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# Recent problems solved by 'lwsshak3'

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 Trigonometry-basics/691970: Evaluate the expression. cot^2(90)-sec(180) both in degrees. The answer is 1 but I don't know how to solve this. Please help.1 solutions Answer 426933 by lwsshak3(6493)   on 2012-12-10 01:20:48 (Show Source): You can put this solution on YOUR website!Evaluate the expression. cot^2(90)-sec(180) both in degrees. The answer is 1 but I don't know how to solve this. Please help ** cot(90)=cos(90)/sin(90)=0/1=0 cot^2(90)=0 .. -sec(180)=-1/cos(180)=-1/-1=1 .. cot^2(90)-sec(180)=0+1=1
 Linear_Equations_And_Systems_Word_Problems/691655: clara is one third her sister's age 4 years later clara will be 3less than half her her sister 's age what was clara s age 3 years ago 1 solutions Answer 426796 by lwsshak3(6493)   on 2012-12-09 02:50:39 (Show Source): You can put this solution on YOUR website!clara is one third her sister's age 4 years later clara will be 3less than half her her sister 's age what was clara s age 3 years ago ** let x=clara's age 3 yrs ago x+3=clara's present age 3(x+3)=sister's present age 4 years later: (x+7=clara's age (3(x+3)+4)=sister's age 3less than half of sister's age=(3(x+3)+4)/2-3 x+7=(3x+9+4)/2)-3 x+10=(3x+13)/2) 2x+20=3x+13 x=7 clara's age 3 yrs ago=7
 Exponential-and-logarithmic-functions/691473: Juan is going to deposit \$1,000 in an account that will earn him compund interest. The account pays 8% interest, compounded monthly. How much money will he have after 10 years?1 solutions Answer 426795 by lwsshak3(6493)   on 2012-12-09 02:12:32 (Show Source): You can put this solution on YOUR website!Juan is going to deposit \$1,000 in an account that will earn him compund interest. The account pays 8% interest, compounded monthly. How much money will he have after 10 years. ** compound interest formula: A=P(1+r)^n, P=initial investment, r=interest rate per period, A=amt after n periods. For given problem: P=1000 r=.08/12 n=10*12=120 .. A=1000(1+.08/12)^120=2219.64 How much money will Juan have after 10 years. \$2219.64
 logarithm/691550: Solve the equation. (x^3)(9^x) -9^x =01 solutions Answer 426794 by lwsshak3(6493)   on 2012-12-09 01:54:47 (Show Source): You can put this solution on YOUR website!Solve the equation. (x^3)(9^x) -9^x =0 9^x(x^3-1)=0 9^x≠0 x^3-1=0 x=1
 Quadratic-relations-and-conic-sections/691313: find vertex, focus, and equation of directrix of y^2 = 36x1 solutions Answer 426791 by lwsshak3(6493)   on 2012-12-09 01:45:25 (Show Source): You can put this solution on YOUR website!find vertex, focus, and equation of directrix of y^2 = 36x This is an equation of a parabola that opens rightwards. Its standard form of equation: (y-k)=4p(x-h), (h,k)=(x,y) coordinates of the vertex. For given equation: y^2=36x vertex:(0,0) axis of symmetry: y=0 or x-axis 4p=36 p=9 focus: (9,0) (p-distance to the right of the vertex on the axis of symmetry) directrix: x=9 (p-distance to the left of the vertex on the axis of symmetry)
 Quadratic-relations-and-conic-sections/691434: Find the standard form of the equation of the hyperbola with the given charateristics and center at the origin Foci:(+-10,0) Asymptotes:y=+-3x1 solutions Answer 426788 by lwsshak3(6493)   on 2012-12-09 01:33:12 (Show Source): You can put this solution on YOUR website!Find the standard form of the equation of the hyperbola with the given charateristics and center at the origin Foci:(+-10,0) Asymptotes:y=+-3x ** This is a hyperbola with horizontal transverse axis. Its standard form of equation: (x-h)^2/a^2-(y-k)^2/b^2=1 For given hyperbola: center: (0,0) slope of asymptotes=±3=b/a (for hyperbolas with horizontal transverse axis) b=±3a c=10 (distance from center to foci) c^2=a^2+b^2=a^2+9a^2=10a^2 100=10a^2 a^2=10 a=√10 b^2=c^2-a^2=100-10=90 equation: x^2/10-y^2/90=1
 Quadratic-relations-and-conic-sections/691496: graph the ellipse using the proper algebra x^2+(y^2/12)=11 solutions Answer 426787 by lwsshak3(6493)   on 2012-12-09 01:01:03 (Show Source): You can put this solution on YOUR website!graph the ellipse using the proper algebra x^2+(y^2/12)=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: x^2+(y^2/12)=1 center: (0,0) a^2=12 a=√12≈3.46 vertices: (0,0±a)=(0,0±√12)=(0,0±3.46)=(0,-3.46) and (0,3.46) (y-intercepts) b^2=1 b=1 x-intercepts:(±1,0)=(-1,0) and (1,0) see graph below as a visual check on the foregoing algebra: y=±(12-12x^2)^.5
 Trigonometry-basics/691292: Find the amplitude, period, and phase shift of the function, and graph one completed period: y=1+COS (3X+pie/2)1 solutions Answer 426750 by lwsshak3(6493)   on 2012-12-08 20:16:52 (Show Source): You can put this solution on YOUR website!Find the amplitude, period, and phase shift of the function, and graph one completed period: y=1+COS (3X+pie/2) ** Equation for the cos function: y=A)cos(Bx-C), A=amplitude, period=2π/B, phase shift=C/B For given equation: 1+cos(3x+π/2) Amplitude=1 B=3 period: 2π/B=2π/3 1/4 period=2π/12=π/6 phase shift=C/B=(π/2)/3=π/6 (to the left) .. Graphing: I don't have the means to graph the function for you, but I can show you how to develop the coordinates with which you can draw the graph: Start with the basic function: y=cos x with a period of 2π/3 coordinates: (0,1), (π/6,0), (π/3,-1), (π/2,0), (2π/3,1) shift to left (π/6): (-π/6,1), (0,0), (π/6,-1), (π/3,0), (π/2,1) bump up 1 unit: (-π/6,2), (0,1), (π/6,0), (π/3,1), (π/2,2) (final configuration
 Trigonometry-basics/691480: Find sin x/2, cos x/2, and tan x/2 from the given information. sec x = 10/9 ,270° < x < 360° sin x/2= cos x/2= tan x/2 = 1 solutions Answer 426739 by lwsshak3(6493)   on 2012-12-08 19:23:13 (Show Source): You can put this solution on YOUR website!Find sin x/2, cos x/2, and tan x/2 from the given information. sec x = 10/9 ,270° < x < 360° sin x/2= cos x/2= tan x/2= ** sec x=10/9=hypotenuse/adjacent side hypotenuse=10 adjacent side=9 opposite side=√(10^2-9^2)=√(100-81)=√19 sin x=-√19/10 (in quadrant IV where sin<0) cos x=9/10 .. Identities: sin x/2=±√[(1-cos x)/2] choose positive root because x/2 is in quadrant II where sin>0 sin x/2=√[(1-cos x)/2]=√[(1-9/10/2]=√.1/2=√.05 .. cos x/2=±√[(1+cos x)/2] choose negative root because x/2 is in quadrant II where cos<0 cos x/2=-√[(1+cos x)/2]=-√[(1+9/10/2]=-√1.9/2=-√.95 .. tan x/2=sin x/(1+cos x)=-(√19/10)/(1+9/10)=-(√19/10)/1.9 .. How to check answers with calculator: sec x=10/9 cos x=9/10 cos^-1=(9/10)≈25.84º (reference angle in specified quadrant IV) standard position of angle=360-25.84=334.16 x/2=334.16/2=167.08 reference angle=180-167.08=12.92º .. sin x/2=sin 12.92≈0.2236..(in quadrant II where sin>0) √.05=0.2236.. .. you can check cos x/2 and tan x/2 in the same way
 logarithm/690518: Can someone assist me to express log[2](A)+log[2](B)-3log[2](C)as a single logarithm. Thank you very much!1 solutions Answer 426635 by lwsshak3(6493)   on 2012-12-08 02:50:40 (Show Source): You can put this solution on YOUR website!Can someone assist me to express log[2](A)+log[2](B)-3log[2](C)as a single logarithm. log2[AB/C^3]
 logarithm/691065: Suppose that ln5 =S and ln11 =T. Use the properties of logarithm in terms of S and T1 solutions Answer 426634 by lwsshak3(6493)   on 2012-12-08 02:46:45 (Show Source): You can put this solution on YOUR website!Suppose that ln5 =S and ln11 =T. Use the properties of logarithm in terms of S and T log5+log11=log[5*11]=S+T log(55)=S+T
 logarithm/691251: Find the domain of the function. (The 8 is supposed to be a subscript, for base 8) f(x)= log8 (x+3)1 solutions Answer 426633 by lwsshak3(6493)   on 2012-12-08 02:37:02 (Show Source): You can put this solution on YOUR website!Find the domain of the function. f(x)= log8 (x+3) domain: (-3,∞)
 logarithm/691298: g(x) = In (x-6) find the domain of the function.1 solutions Answer 426632 by lwsshak3(6493)   on 2012-12-08 02:35:14 (Show Source): You can put this solution on YOUR website!g(x) = In (x-6) find the domain of the function. domain: (6,∞)
 logarithm/691255: Use the Laws of Logarithms to combine the expression. (They are all base ten, because they have no base listed, but you probably know that already) 5log x - (1/3)log (x^2+1) + 4log (x-1)1 solutions Answer 426631 by lwsshak3(6493)   on 2012-12-08 02:32:12 (Show Source): You can put this solution on YOUR website!5log x - (1/3)log (x^2+1) + 4log (x-1) log[(5(x-1)^4)/(x^2+1)^1/3]
 Trigonometry-basics/690972: 1.) Solve each equation. a.) log7(2x)^2+log7 (3)=2 b.) log4 (8)+log4 (x^2-8)=8 c.) log12 (-3a-8)+log12 (a+4) d.) ln-3k=ln(-2k-4) 1 solutions Answer 426576 by lwsshak3(6493)   on 2012-12-07 16:56:09 (Show Source): You can put this solution on YOUR website!1.) Solve each equation. a.) log7(2x)^2+log7 (3)=2 log7[(2x)^2*3]=2 convert to exponential form 7^2=4x^2*3=12x^2 12x^2=49 x^2=49/12 x=±√(49/12)≈±2.02 .. b.) log4 (8)+log4 (x^2-8)=8 log4[8*(x^2-8)]=8 4^8=8*(x^2-8)=8x^2-64 8x^2-64=65536 8x^2=65536+64=65600 x^2=8200 x≈±90.55 .. c.) log12 (-3a-8)+log12 (a+4) log12[(-3a-8)(a+4)]=log12[(-3a^2-20a-32)]=log12[-(3x+8)(x+4)] .. d.) ln-3k)=ln(-2k-4)=? ln argument mssing
 Trigonometry-basics/691184: The following problems give an expression for one of the three functions sin , cos, or tan ,with (theta) in the fi rst quadrant. Find expressions for the other two functions. Your answers will be algebraic expressions of x. (a) sin (theta) = 3/x (b) x = 4 cos (theta) (c) tan (theta) = x1 solutions Answer 426554 by lwsshak3(6493)   on 2012-12-07 15:47:06 (Show Source): You can put this solution on YOUR website!The following problems give an expression for one of the three functions sin , cos, or tan ,with (theta) in the fi rst quadrant. Find expressions for the other two functions. Your answers will be algebraic expressions of x. (a) sin (theta) = 3/x (b) x = 4 cos (theta) (c) tan (theta) = x ** Use the following notation: O=opposite side, A=adjacent side, H=hypotenuse .. (a) sin (theta) = 3/x=O/H O=3 H=x A=√(H^2-O^2)=√(x^2-3^2)=√(x^2-9) cos (theta)=A/H=√(x^2-9)/x tan (theta)=O/A=3/√(x^2-9) .. b) x = 4 cos (theta) cos (theta)=x/4=A/H A=x H=4 O=√(H^2-A^2)=√(4^2-x^2)=√(16-x^2) sin(theta)=O/H=√(16-x^2)/4 tan (theta)=O/A=√(16-x^2)/x (c) tan (theta) = x=O/A O=x A=1 H=√(O^2+A^2)=√(x^2+1^2)=√(x^2+1) cos (theta)=A/H=1/√(x^2+1) sin(theta)=O/H=x/√(x^2+1)
 Exponential-and-logarithmic-functions/690402: solve: log base a of x + log base a of (x-2)= log base a of (x+4)1 solutions Answer 426420 by lwsshak3(6493)   on 2012-12-06 20:41:32 (Show Source): You can put this solution on YOUR website!solve: log base a of x + log base a of (x-2)= log base a of (x+4) loga[x(x-2)]=loga(x+4) x(x-2)=(x+4) x^2-2x=x+4 x^2-3x-4=0 (x-4)(x+1)=0 x=4 or x=-1 (reject, (x-2)>0)
 Exponential-and-logarithmic-functions/690548: Solve log 2x + log 12 = 3. Round to the nearest hundredth if necessary. 1 solutions Answer 426419 by lwsshak3(6493)   on 2012-12-06 20:33:47 (Show Source): You can put this solution on YOUR website!Solve log 2x + log 12 = 3. Round to the nearest hundredth if necessary. log 2x + log 12 = 3 log(2x*12)=log(10^3) 2x*12=10^3=1000 2x=1000/12 x=1000/24 x≈41.67
 Exponential-and-logarithmic-functions/690550: Write the expression as a single natural logarithm. 2 ln x – 5 ln c 1 solutions Answer 426418 by lwsshak3(6493)   on 2012-12-06 20:27:51 (Show Source): You can put this solution on YOUR website!Write the expression as a single natural logarithm. 2 ln x – 5 ln c ln[x^2/c^5]
 Exponential-and-logarithmic-functions/690554: Solve the exponential equation. 1/16= 64^(4x–3) 1 solutions Answer 426417 by lwsshak3(6493)   on 2012-12-06 20:24:58 (Show Source): You can put this solution on YOUR website!Solve the exponential equation. 1/16= 64^(4x–3) 1/4^2=4^3^(4x-3) 4^-2)=4^(12x-9) 12x-9=-2 12x=7 x=7/12
 Exponential-and-logarithmic-functions/690557: Solve the logarithmic equation. Round to the nearest ten-thousandth if necessary. 3 log 2x = 4 1 solutions Answer 426415 by lwsshak3(6493)   on 2012-12-06 20:19:19 (Show Source): You can put this solution on YOUR website!Solve the logarithmic equation. Round to the nearest ten-thousandth if necessary. 3 log(2x) = 4 log(2x)=4/3 convert to exponential form 10^(4/3)=2x 2x≈21.5443 x≈21.5443/2 x≈10.7722
 Exponential-and-logarithmic-functions/690558: Solve ln(5x + 7) = 8. Round to the nearest thousandth.1 solutions Answer 426412 by lwsshak3(6493)   on 2012-12-06 20:12:15 (Show Source): You can put this solution on YOUR website!Solve ln(5x + 7) = 8. Round to the nearest thousandth. ln(5x + 7) = 8 convert to exponential form: base(e) raised to log of the number(8)=number(5x+7) e^8=5x+7 5x=e^8-7 x=(e^8-7)/5 x≈594.792
 Exponential-and-logarithmic-functions/690560: Use natural logarithms to solve the equation. Round to the nearest thousandth. 5e^(2x + 11)=30 1 solutions Answer 426408 by lwsshak3(6493)   on 2012-12-06 20:02:31 (Show Source): You can put this solution on YOUR website!Use natural logarithms to solve the equation. Round to the nearest thousandth. 5e^(2x + 11)=30 divide by 5 e^(2x + 11)=6 (2x+11)lne=ln6 lne=1 2x+11=ln6 2x=ln6-11 x=(ln6-11)/2 x≈-4.604
 Exponential-and-logarithmic-functions/690685: solve each equation for x by first rewriting both sides as powers of the same base. 32^x=641 solutions Answer 426402 by lwsshak3(6493)   on 2012-12-06 19:53:12 (Show Source): You can put this solution on YOUR website!solve each equation for x by first rewriting both sides as powers of the same base. 32^x=64 (2^5)^x=2^6 2^5x=2^6 5x=6 x=6/5
 Exponential-and-logarithmic-functions/690845: Whats (5x)^(-1/2)=10? Steps please1 solutions Answer 426401 by lwsshak3(6493)   on 2012-12-06 19:50:13 (Show Source): You can put this solution on YOUR website!Whats (5x)^(-1/2)=10 (5x)^(-1/2)=10 1/(5x)^(1/2)=10 √(5x)=1/10 square both sides 5x=1/100 x=1/500
 logarithm/690406: Log(base 9)(x-6)+log (base 9)(x-6)=11 solutions Answer 426298 by lwsshak3(6493)   on 2012-12-06 03:24:54 (Show Source): You can put this solution on YOUR website!Log(base 9)(x-6)+log (base 9)(x-6)=1 log9[(x-6)(x-6)]=log9(9) x^2-12x+36=9 x^2-12x+25=0 solve for x by completing the square x^2-12x+25=0 (x^2-12x+36)=-25+36 (x-6)^2=11 x-6=±√11 x=6±√11 x=6±3.317 x≈2.683(reject, (x-6)>0) or x≈9.317