Algebra ->  Tutoring on algebra.com -> See tutors' answers!      Log On

 Tutoring Home For Students Tools for Tutors Our Tutors Register Recently Solved
 By Tutor
| By Problem Number |

Tutor:

# Recent problems solved by 'lwsshak3'

Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289 , 1290..1319 , 1320..1349 , 1350..1379 , 1380..1409 , 1410..1439 , 1440..1469 , 1470..1499 , 1500..1529 , 1530..1559 , 1560..1589 , 1590..1619 , 1620..1649 , 1650..1679 , 1680..1709 , 1710..1739 , 1740..1769 , 1770..1799 , 1800..1829 , 1830..1859 , 1860..1889 , 1890..1919 , 1920..1949 , 1950..1979 , 1980..2009 , 2010..2039 , 2040..2069 , 2070..2099 , 2100..2129 , 2130..2159 , 2160..2189 , 2190..2219 , 2220..2249 , 2250..2279 , 2280..2309 , 2310..2339 , 2340..2369 , 2370..2399 , 2400..2429 , 2430..2459 , 2460..2489 , 2490..2519 , 2520..2549 , 2550..2579 , 2580..2609 , 2610..2639 , 2640..2669 , 2670..2699 , 2700..2729 , 2730..2759 , 2760..2789 , 2790..2819 , 2820..2849 , 2850..2879 , 2880..2909 , 2910..2939 , 2940..2969 , 2970..2999 , 3000..3029 , 3030..3059 , 3060..3089 , 3090..3119 , 3120..3149 , 3150..3179 , 3180..3209 , 3210..3239 , 3240..3269 , 3270..3299 , 3300..3329 , 3330..3359 , 3360..3389 , 3390..3419 , 3420..3449 , 3450..3479 , 3480..3509 , 3510..3539 , 3540..3569 , 3570..3599 , 3600..3629 , 3630..3659 , 3660..3689 , 3690..3719 , 3720..3749 , 3750..3779 , 3780..3809 , 3810..3839 , 3840..3869 , 3870..3899 , 3900..3929 , 3930..3959 , 3960..3989 , 3990..4019 , 4020..4049 , 4050..4079 , 4080..4109 , 4110..4139 , 4140..4169 , 4170..4199 , 4200..4229 , 4230..4259 , 4260..4289 , 4290..4319 , 4320..4349 , 4350..4379 , 4380..4409 , 4410..4439 , 4440..4469 , 4470..4499 , 4500..4529 , 4530..4559 , 4560..4589 , 4590..4619 , 4620..4649 , 4650..4679 , 4680..4709 , 4710..4739 , 4740..4769 , 4770..4799 , 4800..4829 , 4830..4859 , 4860..4889 , 4890..4919 , 4920..4949 , 4950..4979 , 4980..5009 , 5010..5039 , 5040..5069 , 5070..5099 , 5100..5129 , 5130..5159 , 5160..5189 , 5190..5219 , 5220..5249 , 5250..5279 , 5280..5309 , 5310..5339 , 5340..5369 , 5370..5399 , 5400..5429 , 5430..5459 , 5460..5489 , 5490..5519 , 5520..5549 , 5550..5579 , 5580..5609 , 5610..5639 , 5640..5669 , 5670..5699 , 5700..5729 , 5730..5759 , 5760..5789 , 5790..5819 , 5820..5849 , 5850..5879 , 5880..5909 , 5910..5939 , 5940..5969 , 5970..5999 , 6000..6029 , 6030..6059 , 6060..6089 , 6090..6119 , 6120..6149 , 6150..6179 , 6180..6209 , 6210..6239 , 6240..6269 , 6270..6299 , 6300..6329 , 6330..6359 , 6360..6389 , 6390..6419 , 6420..6449 , 6450..6479 , 6480..6509, >>Next

 Quadratic-relations-and-conic-sections/744640: write the standard form of an ellipse. x^2+4y^2+6x-91=01 solutions Answer 453805 by lwsshak3(6505)   on 2013-05-05 20:20:59 (Show Source): You can put this solution on YOUR website!write the standard form of an ellipse: x^2+4y^2+6x-91=0 x^2+6x+4y^2=91 complete the square: (x^2+6x+9)4y^2=9+91 (x+3)^2+4y^2=100 (x+3)^2/100+y^2/25=1 This is and ellipse with horizontal major axis Its standard form:(x-h)^2/a^2+(y-k)^2/b^2 a>b,(h,k)=(x,y) coordinates of the center:
