 solutions Answer 457311 by lwsshak3(6522)   on 2013-05-22 15:25:59 (Show Source): You can put this solution on YOUR website! sec^-1 (-1) let x=angle whose sec=(-1) sec(x)=-1 cos(x)=1/sec(x)=1/-1=-1 x=π

Exponents/751565: x^5/2=32 1 solutions Answer 457310 by lwsshak3(6522)   on 2013-05-22 15:20:35 (Show Source): You can put this solution on YOUR website! x^5/2=32 raise both sides to the (2/5) power. x=32^2/5 x=4 (5th root of 32=2(squared)=4)

Coordinate-system/751611: Determine the equation of the line passing through the points (1, 2) and (6, -4). Please show work. 1 solutions Answer 457309 by lwsshak3(6522)   on 2013-05-22 15:15:16 (Show Source): You can put this solution on YOUR website! Determine the equation of the line passing through the points (1, 2) and (6, -4). Please show work. *** Equation for a line: y=mx+b, m=slope, b=y-intercept slope=∆y/∆x=(-4-2)/6-1)=-6/5 y=-6x/5+b solve for b using coordinates of one of given points(1,2) on the line 2=-6/5+b b=2+6/5=16/5 Equation of given line:y=-6x/5+16/5

Equations/751531: this is my problem P = A((1+r)-n P= 20,000(1+ 0.06)-20 P = 22,000(1.06)-20 this is my question do i divide or multiply here to get my answer 1 solutions Answer 457308 by lwsshak3(6522)   on 2013-05-22 15:05:38 (Show Source): You can put this solution on YOUR website! this is my problem P = A((1+r)-n P= 20,000(1+ 0.06)-20 P = 22,000(1.06)-20 *** The formula is not written correctly for compound interest. It should be written as follows: A=P(1+r)^n, P=initial investment, r=interest rate per period, n=number of periods, A=amount after n periods.

Trigonometry-basics/751312: Verify: (1-sin(x)/1+sin(x))=(sec(x)-tan(x))^2 1 solutions Answer 457208 by lwsshak3(6522)   on 2013-05-22 03:06:45 (Show Source): You can put this solution on YOUR website! Verify: start with LHS verified: LHS=RHS

?t?e?s?t/751425: graph this equation: x^2-4y=0 and state vertex & focus 1 solutions Answer 457199 by lwsshak3(6522)   on 2013-05-22 02:10:01 (Show Source): You can put this solution on YOUR website! graph this equation: x^2-4y=0 and state vertex & focus x^2-4y=0 x^2=4y Parabola opens upward Basic form of equation:(x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex vertex:(0,0) axis of symmetry: x=0 or y-axis 4p=4 p=1 focus: (0,1) (p-distance above vertex on the axis of symmetry) see graph below: y=x^2/4

Quadratic-relations-and-conic-sections/751431: find and equation for a hyperbola with vertices at (2,3) and (2,-1) and foci at (2,6) and (2,-4) 1 solutions Answer 457198 by lwsshak3(6522)   on 2013-05-22 01:52:52 (Show Source): You can put this solution on YOUR website! find and equation for a hyperbola with vertices at (2,3) and (2,-1) and foci at (2,6) and (2,-4) hyperbola has a vertical transverse axis. Its standard form of equation: , (h,k)=(x,y) coordinates of the center For given hyperbola: x-coordinate of center=2 y-coordinate of center=1 (midpoint of vertical transverse axis) center: (2,1) a=2 (distance from center to vertices) a^2=4 c=5 (distance from center to foci) c^2=25 c^2=a^2+b^2 b^2=c^2-a^2=25-4=21 Equation of given hyperbola: