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 Exponents/616506: please help! im confused!! (1/3)^(5x+2)=(1/9)^3 solving for x i only got (1/3)^(5x+2)=1/729 i still cant figure out what x is1 solutions Answer 387709 by lwsshak3(6456)   on 2012-05-31 03:57:38 (Show Source): You can put this solution on YOUR website!(1/3)^(5x+2)=(1/9)^3 solving for x 3^-(5x+2)=1/9^3 3^-(5x+2=1/(3^2)^3=1/3^6=3^-6 -(5x+2=-6 5x+2=6 5x=4 x=4/5
 Equations/616505: A rectangle is 5 cm longer than it is wide. If the length and breadth are increased by 2 cm each, the area increases by 50 cm squared. Find the dimensions of the original rectangle. I've got a diagram of the two rectangles with the breadth as 'x' and the length as 'x + 5' in the original and then on the second one's breadth as 'x + 2' and the length as 'x + 7' but I just don't know how to put it into an equation because it is under simultaneous equations and I'm completely confused because usually you have what it equals to work it out.1 solutions Answer 387708 by lwsshak3(6456)   on 2012-05-31 03:36:40 (Show Source): You can put this solution on YOUR website!A rectangle is 5 cm longer than it is wide. If the length and breadth are increased by 2 cm each, the area increases by 50 cm squared. Find the dimensions of the original rectangle. ** let x=width of original rectangle x+5=length of original rectangle area of original rectangle=x(x+5)=x^2+5x .. x+2=width of larger rectangle x+5+2=length of larger rectangle area of larger rectangle=(x+2)(x+7)=x^2+9x+14 .. Area of larger rectangle-area of original rectangle=50 x^2+9x+14-x^2-5x=50 4x=36 x=9 x+5=14 .. width of original rectangle=9 cm length of original rectangle=14 cm
 Miscellaneous_Word_Problems/616476: a farmer decides to enclose a rectangular garden using the side of a barn as one side of the rectangle. What is the maximum area that the farmer can enclose with 60 foot of fence? What should the dimensions of the garden be to get this area?1 solutions Answer 387707 by lwsshak3(6456)   on 2012-05-31 03:20:53 (Show Source): You can put this solution on YOUR website!a farmer decides to enclose a rectangular garden using the side of a barn as one side of the rectangle. What is the maximum area that the farmer can enclose with 60 foot of fence? What should the dimensions of the garden be to get this area? ** let x=length of each of the two sides perpendicular to the side of the barn 60-2x=length of third side parallel to side of the barn Area=x(60-2x) =60x-2x^2 =-2x^2+60x complete the square =-2(x^2-30x+225)+450 =-2(x-15)^2+450 maximum area=450 sqft dimensions of garden: 15 ft by 30 ft
 Quadratic_Equations/616485: I am trying to find the solutions to: (x^2-3)^2-43(x^2-3)+42=0 I can not figure out how to turn it into a quadratic and solve it.1 solutions Answer 387706 by lwsshak3(6456)   on 2012-05-31 02:55:10 (Show Source): You can put this solution on YOUR website!(x^2-3)^2-43(x^2-3)+42=0 let u=(x^2-3) u^2-43u+42=0 (u-42)(u-1)=0 u=1=x^2-3 x^2=4 x=±√4=±2 or x=42=x^2-3 x^2=45 x=±√45 solution: x=-√45, -2, 2, √45
 logarithm/616468: solve and simplify. 3log of 27(x-1)=2 1 solutions Answer 387705 by lwsshak3(6456)   on 2012-05-31 02:42:35 (Show Source): You can put this solution on YOUR website!solve and simplify. 3log of 27(x-1)=2 log27(x-1)=2/3 convert to exponential form 10^(2/3)=27(x-1)=27x-27 27x=10^(2/3)+27≈31.64 x≈31.64/27≈1.17
 Exponential-and-logarithmic-functions/616501: Solve for x: log (x - 6) base of 2 = 3 - log 4 base of 21 solutions Answer 387704 by lwsshak3(6456)   on 2012-05-31 02:32:26 (Show Source): You can put this solution on YOUR website!Solve for x: log (x - 6) base of 2 = 3 - log 4 base of 2 log2(x-6)=3-log2(4) log2(x-6)+log2(4)=3 log2[(x-6)*4]=3 convert to exponential form 2^3=4(x-6)=8 4x-24=8 4x=32 x=8
 Quadratic-relations-and-conic-sections/616287: Write the equation of a hyperbola with vertices at (-3, 0) and (3, 0) and co-vertices (0, 5) and (0, -5).1 solutions Answer 387654 by lwsshak3(6456)   on 2012-05-30 19:37:26 (Show Source): You can put this solution on YOUR website!Write the equation of a hyperbola with vertices at (-3, 0) and (3, 0) and co-vertices (0, 5) and (0, -5). Given hyperbola has a horizontal transverse axis. Its standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center For given hyperbola: center:(0,0) length of horizontal transverse axis=6=2a a=3 a^2=9 .. length of co-vertices or conjugate axis=10=2b b=5 b^2=25 .. Equation: x^2/9-y^2/25=1
 Quadratic-relations-and-conic-sections/616343: What is the radius of the circle given by the equation below? x2 - 14x + y2 + 12y = -691 solutions Answer 387652 by lwsshak3(6456)   on 2012-05-30 19:21:43 (Show Source): You can put this solution on YOUR website!What is the radius of the circle given by the equation below? Standard form of equation for a circle: (x-h)^2+(y-k)^2=r^2, (h,k)=(x,y) coordinates of center, r=radius x2 - 14x + y2 + 12y = -69 complete the square. (x^2-14x+49)+(y^2+12y+36) = -69+49+36 (x-7)^2+(y+6)^2=16 r^2=16 radius=√16=4
 Numbers_Word_Problems/616354: The sum of two numbers is 45. The larger number, y, is 6 less than twice the smaller number, x. What is the smaller number x? Also what is the equation?1 solutions Answer 387649 by lwsshak3(6456)   on 2012-05-30 19:08:13 (Show Source): You can put this solution on YOUR website!The sum of two numbers is 45. The larger number, y, is 6 less than twice the smaller number, x. What is the smaller number x? Also what is the equation? ** let x=smaller number y=2x-6=larger number x+y=45 x+2x-6=45 (equation) 3x=39 x=13=smaller number
 Circles/616356: What is the center and the radius of a circle whose equation is x^2+y^2 -4x +14y +28=0 ?1 solutions Answer 387648 by lwsshak3(6456)   on 2012-05-30 19:01:28 (Show Source): You can put this solution on YOUR website!What is the center and the radius of a circle whose equation is x^2+y^2 -4x +14y +28=0 ? complete the square x^2 -4x+y^2+14y +28=0 (x^2-4x+4)+(y^2+14y+49)=-28+4+49 (x-2)^2+(y+7)^2=25 Standard form of equation for a circle: (x-h)^2+(y-k)^2=r^2, (h,k)=(x,y) coordinates of center, r=radius For given equation: center:(2,-7) r^2=251 r=√25=5
 Rational-functions/616288: I need help with the following question: Find all rational zeros of the polynomial. 1 solutions Answer 387611 by lwsshak3(6456)   on 2012-05-30 16:29:46 (Show Source): You can put this solution on YOUR website!Find all rational zeros: y=2x^3+2x^2-50x-50 y=2x^2(x+1)-50(x+1) (x+1)(2x^2-50)=0 x+1=0 x=-1 or 2x^2-50=0 x^2-25=0 x^2=25 x=±√25=±5 zeros: -5, -1, and 5
 Trigonometry-basics/616295: Solve 2tan theta=2cot theta for all values of theta if theta is measured in degrees1 solutions Answer 387607 by lwsshak3(6456)   on 2012-05-30 16:19:07 (Show Source): You can put this solution on YOUR website!Solve 2tan theta=2cot theta for all values of theta if theta is measured in degrees 2tanx=2cotx tanx=cotx tan45=cot45=1 x=45º+360º*n and 225º+360º*n, n=integer
 Trigonometry-basics/616291: Simplify 7sin theta sec theta1 solutions Answer 387602 by lwsshak3(6456)   on 2012-05-30 16:08:33 (Show Source): You can put this solution on YOUR website!Simplify 7sin theta sec theta =7sinxsecx =7sinx*1/cosx =7sinx/cosx =7tanx
 Linear-equations/616270: points (1,4) and (4,2) lie on line k. what is the slope of the line that is perpendicular to k?1 solutions Answer 387599 by lwsshak3(6456)   on 2012-05-30 16:04:38 (Show Source): You can put this solution on YOUR website!points (1,4) and (4,2) lie on line k. what is the slope of the line that is perpendicular to k? slope of straight line=∆y/∆x slope of line k=(2-4)/(4-1)=-2/3 slope of line perpendicular to line k =negative reciprocal of slope of line k =3/2
 Graphs/616282: What is the slope of the line through the points (2,-6)(9,13) 1 solutions Answer 387593 by lwsshak3(6456)   on 2012-05-30 15:55:48 (Show Source): You can put this solution on YOUR website!What is the slope of the line through the points (2,-6)(9,13) slope of a straight line=∆y/∆x =change in y/change in x =(13-(-6))/(9-2) =19/7
 Quadratic-relations-and-conic-sections/616273: the ellipse is center at the origin. the legnth of its horizontal axis is 8,and the length of its vertical axis is 6.what is the equation for the ellipse1 solutions Answer 387590 by lwsshak3(6456)   on 2012-05-30 15:50:15 (Show Source): You can put this solution on YOUR website!the ellipse is center at the origin. the legnth of its horizontal axis is 8,and the length of its vertical axis is 6.what is the equation for the ellipse ** standard form of equation for an ellipse with horizontal major axis: (x-h)^2/a^2+(y-k)^2/b^2=1, (a>b), (h,k)=(x,y) coordinates of center For given problem: center: (0,0) length of horizontal major axis=8=2a a=4 a^2=16 .. length of vertical or minor axis=6=2b b=3 b^2=9 .. equation: x^2/16+y^2/9=1
 Exponential-and-logarithmic-functions/616251: Solve the following exponential equation. Exact answers only. π^(1-8x)=e^(3x) I know that to start I need to lake the natural log of both sides but I'm not even sure of how to do that.1 solutions Answer 387588 by lwsshak3(6456)   on 2012-05-30 15:38:52 (Show Source): You can put this solution on YOUR website!Solve the following exponential equation. Exact answers only. π^(1-8x)=e^(3x) take log of both sides (1-8x)ln(π)=3xlne lne=1(log of base=1) ln(π)-8ln(π)x=3x 3x+8ln(π)x=ln(π) x(3+8ln(π))=ln(π) x=ln(π)/(3+8ln(π))
 logarithm/616257: How do I condense the following expression: 4(log12-log3)-5(log16-log4)1 solutions Answer 387582 by lwsshak3(6456)   on 2012-05-30 15:16:35 (Show Source): You can put this solution on YOUR website!How do I condense the following expression: 4(log12-log3)-5(log16-log4) =4(log12/3)-5(log16/4) =4(log4)-5(log4) =-log4
 logarithm/616145: Simplify the expression below and write it as a single logarithim 3log(x+4)-2log(x-7)+5log(x-2)-log(x^2)1 solutions Answer 387580 by lwsshak3(6456)   on 2012-05-30 15:10:52 (Show Source): You can put this solution on YOUR website!Simplify the expression below and write it as a single logarithim 3log(x+4)-2log(x-7)+5log(x-2)-log(x^2) 3log(x+4)+5log(x-2)-[2log(x-7)+log(x^2)] log[((x+4)^3*(x-2)^5)/((x-7)^2*x^2)]
 Quadratic-relations-and-conic-sections/616210: For the conic section described by the equation 9x^2+4y^2-18x+16y-11=0 What is the center? What is the length of the semimajor axes? What is the focal length?1 solutions Answer 387578 by lwsshak3(6456)   on 2012-05-30 14:59:07 (Show Source): You can put this solution on YOUR website!For the conic section described by the equation 9x^2+4y^2-18x+16y-11=0 What is the center? What is the length of the semimajor axes? What is the focal length? ** 9x^2+4y^2-18x+16y-11=0 complete the square 9x^2-18x+4y^2+16y-11=0 9(x^2-2x+1)+4(y^2+4y+4)=11+9+16 9(x-1)^2+4(y+2)^2=36 (x-1)^2/4+(y+2)^2/9=1 This is an equation for an ellipse with vertical major axis. Its standard form: ((x-h)^2/b^2+(y-k)^2/a^2=1, (a>b), (h,k)=(x,y) coordinates of the center For given equation: center: (1,-2) b^2=4 b=2 length of semi major or minor axis=2b=4 .. a^2=9 c^2=a^2-b^2=9-4=5 c=√5 focal length=22c=2√5
 Quadratic-relations-and-conic-sections/615731: Complete the square, then graph and identify the vertex, focus, directrix, and endpoints of the latus rectum for the equation -14x+2y^2-8y=201 solutions Answer 387486 by lwsshak3(6456)   on 2012-05-30 03:40:28 (Show Source): You can put this solution on YOUR website!Complete the square, then graph and identify the vertex, focus, directrix, and endpoints of the latus rectum for the equation -14x+2y^2-8y=20 -14x+2y^2-8y=20 complete the square 2(y^2-4y+4)=20+14x+8 2(y-2)^2=14x+28 divide by 2 (y-2)^2=7x+14 (y-2)^2=7(x+2) This is an equation of a parabola that opens rightwards. Its standard form: (y-k)^2=4px, (h,k)=(x,y) coordinates of the vertex For given equation:(y-2)^2=7(x+2) vertex:(-2,2) axis of symmetry: y=2 4p=7 p=7/4 Focus: (-2+p,2)=(-2+7/4,2)=(-1/4,2) (p distance to the right of the vertex on the axis of symmetry) Directrix: x=(-2-p)=(-2-7/4)=-15/4 (p distance to the left of the vertex on the axis of symmetry) latus rectum: length of latus rectum=4p=7 2p=7/2 end points: (-1/4,2±2p) =(-1/4,2±7/2) =(-1/4,-1.5) and (-1/4,5.5) see graph below: y=(7(x+2))^.5+2
 Quadratic-relations-and-conic-sections/616012: What is the standard form of the ellipse with the equation 16x^2+25y^2=150y+175?1 solutions Answer 387485 by lwsshak3(6456)   on 2012-05-30 03:21:31 (Show Source): You can put this solution on YOUR website!What is the standard form of the ellipse with the equation 16x^2+25y^2=150y+175? complete the square: 16x^2+25(y^2-6y+9)=175+225 16x^2+25(y-3)^2=400 x^2/25+(y-3)^2/16=1 This is an equation of an ellipse with horizontal major axis. Its standard form: (x-h)^2/a^2+(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center
 logarithm/607923: logx^e^5 = 51 solutions Answer 387483 by lwsshak3(6456)   on 2012-05-30 03:02:23 (Show Source): You can put this solution on YOUR website!logx^e^5 = 5 e^5logx=5 logx=5/e^5 x=10^(5/e^5) x≈1.08
 logarithm/611444: Solve the equation: ln x + ln 2x=2.41 solutions Answer 387482 by lwsshak3(6456)   on 2012-05-30 02:46:27 (Show Source): You can put this solution on YOUR website!Solve the equation: ln x + ln 2x=2.4 ln(x*2x)=2.4 ln(2x^2)=2.4 e^2.4=2x^2 x^2=e^(2.4)/2 x=√(e^(2.4)/2) x≈2.348
 logarithm/612238: 5^4x+3=291 solutions Answer 387481 by lwsshak3(6456)   on 2012-05-30 02:27:10 (Show Source): You can put this solution on YOUR website!5^4x+3=29 5^4x=29-3=26 4xlog5=log26 x=log26/4log5 x≈.506
 logarithm/612244: Log base 3(x-4)=21 solutions Answer 387480 by lwsshak3(6456)   on 2012-05-30 02:15:00 (Show Source): You can put this solution on YOUR website!Log base 3(x-4)=2 convert to exponential form 3^2=x-4=9 x=13
 logarithm/614264: 1.) e^x=148.413 2.) 25=e^3x 3.) 9e^5x=5 4.) 1/4e^x=30 1 solutions Answer 387479 by lwsshak3(6456)   on 2012-05-30 02:08:40 (Show Source): You can put this solution on YOUR website!lne=1 1.) e^x=148.413 xlne=ln(148.413 x=ln(148.413)≈4.999...≈5 .. 2.) 25=e^3x 3xlne=ln25 3x=ln25 x=ln25/3≈1.073 3.) 9e^5x=5 e^5x=5/9 5xlne=ln(5/9) x=ln(5/9)/5≈-0.117 .. 4.) 1/4e^x=30 e^x=120 xlne=ln120 x=ln120≈4.787
 logarithm/616058: Can you please help me solve and simplify? 3log of 27(x-1)=21 solutions Answer 387478 by lwsshak3(6456)   on 2012-05-30 01:46:54 (Show Source): You can put this solution on YOUR website!Can you please help me solve and simplify? 3log of 27(x-1)=2 log of 27(x-1)=2/3 convert to exponential form 10^2/3=27(x-1)=27x-27 27x=10^2/3+27 x=(10^2/3+27)/27 using calculator x≈1.172