 Quadratic-relations-and-conic-sections/745114: Which is the equation with focus of (6,0) A) Y=-1/6^2 B) y=1/12x^2 C) x=1/24y^2 D) x=-1/6y^21 solutions Answer 453796 by lwsshak3(6505)   on 2013-05-05 19:51:19 (Show Source): You can put this solution on YOUR website!Which is the equation with focus of (6,0) A) Y=-1/6^2 B) y=1/12x^2 C) x=1/24y^2 D) x=-1/6y^2 *** Parabola opens rightward: Its basic equation: (y-k)^2=4p(x-h), (hk)=(x,y) coordinates of the vertex For given equation: vertex: (0,0) axis of symmetry: y=0 p=6 (distance from vertex to focus on the axis of symmetry) 4p=24 equation: y^2=4px=24x x=y^2/24 ans. C)
 logarithm/744933: For the functions ƒ(x) = log_6(x) and g(x)=log_1/4(x), for what values of x is g(x) > ƒ(x)?1 solutions Answer 453673 by lwsshak3(6505)   on 2013-05-05 03:06:56 (Show Source): You can put this solution on YOUR website!For the functions ƒ(x) = log_6(x) and g(x)=log_1/4(x), for what values of x is g(x) > ƒ(x)? .. .. At x=1, g(x)=f(x) At x<1, g(x)>f(x) Between (0 and 1), f becomes more negative as x becomes more negative; whereas, g becomes more positive as x becomes more negative.
 logarithm/744936: Write as a single logarithm: 2(log_2(x-1)^3+7log_2(x+4)^5)1 solutions Answer 453672 by lwsshak3(6505)   on 2013-05-05 02:06:27 (Show Source): You can put this solution on YOUR website!Write as a single logarithm: 2(log_2(x-1)^3+7log_2(x+4)^5)
 logarithm/744937: Identify the transformations of the function ƒ(x) = log_2(2(x + 6)) from the base graph g(x) = log_2(x).1 solutions Answer 453669 by lwsshak3(6505)   on 2013-05-05 01:50:52 (Show Source): You can put this solution on YOUR website!Identify the transformations of the function ƒ(x) = log_2(2(x + 6)) from the base graph g(x) = log_2(x). *** transformations: moves x-intercept 6 units left, from (0,1) to (0,-5) and stretches basic curve up vertically. Asymptote moves from y-axis to x=-6, 6 units to the left.
 Trigonometry-basics/744248: Find solutions in radians for 0 less then or equal to x less then 2pi: sin^2(3x)-3sin(3x)-4=01 solutions Answer 453569 by lwsshak3(6505)   on 2013-05-04 04:16:01 (Show Source): You can put this solution on YOUR website!Find solutions in radians for 0 less then or equal to x less then 2pi: sin^2(3x)-3sin(3x)-4=0 sin((3x)-4)(sin(3x)+1)=0 .. sin((3x)-4)=0 sin(3x)=4 (reject,(-1 ≤ sin(3x) ≤ 1)) .. sin(3x)+1=0 sin(3x)=-1 3x=3π/2 x=3π/6=π/2
 Trigonometry-basics/744764: Evaluate the given expression without using a calculator. (Assume any variables represent positive numbers) Sin( arccos 3/5 + arctan 1/2)1 solutions Answer 453537 by lwsshak3(6505)   on 2013-05-03 21:09:12 (Show Source): You can put this solution on YOUR website!Evaluate the given expression without using a calculator. (Assume any variables represent positive numbers) Sin( arccos 3/5 + arctan 1/2) *** let x=arccos 3/5 let y=arctan 1/2 .. .. .. Sin( arccos 3/5 + arctan 1/2)=sin(x+y) =sin(x) cos(y)+cos(x) sin(y) =4/5*2/√5+3/5*1/√5 =8/5√5+3/5√5=11/(5√5) =(11√5)/25 .. Check: (with calculator) cos(x)=3/5 x≈53.13º tan(y)=1/2 y≈26.57º x+y≈79.7º sin(x+y)=sin79.7º≈0.9839 exact answer=(11√5)/25≈0.9839
