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# Recent problems solved by 'lwsshak3'

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 Trigonometry-basics/690398: if cos(a) is 2/3 and sin(b) is 4/5 what is cos(a+b)1 solutions Answer 426278 by lwsshak3(6761)   on 2012-12-06 00:39:20 (Show Source): You can put this solution on YOUR website!if cos(a) is 2/3 and sin(b) is 4/5 what is cos(a+b) .. cos a=2/3=adj side/hypotenuse opp side=√(3^2-2^2)=√(9-4)=√5 sin a=√5/3 .. sin b=4/5 (3-4-5 right triangle) cos b=3/5 .. cos(a+b)=cos a cos b -sin a sin b =2/3*3/5 -√5/3*4/5 =6/15-4√5/15 =(6-4√5)/15 .. calculator check: a=cos^1(2/3)≈.8411 b=sin^-1(4/5)=.9273 cos(a+b)=cos(1.7684)=-0.1962.. (6-4√5)/15=-0.1962..
 Quadratic-relations-and-conic-sections/689855: Find the equation of the ellipse with the given conditions. Center (1,1), vertex (1,3), passing through (0,0). I need your help.1 solutions Answer 426224 by lwsshak3(6761)   on 2012-12-05 19:33:25 (Show Source): You can put this solution on YOUR website!Find the equation of the ellipse with the given conditions. Center (1,1), vertex (1,3), passing through (0,0). ** This is an ellipse with vertical major axis. Its standard form of equation:, a>b, (h,k)=(x,y) coordinates of center For given ellipse: center: (1,1) length of major axis=4=2a a=2 a^2=4 plug in coordinates of given point on ellipse: (0,0) (x-1)^2/b^2+(y-1)^2/4=1 (0-1)^2/b^2+(0-1)^2/4=1 1/b^2+1/4=1 1/b^2=3/4 b^2=4/3 Equation of given ellipse:
 Quadratic-relations-and-conic-sections/689806: Find an equation in standard form of the parabola passing through the following points (-1,1) (0,5) (2,7) Thank You.1 solutions Answer 426065 by lwsshak3(6761)   on 2012-12-05 03:02:04 (Show Source): You can put this solution on YOUR website!Find an equation in standard form of the parabola passing through the following points (-1,1) (0,5) (2,7) y=Ax^2+Bx+C (1) 1=A-B+C (-1,1) (2) 5=C (0,5) (3) 7=4A+2B+C (2,7) .. 1=A-B+5 7=4A+2B+5 2=2A-2B+10 add 9=6A+15 6A=-6 A=-1 .. B=A+5-1=3 Equation: y=Ax^2+Bx+C y=-x^2+3x+5 complete the square y=-(x^2-3x+9/4)+9/4+5 y=-(x-3/2)^2+29/4 This is an equation of a parabola that opens downwards with vertex at(3/2,29/4)
 Trigonometry-basics/689562: I need help on trig identities. Verify the following equation: cosx(cosx-secx)=-sin^2x So far I used the foil method to get cos^2x-cosxsecx but I'm not sure if that is right1 solutions Answer 426063 by lwsshak3(6761)   on 2012-12-05 02:23:40 (Show Source): You can put this solution on YOUR website!Verify the following equation: cosx(cosx-secx)=-sin^2x start with left side cosx(cosx-secx) =cosx(cosx-1/cosx) =cos^2x-1 =-(1-cos^2x) =-sin^2x verified: left side=right side
 Trigonometry-basics/689731: given sec theta equals -10 with 180< theta< 270 find and simplify sin theta/2 as an exact value1 solutions Answer 426062 by lwsshak3(6761)   on 2012-12-05 02:15:58 (Show Source): You can put this solution on YOUR website!given sec theta equals -10 with 180< theta< 270 find and simplify sin theta/2 as an exact value use x for theta secx=-10=1/cos cosx=-1/10=adjacent side/hypotenuse opposite side=√(10^2-1^2)=√(100-1)=√99 sinx=-√99/10 (in quadrant III where sin<0) sin(x/2) =√[(1-cosx)/2] =√[(1-(-1/10)/2] =√[(1+(1/10)/2] =√(1.1/2) .. check with calculator: sin^-1(√99/10)=84.26º+180=264.26º(in quadrant III as given) sin(x/2)=132.13 (in quadrant II where sin>0) reference angle=132.13-180=47.87º sin 47.87º≈0.7416.. √(1.1/2)≈0.7416..
 Trigonometry-basics/689725: how do I found the exact answer of sec(-5 pi/8)? I tried to add or subtract 2pi so that it is a value on the chart but that does not work and I don't know what else to try. I think maybe a double angle therom or something? 1 solutions Answer 426037 by lwsshak3(6761)   on 2012-12-04 21:26:16 (Show Source): You can put this solution on YOUR website!how do I found the exact answer of sec(-5 pi/8)? use identity for cos half-angle formula: cos (s/2)=±√[(1+cos s)/2] -sec(-5π/8)=-1/cos(-5π/8) solve for cos(-5π/8), then take the negative reciprocal. cos(-5π/8)=cos(-5π/4)/2)=-√[1+cos(-5π/4)/2] cos(-5π/4)=cos π/4=-√2/2 in quadrant II where cos<0 -√[1+cos(-5π/4)/2]=-√[1-√2/2/2]=-√(2-√2)/2 -sec(-5π/8)=-1/cos(-5π/8) -sec(-5π/8)=-1/-√(2-√2)/2 -sec(-5π/8)=2/√(2-√2) check with computer: cos(-5π/8)=-.3836 -sec(-5π/8)≈-1/-.3836=2.6131.. 2/√(2-√2)≈2.6131..
 Trigonometry-basics/689720: What is the reference angle for 5pi/8 and -490 degrees? 1 solutions Answer 426024 by lwsshak3(6761)   on 2012-12-04 20:31:11 (Show Source): You can put this solution on YOUR website!What is the reference angle for 5pi/8 and -490 degrees? 5π/8=1.9635 radians in quadrant II reference angle=π-1.9635≈1.1781 .. -490º+360º=-130º in quadrant III reference angle=-130+180=50º
 Trigonometry-basics/689602: find period and phase shift, y= -4 tan(5x + pi/2) 1 solutions Answer 426018 by lwsshak3(6761)   on 2012-12-04 20:20:51 (Show Source): You can put this solution on YOUR website!find period and phase shift, y= -4 tan(5x + pi/2) Equation of tan function: y=tan(Bx-C), period=π/B, phase shift=C/B For given tan function: B=5 period=π/B=π/5 C=π/2 phase shift=C/B=(π/2)/5=π/10 (shift to the left)
 Trigonometry-basics/689603: Find the exact value of the six trig functions. (5,9) is a point on the terminal side of theta1 solutions Answer 426011 by lwsshak3(6761)   on 2012-12-04 20:06:53 (Show Source): You can put this solution on YOUR website!Find the exact value of the six trig functions. (5,9) is a point on the terminal side of theta use x for theta you are working with a reference angle in quadrant I where all 6 functions>0 tanx=9/5=opposite side/adjacent side hypotenuse=√(9^2+5^2)=√(81+25)=√106 sinx=9/√106 cosx=5/√106 tanx=9/5 cscx=√106/9 secx=√106/5 cotx=5/9
 Quadratic-relations-and-conic-sections/689523: how do you Write the standard equation for the ellipse with the given characteristics Foci: (5, 0), (-5, 0) Vertices: (9, 0), (-9, 0)1 solutions Answer 426008 by lwsshak3(6761)   on 2012-12-04 19:48:32 (Show Source): You can put this solution on YOUR website!Write the standard equation for the ellipse with the given characteristics Foci: (5, 0), (-5, 0) Vertices: (9, 0), (-9, 0) ** This is an ellipse with horizontal major axis. Its standard form of equation: , a>b, (h,k)=(x,y) coordinates of center For given ellipse: center: (0,0) a=9 (center to vertex on the major axis) a^2=81 c=5 (center to foci on the major axis) c^2=25 c^2=a^2-b^2 b^2=a^2-c^2=81-25=56 equation:
 Rate-of-work-word-problems/689466: bOB OWN A WATCH REPAIR SHOP. hE HAS found that the cost of operating his shop is given by c(x) = 2x^2-112x+58, where c is cost and x is the number of watches repaired. how many watches must he repari to have the lowest cost?1 solutions Answer 425972 by lwsshak3(6761)   on 2012-12-04 15:49:25 (Show Source): You can put this solution on YOUR website!bOB OWN A WATCH REPAIR SHOP. hE HAS found that the cost of operating his shop is given by c(x) = 2x^2-112x+58, where c is cost and x is the number of watches repaired. how many watches must he repair to have the lowest cost? ** c(x) = 2x^2-112x+58 This is an equation of a parabola that opens upwards, (curve has a minimum) Its standard form: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex. (y-value is the minimum in this case. c(x) = 2x^2-112x+58 complete the square c(x)=2(x^2-56x+28^2)-2*28^2+58 c(x)=2(x^2-56x+784)-1568+58 c(x)=2(x-28)^2-1510 vertex: (28,-1510) how many watches must he repair to have the lowest cost? 28 see graph below as a visual check
 Graphs/689459: Solve the following rational inequality. x>= 3/x+2 write your answer in interval notation 1 solutions Answer 425965 by lwsshak3(6761)   on 2012-12-04 15:07:26 (Show Source): You can put this solution on YOUR website!Solve the following rational inequality. x>= 3/(x+2) write your answer in interval notation x+2=0 x≠-2 number line: <.....-.....-2......+.......> solution: (-2,∞)
 Polynomials-and-rational-expressions/689194: I have to find the coordinates of the vertical and the equation of the axis of symmetry for the parabola given by the q Equation ; Y= -1/4 x squared -11 solutions Answer 425958 by lwsshak3(6761)   on 2012-12-04 14:31:50 (Show Source): You can put this solution on YOUR website!Find the coordinates of the vertex and the equation of the axis of symmetry for the parabola given. Y= -1/4 x squared -1 y=-(1/4)x^2-1 4y=-x^2-4 x^2=-4y-4 x^2=-4(y+1) This is an equation of a parabola that opens downwards Its standard form: (x-h)^2=-4p(y-k), (h,k)=(x,y) coordinates of vertex For given equation:x^2=-4(y+1) vertex: (0,-1) axis of symmetry: x=0 or y-axis
 Trigonometry-basics/689299: how do you find the vertices, foci, of the ellipse. 56x^2 + 7y^2 = 8 56x^2/8+7y^2/8=8/8 x^2/sqrt7+y^2/(sqrt14/4)^2=1 eccentricity=sqrt14/41 solutions Answer 425952 by lwsshak3(6761)   on 2012-12-04 14:12:21 (Show Source): You can put this solution on YOUR website!how do you find the vertices, foci, of the ellipse. 56x^2 + 7y^2 = 8 ** 56x^2+7y^2 = 8 7x^2+7y^2/8 = 1 This is an equation of an ellipse with vertical major axis. Its standard form:, a>b, (h,k)=(x,y) coordinates of center For given equation:7x^2+7y^2/8 = 1 center: (0,0) a^2=8/7 a=√(8/7)≈1.07 vertices:(0,0±a)=(0,0±1.07)=(0,-1.07) and (0,1.07 ) b^2=1/7 b=√7/7≈.38 c^2=a^2-b^2=8/7-1/7=7/7=1 c=1 foci: (0,0±c)=(0,0±1)=(0,-1) and (0,1)
 Rational-functions/689215: How do I write an equation for a parabola whose vertex is (4,1) and passes through the point (2,13)?1 solutions Answer 425871 by lwsshak3(6761)   on 2012-12-04 02:16:15 (Show Source): You can put this solution on YOUR website!How do I write an equation for a parabola whose vertex is (4,1) and passes through the point (2,13)?Standard form of equation for a parabola: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex plug in given coordinates of the vertex(4,1) (x-4)^2=4p(y-1) plug in coordinates of given point (2,13) (2-4)^2=4p(13-1) 4=4p(12) 4p=4/12=1/3 equation of parabola: (x-4)^2=(y-1)/3
 logarithm/689176: what is the solution to this problem ln(x-9)-ln(x+1)=ln(x-6)-ln(x+5) ?1 solutions Answer 425869 by lwsshak3(6761)   on 2012-12-04 01:46:32 (Show Source): You can put this solution on YOUR website!what is the solution to this problem ln(x-9)-ln(x+1)=ln(x-6)-ln(x+5) ? ln(x-9)-ln(x+1)=ln(x-6)-ln(x+5) ln[(x-9)/(x+1)]=ln[(x-6)/(x+5)] (x-9)/(x+1)=(x-6)/(x+5) (x+1)(x-6)=(x-9)(x+5) x^2-5x-6=x^2-4x-45 x=39
 Quadratic-relations-and-conic-sections/689066: Can someone please help me find the foci and asymptotes to (y-1)^2/9-(x-3)^2/4=1 1 solutions Answer 425868 by lwsshak3(6761)   on 2012-12-04 01:19:20 (Show Source): You can put this solution on YOUR website!find the foci and asymptotes to (y-1)^2/9-(x-3)^2/4=1 This is an equation of a hyperbola with vertical transverse axis (y-term listed first) Its standard form of equation: , (h,k)=(x,y) coordinates of center For given equation: center: (3,1) a^2=9 a=√9=3 b^2=4 b=√4=2 c^2=a^2+b^2=9+4=13 c=√13≈3.6 foci: (3,1±c)=(3,1±3.6)=(3,-2.6) and (3,4.6) Asymptotes: Straight line equations that go thru center, y=mx+b, m=slope, b=y-intercept slopes of asymptotes for hyperbolas with vertical transverse axis=±a/b=±3/2 equation of asymptote with negative slope: y=-3x/2+b solve for b using coordinates of center(3,1) 1=-3*3/2+b b=1+9/2=11/2 equation of asymptote: y=-3x/2+11/2 .. equation of asymptote with positive slope: y=3x/2+b solve for b using coordinates of center(3,1) 1=3*3/2+b b=1-9/2=-7/2 equation of asymptote: y=3x/2-7/2
 Quadratic-relations-and-conic-sections/689067: Can someone please help me find the vertices and foci of the ellipse 3x^2+7y^2=21. I have vertices (0,sqrt -3) and (0, sqrt 3) and foci (0,2) and (0,-2) but not sure if it's right.1 solutions Answer 425865 by lwsshak3(6761)   on 2012-12-04 00:45:13 (Show Source): You can put this solution on YOUR website!Can someone please help me find the vertices and foci of the ellipse 3x^2+7y^2=21 x^2/7+y^2/3=1 This is an equation of an ellipse with horizontal major axis. Its standard form of equation: , a>b, (h,k)=(x,y) coordinates of center. center: 0,0) a^2=7 a=√7 vertices: (0±a,0)=(0±√7,0)=(-√7,0) and (√7,0) b^2=3 c^2=a^2-b^2=7-3=4 c=2 foci: (0±c,0)=(0±2,0)=(-2,0) and (2,0)