 Quadratic-relations-and-conic-sections/616031: Can someone help me with this question for my review? x^2/36-y^2/16=1 center: vertices: conjugate points: foci: asymptotes: I don't know how to start, it's overwhelming X_X IM STUCK....1 solutions Answer 387468 by lwsshak3(6456)   on 2012-05-29 22:38:22 (Show Source): You can put this solution on YOUR website!x^2/36-y^2/16=1 This is an equation of a hyperbola with horizontal transverse axis. Its standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of the center For given equation:x^2/36-y^2/16=1 center: (0,0) .. a^2=36 a=√36=6 vertices: (0±a,0)=(0±6,0)=(-6,0) and (6,0) .. b^2=16 b=√16=4 conjugate points: (0,0±b)=(0,0±4)=(0-4) and (0,4) .. c^2=a^2+b^2=36+16=52 c=√52≈7.2 foci: (0±c,0)=(0±7.2,0)=(-7.2,0) and (7.2,0) .. slope of asymptotes for hyperbolas with horizontal transverse axis=±b/a=±4/6=±2/3 asymptotes are straight lines that intersect at the center. Equation for asymptote with negative slope: y=-2x/3+b since y-intercept is at 0, b=0 equation of asymptote: y=-2x/3 .. Equation for asymptote with positive slope: y=2x/3+b since y-intercept is at 0, b=0 equation of asymptote: y=2x/3
 Quadratic-relations-and-conic-sections/616029: If you're given x^2/25+y^2/16=1 What is the foci? - - Given 36x^2+9y^2-324=0 Find the center: vertices: co-vertices: foci: I know it's c^2=a^2+b^2 or subtraction? Sorry, These questions I have trouble with on the review, please help!1 solutions Answer 387461 by lwsshak3(6456)   on 2012-05-29 22:00:29 (Show Source): You can put this solution on YOUR website!x^2/25+y^2/16=1 What is the foci? This is an equation of an ellipse with horizontal major axis Its standard form: (x-h)^2/a^2+(y-k)^2/b^2=1, (a>b), (h,k)=(x,y) coordinates of the center For given equation: center: (0,0) a^2=25 b^2=16 c^2=a^2-b^2=25-16=9 c=√9=3 foci: (0±c,0)=(0±3,0)=(-3,0) and (3,0) - - Given 36x^2+9y^2-324=0 36x^2+9y^2=324 divide by 324 x^2/9+y^2/36=1 This is an ellipse with vertical major axis center: (0,0) .. a^2=36 a=√36=6 vertices:(0,0±a)=(0,0±6)=(0,-6) and (0,6) .. b^2=9 b=3 co-vertices:(0±b,0)=(0±3,0)=(-3,0) and (3,0) .. c^2=a^2-b^2=36-9=25 c=√25=5 foci:(0,0±c)=(0,0±5)=(0,-5) and (0,5)
 Proportions/615969: I am 62 and am tutoring a 50-year old woman who needs to pass the entrance exam for a college algebra class at the University of Maine in Augusta this Fall. I have done this question by calulator for an answer of B=1.08, but she reduced the fractions longhand and got B=0.1. Doing it her way step-by-step I see the problem is in the cross-products, but I cannot write it out in a step-by-step fashion to get the right answer either. Would you please write out the step-by-step reduction for me? The question is: 3¼/2 = B/two-thirds, that is 3 and one-quarter over 2 = B over two-thirds. Thank you. I see my student again next Tuesday, June 5th, Bob Gross1 solutions Answer 387422 by lwsshak3(6456)   on 2012-05-29 20:03:51 (Show Source): You can put this solution on YOUR website!3¼/2 = B/two-thirds 3 and 1/4=13/4 rewrite equation: (13/4)/2=B/(2/3) cross-multiply 2B=(2/3)(13/4=26/12 divide by 2 B=26/24=13/12 If an exact answer is wanted, you must leave it in this fraction form. If a decimal answer is wanted, you will get 13/12=1.08333... so, you would probably round it off depending on how many decimal places are wanted. one decimal place=1.1 two decimal places=1.08 three decimal places=1.083 etc. hope this helps