 Trigonometry-basics/744514: Prove the following identity: 2 sec^2 = 1/1- sin x + 1/1+ sin x,1 solutions Answer 453534 by lwsshak3(6505)   on 2013-05-03 21:02:26 (Show Source): You can put this solution on YOUR website!Prove the following identity: 2 sec^2 = 1/1- sin x + 1/1+ sin x, *** start with right side LCD:(1+sin(x))(1+sin(x))=1-sin^2(x) verified: right side=left side
 Trigonometry-basics/744730: If sec x = -sqrt10 with 180 degrees less than or equal to x less than or equal to 270 degrees, find the following and write your answer in an exact form. A. Sin(2x) B. Cos(x/2) I appreciate all the help. :)1 solutions Answer 453530 by lwsshak3(6505)   on 2013-05-03 20:43:37 (Show Source): You can put this solution on YOUR website!If sec x = -sqrt10 with 180 degrees less than or equal to x less than or equal to 270 degrees, find the following and write your answer in an exact form. A. Sin(2x) B. Cos(x/2) *** (180º ≤ x ≤ 270º) .. A. .. B. =
 Trigonometry-basics/744624: Given that sin A= -4/5 where A is in quadrant 3 and sin B= 12/13 where B is in quadrant 2. Find: A. cosA B. cosB C. sec(A-B) D. tan(2A) E. cos(B/2) Help!1 solutions Answer 453509 by lwsshak3(6505)   on 2013-05-03 16:54:18 (Show Source): You can put this solution on YOUR website!Given that sin A= -4/5 where A is in quadrant 3 and sin B= 12/13 where B is in quadrant 2. Find: A. cosA= .. B. cosB = .. C. sec(A-B)=1/cos(A-B) cos(A-B)=cosA cosB+SinA sinB=-3/5*-5/13+(-4/5)*12/13=15/65-48/65=-33/65 sec(A-B)=-65/33 .. D. tan(2A)=(2tanA)/(1-tan^2A) tanA=sinA/cosA=(-4/5)/(-3/5)=4/3 tan(2A)=(8/3)/(1-16/9)=(8/3)/(-7/9)=-72/21 .. E. cos(B/2)==√((8/13)/2))=√(8/26)
 Trigonometry-basics/744674: if tan theta equals 2 with theta in quadrant III find sec theta. 1 solutions Answer 453498 by lwsshak3(6505)   on 2013-05-03 15:38:53 (Show Source): You can put this solution on YOUR website!if tan theta equals 2 with theta in quadrant III find sec theta. Identity: sec^2(x)-tan^2(x)=1 sec^2(x)=4+1=5 sec(x)=-√5 (x in Q3 where sec<0)
 logarithm/744486: 1. Solve the following equations a) log(base 2) x+1 + log(base 2) x-1 = 3 b) log(base 2) x^2-1 =3 c) explain why the equations log(base 2) x+1 + log(base 2) x-1 = 3 and log(base 2) x^2-1 =3 are not equal. Can you please help me out? Thanks so much in advance:) Can you please show all the steps it would really help me understand:)1 solutions Answer 453356 by lwsshak3(6505)   on 2013-05-03 03:10:24 (Show Source): You can put this solution on YOUR website!1. Solve the following equations a) log(base 2) x+1 + log(base 2) x-1 = 3 (x+1)(x-1)=8 x^2-1=8 x^2=9 x=±3 .. b) log(base 2) x^2-1 =3 x^2-1=8 x^2=9 x=±3 .. c) explain why the equations log(base 2) x+1 + log(base 2) x-1 = 3 and log(base 2) x^2-1 =3 are not equal. The equations of a) and b) are equal as shown above.
 Quadratic-relations-and-conic-sections/743371: Graph the ellipse and identify the center,vertrices,foci and endpoints of 25x^2+9y^2=2251 solutions Answer 453355 by lwsshak3(6505)   on 2013-05-03 02:47:54 (Show Source): You can put this solution on YOUR website!Graph the ellipse and identify the center,vertrices,foci and endpoints of This ellipse has a vertical major axis. Its standard form of equation: , a>b, (h,k)=(x,y) coordinates of center center: (0,0) a^2=25 a=5 vertices: (0,0±a)=(0,0±5)=(0,-5) and (0,5) .. b^2=9 b=3 end points of minor axis: (0±b,0)=(0±3,0)=(-3,0) and (3,0) .. c^2=a^2-b^2=25-9=16 c=4 foci: (0,0±c)=(0,0±4)=(0,-4) and (0,4)
 Quadratic-relations-and-conic-sections/743930: consider the ellipse 9x^2+49y^2=9 Its vertices are (0,+or-A) Its Foci are (0,+or-B) Its eccentricity is? The length of its major and minor axis are?1 solutions Answer 453354 by lwsshak3(6505)   on 2013-05-03 02:27:13 (Show Source): You can put this solution on YOUR website!consider the ellipse 9x^2+49y^2=9 Its vertices are (0,+or-A) Its Foci are (0,+or-B) Its eccentricity is? The length of its major and minor axis are? *** I will use standard notation which you can convert to your own notation: Given ellipse has a horizontal major axis. Its standard form of equation: ,(a>b), (h,k)=(x,y) coordinates of center. a^2=1 a=1 length of major axis=2a=2 b^2=9/49 b=3/7 length of minor axis=2b=6/7≈0.8571 c^2=a^2-b^2=1-9/49=49/49-9/49=40/49 c=√(40/49)≈0.9035 eccentricity=c/a=0.9035/1≈0.9035
 Quadratic-relations-and-conic-sections/744434: focus (2,5) and directrix y=3 write standard form of the equation of the parabola with the given criteria.1 solutions Answer 453351 by lwsshak3(6505)   on 2013-05-03 01:28:43 (Show Source): You can put this solution on YOUR website!focus (2,5) and directrix y=3 write standard form of the equation of the parabola with the given criteria. *** Given parabola opens upward. (directrix below focus) Its basic form of equation: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex. y-coordinate of vertex=midpoint between focus and directrix on the axis of symmetry=(5+3)/2=4 x-coordinate of vertex=2 vertex: (2,4) axis of symmetry: x=2 p=1 (distance from vertex to focus or directrix on the axis of symmetry) 4p=4 Equation of given parabola: (x-2)^2=4(y-4)
 Quadratic-relations-and-conic-sections/743933: Identify the vertex, focus, and directrix of the graph of y=1/12(x+5)^2-31 solutions Answer 453312 by lwsshak3(6505)   on 2013-05-02 19:56:29 (Show Source): You can put this solution on YOUR website!Identify the vertex, focus, and directrix of the graph of y=1/12(x+5)^2-3 rewrite equation to basic form: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex 1/12(x+5)^2=y+3 (x+5)^2=12(y+3) vertex:(-5,-3) axis of symmetry: x=-5 4p=12 p=3 focus: (-5,0) directrix:(-5,-6)
 Quadratic-relations-and-conic-sections/744040: Classify the conic section and write its ezuation in standard form. Then graph the equation. 4x^2+9y^2+40x+72y+208=0 It is an ellipse. I Completed the square and all that but I can't factor it after that..1 solutions Answer 453306 by lwsshak3(6505)   on 2013-05-02 19:30:24 (Show Source): You can put this solution on YOUR website!Classify the conic section and write its equation in standard form. Then graph the equation. 4x^2+9y^2+40x+72y+208=0 complete the square: 4x^2+40x+9y^2+72y+208=0 4(x^2+10x+25)+9(y^2+8y+16)=-208+100+144 4(x+5)^2+9(y+4)^2=36 Given conic section is an ellipse with horizontal major axis with center at (-5,-4) y=±(4-4(x+5)^2/9)^.5-4 see graph below as a visual check:
 Quadratic-relations-and-conic-sections/744051: A satellite dish has a parabolic cross section and is 6 ft deep. The focus is 4 ft from the vertex. Find the width of the satellite dish at the opening. Round your answer to the nearest foot.1 solutions Answer 453301 by lwsshak3(6505)   on 2013-05-02 19:06:30 (Show Source): You can put this solution on YOUR website!A satellite dish has a parabolic cross section and is 6 ft deep. The focus is 4 ft from the vertex. Find the width of the satellite dish at the opening. Round your answer to the nearest foot. *** place vertex at (0,0) basic equation: x^2=4py p=4 (distance from vertex to focus) 4p=16 equation: x^2=16y y=6 at ends of parabolic cross section x^2=16*6=96 x=±√96 width of the satellite dish at the opening=2*√96≈20 ft
 Quadratic-relations-and-conic-sections/744406: write an equation for the conic section parabola with vertex at(0,0) and directrix x = 51 solutions Answer 453298 by lwsshak3(6505)   on 2013-05-02 18:49:34 (Show Source): You can put this solution on YOUR website!write an equation for the conic section parabola with vertex at(0,0) and directrix x = 5 *** This is a parabola that opens rightward. Its basic form of equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex For given parabola: given vertex: (0,0) axis of symmetry: y=0 or x-axis p=5 (distance from vertex to directrix on the axis of symmetry) 4p=20 Equation: y^2=20x
 Quadratic-relations-and-conic-sections/744099: Write the standard for equation of an ellipse. 1. x^2+4y^2-6x-64y+165=0 1 solutions Answer 453292 by lwsshak3(6505)   on 2013-05-02 18:14:14 (Show Source): You can put this solution on YOUR website!Write the standard for equation of an ellipse. 1. x^2+4y^2-6x-64y+165=0 complete the square: x^2-6x+4y^2-64y+165=0 (x^2-6x+9)+4(y^2-16y+64)=-165+9+256 (x-3)^2+4(y-8)^2=100
 Quadratic-relations-and-conic-sections/744390: parabola with vertex at (0,0) and directrix at x=-31 solutions Answer 453290 by lwsshak3(6505)   on 2013-05-02 18:04:11 (Show Source): You can put this solution on YOUR website!parabola with vertex at (0,0) and directrix at x=-3 Given parabola opens rightward. Its basic form of equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex given vertex: (0,0) axis of symmetry: y=0 p=3 (distance from vertex to directrix on the axis of symmetry 4p=12 Equation of given parabola: y^2=12x
 Quadratic-relations-and-conic-sections/744237: How do i find the vertices and foci of; (x^2/64)-(9y^2/4)=11 solutions Answer 453265 by lwsshak3(6505)   on 2013-05-02 15:30:26 (Show Source): You can put this solution on YOUR website!How do i find the vertices and foci of; (x^2/64)-(9y^2/4)=1 Given ellipse has a horizontal major axis. Its standard form of equation: ,a>b, (h,k)=(x,y) coordinates of center (x^2/64)-(9y^2/4)=1 (x^2/64)-y^2/(4/9)=1 center: (0,0) a^2=64 a=8 vertices: (0±a,0)=(0±8,0)=(-8,0) and (8,0) b^2=4/9 b=2/3 c^2=a^2-b^2=64-4/9=(576/9)-(4/9)=572/9 c=√(572/9)≈7.97 foci: (0±c,0)=(0±7.97,0)=(-7.97,0) and (7.97,0)
 Quadratic-relations-and-conic-sections/744231: I need to write an equation for an ellipse with the endpoints of the major axis (7,3) (7,9) , endpoints of the minor axis at (5,6) , (9,6).1 solutions Answer 453256 by lwsshak3(6505)   on 2013-05-02 15:02:06 (Show Source): You can put this solution on YOUR website!I need to write an equation for an ellipse with the endpoints of the major axis (7,3) (7,9) , endpoints of the minor axis at (5,6) , (9,6). *** Given ellipse has a vertical major axis. (y-coordinates of major axis change but x-coordinates do not) Its standard form of equation: , a>b, (h,k)=(x,y) coordinates of center y-coordinate of center=6 (midpoint of vertical major axis, (9+3)/2=12/2=6 x-coordinate of center=7 center:(7,6) a=6 (distance from center to end points of major axis) a^2=36 b=2 (distance from center to end points of minor axis) b^2=4 Equation for given ellipse:
 Quadratic-relations-and-conic-sections/744294: Find the equation for the hyperbola with foci (4,6) and (-8,6) and a transverse axis length of 81 solutions Answer 453247 by lwsshak3(6505)   on 2013-05-02 14:23:36 (Show Source): You can put this solution on YOUR website!Find the equation for the hyperbola with foci (4,6) and (-8,6) and a transverse axis length of 8 *** Given hyperbola has a horizontal transverse axis. Its standard form of equation: , (h,k)=(x,y) coordinates of center. For given hyperbola: center: (-2, 6) length of transverse axis=8=2a a=4 a^2=16 c=6 (distance from center to foci on the transverse axis) c^2=36 c^2=a^2+b^2 b^2=c^2-a^2=36-16=20 Equation of given hyperbola: