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lwsshak3 answered: 6461 problems
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 logarithm/592064: What is the solution(s) to log base 4 of 4x + log base 4 of (x+3) = 2 1 solutions Answer 375836 by lwsshak3(6463)   on 2012-03-27 03:36:41 (Show Source): You can put this solution on YOUR website!What is the solution(s) to log base 4 of 4x + log base 4 of (x+3) = 2 log4[(4x)(x+3)]=2 convert to exponential form: base(4) raised to log of number(2)=number(4x)(x+3) 4^2=(4x)(x+3)=16 4x^2+12x=16 4x^2+12x-16=0 x^2+3x-4=0 (x+4)(x-1)-0 x=-4 (reject,x>0) or x=1
 Quadratic-relations-and-conic-sections/591303: Write an equation in standard form for the circle that statisfies these conditions center(0,7) tangent to x-axis1 solutions Answer 375835 by lwsshak3(6463)   on 2012-03-27 03:12:24 (Show Source): You can put this solution on YOUR website!Write an equation in standard form for the circle that statisfies these conditions center(0,7) tangent to x-axis ** If you make a rough sketch of given points, you will see that the radius=7 Standard form of equation for a circle: (x-h)^2+(y-k)^2=r^2, (h,k)=(x,y) coordinates of center, r=radius. Equation of given circle x^2+(y-7)^2=7^2 x^2+(y-7)^2=49
 Quadratic-relations-and-conic-sections/591913: write an equation in vertex form then graph it. Label center,verticies,co-verticies and the foci> x^2+4y^2=161 solutions Answer 375834 by lwsshak3(6463)   on 2012-03-27 03:01:45 (Show Source): You can put this solution on YOUR website!write an equation in vertex form then graph it. Label center,verticies,co-verticies and the foci> x^2+4y^2=16 divide by 16 x^2/16+y^2/4=1 This is an equation of an ellipse with horizontal major axis of the standard form: (x-h)^2/a^2=(y-k)^2/b^2=1,a>b, (h,k)=(x,y) coordinates of center. For given ellipse: center:(0,0) a^2=16 a=√16=4 vertices: (0±a,0)=(0±4,0)=(-4,0) and (4,0) .. b^2=4 b=√4=2 co-vertices: (0,0±b)= (0,0±2)=(0,-2) and (0,2) .. c^2=a^2-b^2=16-4=12 c=√12≈3.46 Foci:(0±c,0)=(0±√12,0)=(-3.46,0) and (3.46,0) see graph below: y=±(4-x^2/4)^.5
 Quadratic-relations-and-conic-sections/591939: In which direction does the parabola that is given by the equation below open? x = 1.5(y - 7)2 + 5 1 solutions Answer 375833 by lwsshak3(6463)   on 2012-03-27 02:38:59 (Show Source): You can put this solution on YOUR website!In which direction does the parabola that is given by the equation below open? x = 1.5(y - 7)2 + 5 ** Standard form of equation for parabola: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex. For given equation: vertex:(7,5) A=1.5 If A>0, parabola opens upwards. if A<0, parabola opens downwards. So given parabola opens upwards
 Quadratic-relations-and-conic-sections/592032: The vertex of the parabola below is at the point (4, 6.5) and the point (5, 4.5) is on the parabola. What is the equation of the parabola? a. y=-2.5(x+2)^2+4.5 b. x=-2(y-4)^2+6.5 c. y=-2(x-4)^2+6.5 d. y=-2(x-4)^2+6.5 1 solutions Answer 375832 by lwsshak3(6463)   on 2012-03-27 02:30:27 (Show Source): You can put this solution on YOUR website!The vertex of the parabola below is at the point (4, 6.5) and the point (5, 4.5) is on the parabola. What is the equation of the parabola? ** Standard form of equation for a parabola: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex. y=A(x-4)^2+6.5 Solving for A using coordinates of given point (5,4.5) 4.5=A(5-4)+6.5 4.5=A(1)+6.5 A=-2 Equation: y=-2(x-4)^2+6.5 answers are c and d which are duplicates.
 Quadratic-relations-and-conic-sections/591655: write an equation in standard form of the hyperpola with foci at (+3, 0) (-3,0) if the difference in the distances from a poin (x,y) on the hyperbola to the foci is 41 solutions Answer 375780 by lwsshak3(6463)   on 2012-03-26 18:42:18 (Show Source): You can put this solution on YOUR website!write an equation in standard form of the hyperpola with foci at (+3, 0) (-3,0) if the difference in the distances from a poin (x,y) on the hyperbola to the foci is 4. ** Standard form of equation for a hyperbola with horizontal transverse axis: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center For given hyperbola: center:(0,0) Difference in distances from a point (x,y) on the hyperbola to each of the foci is 4=2a a=2 a^2=4 c=3 c^2=9 c^2=a^2+b^2 b^2=c^2-a^2=9-4=5 Equation of given hyperbola: x^2/4-y^2/5=1
 Inequalities/591571: Solve the inequality. Write the solution set in interval notation. 2/3y-1/2>_ y+ 11/3 The >_ means greater than or equal to.1 solutions Answer 375675 by lwsshak3(6463)   on 2012-03-26 03:35:31 (Show Source): You can put this solution on YOUR website!Solve the inequality. Write the solution set in interval notation. 2/3y-1/2>_ y+ 11/3 2y/3-1/2≥ y+ 11/3 LCD:6 4y-3≥6y+22 -2y≥25 divide both sides by -2 and reverse inequality sign y≤-25/2 solution: (-∞,-25/2)
 Inequalities/591162: 5/9(t+2)<= 1/4 (2t-6)1 solutions Answer 375523 by lwsshak3(6463)   on 2012-03-25 03:23:19 (Show Source): You can put this solution on YOUR website!5/9(t+2)<= 1/4 (2t-6) 5(t+2)/9≤(2t-6)/4 5(t+2)/9-(2t-6)/4≤0 LCD:36 20(t+2)-9(2t-6)=0 20t+40-18t+54=0 2t+94=0 2t=-94 t=-47 number line <.....-.....-47......+......> solution: (-∞,-47]
 Trigonometry-basics/591178: find the graph of the equation, and the period and phase shift. y=sin(1/4x+pi/3)1 solutions Answer 375522 by lwsshak3(6463)   on 2012-03-25 02:02:15 (Show Source): You can put this solution on YOUR website!find the graph of the equation, and the period and phase shift. y=sin(1/4x+pi/3) ** Equation used for graphing sin functions: y=Asin(Bx-C), A=amplitude, Period=2π/B, Phase shift=C/B. For given equation: y=sin(1/4x+pi/3): A=1 B=1/4 Period: 2π/B=2π/(1/4)=8π 1/4 period=8/4=2π phase shift: C/B=(π/3)/(1/4)=4π/3 (shift to the left) .. Graph for one period: scale of x-axis in radians With no phase shift, (x,y) coordinates for the given sin function would be as follows: (0,0), (2π,1), (4π,0), (6π,-1), (8π,0) .. A phase shift of 4π/3 radians would shift x-coordinates to the left as follows: (-4π/3,0), (2π/3,1), (8π/3,0), (14π/3,-1), (20π/3,0) .. y-intercept set x=0 y=sin(π/3) y=√3/2 You now have all the points you need including the y-intercept to plot given sin function.
 logarithm/588829: If log_x^2(9)-log_x(36)=1 What does x equal?1 solutions Answer 375521 by lwsshak3(6463)   on 2012-03-25 01:06:11 (Show Source): You can put this solution on YOUR website!If log_x^2(9)-log_x(36)=1 What does x equal? ** log(9x^2)-log(36x)=1 log[(9x^2)/(36x)]=1 convert to exponential form 10^1=(9x^2)/(36x)=10 x/4=10 x=40
 logarithm/588468: how do you solve log7 x+2 log7 x - log7 3=log7 721 solutions Answer 375520 by lwsshak3(6463)   on 2012-03-25 00:55:29 (Show Source): You can put this solution on YOUR website!how do you solve log7 x+2 log7 x - log7 3=log7 72 log7(x)+log7(x^2)-log7(3)=log7(72) log7(x)+log7(x^2)-log7(3)-log7(72)=0 place under single log log[x(x^2)/(3*72)]=0 convert to exponential form 10^0=x^3/216=1 x^3=216 x=6
 logarithm/588859: Solve Equation ln(3x+1)/ln(2x-1)=2 1 solutions Answer 375509 by lwsshak3(6463)   on 2012-03-24 21:51:13 (Show Source): You can put this solution on YOUR website!Solve Equation ln(3x+1)/ln(2x-1)=2 ln(3x+1)=2ln(2x-1) ln(3x+1)=ln(2x-1)^2 3x+1=(2x-1)^2 3x+1=4x^2-4x+1 4x^2-7x=0 x(4x-7)=0 x=0 (reject,(2x-1)>0) x=7/4
 logarithm/589004: Solve Equation(Do no use mixed numbers in your answer). ln(5x-6)+ln(x)=ln(e) please help me with this problem.I would appreciate it 1 solutions Answer 375508 by lwsshak3(6463)   on 2012-03-24 21:42:04 (Show Source): You can put this solution on YOUR website!Solve Equation(Do no use mixed numbers in your answer). ln(5x-6)+ln(x)=ln(e) ln[(5x-6)(x)]=1 convert to exponential form e^1=5x^2-6x 5x^2-6x-e=0 solve by quadratic formula: a=5, b=-6, c=e x=[6±√(36-4*5*e)]/2*10 x=[6±√(36-20e)]/20 plug in e=2.7183 to solve for x
 logarithm/589165: log base X(4)+log base X(9)=21 solutions Answer 375504 by lwsshak3(6463)   on 2012-03-24 21:24:07 (Show Source): You can put this solution on YOUR website!log base X(4)+log base X(9)=2 logx(4)+logx(9)=2 logx(4*9)=2 logx(36)=2 base(x) raised to log of number(2)=number(36) x^2=36 x=-6(reject, base>1) x=6 (ans)
 logarithm/590586: Assume that log 2 = .301 and log 3 = .477, Solve 1)log 1/9 2)log 30 3)log 5 4)log 0.0061 solutions Answer 375498 by lwsshak3(6463)   on 2012-03-24 21:10:56 (Show Source): You can put this solution on YOUR website!Assume that log 2 = .301 and log 3 = .477, Solve 1)log 1/9 =log 1/3^2 =log 3^-2 =-2log3 =-2*.477 =-0.954 .. 2)log 30 =log 10*3 =log 10+log 3 =1+.477 =1.477 .. 3)log 5 =log(30/6) =log 30-log 6 =log 10+log3-(log 3+log 2) =1+.477-.477-.301 =1-.301 =.699 .. 4)log 0.006 =log 6*10^-3 =10^-3*log6 =log(.001)+(log3+log2) =-3+(.301+.477) =7.78*10^-4 =-2.222
 Quadratic-relations-and-conic-sections/590524: the center of the hyperbola is at (2,5). One vertex of the hyperbola is at (0,5) where is the other vertex1 solutions Answer 375474 by lwsshak3(6463)   on 2012-03-24 17:39:27 (Show Source): You can put this solution on YOUR website!the center of the hyperbola is at (2,5). One vertex of the hyperbola is at (0,5) where is the other vertex ** Note that the horizontal transverse axis and one vertex of given hyperbola lie on the line y=5. Given vertex is 2 units to the left of center at (0,5) so the other vertex must be 2 units to the right of center at (4,5).
 Quadratic-relations-and-conic-sections/590695: x^2/16+y^2/9=1 What is the foci and vertices of this ellipse equatioin1 solutions Answer 375470 by lwsshak3(6463)   on 2012-03-24 17:16:20 (Show Source): You can put this solution on YOUR website!x^2/16+y^2/9=1 What is the foci and vertices of this ellipse equation ** Given equation is that of an ellipse with horizontal major axis of the standard form: (x-h)^2/a^2+(y-k)^2/b^2=1,a>b, (h,k)=(x,y) coordinates of center. center:0,0) a^2=16 a=√16=4 vertices:(0±a,0)=(0±4,0)=(-4,0) and ((4,0) b^2=9 b=√9=3 c^2=a^2-b^2=16-9=7 c=√7 Foci: (0±c,0)=(0±√7,0)=(-√7,0) and (√7,0)
 Quadratic-relations-and-conic-sections/590828: What is the center of the ellipse whose equation is (x2)/25 + (y2)/4 = 1?1 solutions Answer 375464 by lwsshak3(6463)   on 2012-03-24 16:48:07 (Show Source): You can put this solution on YOUR website!What is the center of the ellipse whose equation is (x2)/25 + (y2)/4 = 1? ** Standard form of equation for an ellipse with horizontal major axis (a^2 under x^2): (x-h)^2/a^2+(y-k)^2/b^2=1,a>b, (h,k)=(x,y) coordinates of the center. .. For given equation, h and k do not appear in the equation, so it is assumed they are=0, that is, the coordinates of the center are (0,0)
 Quadratic-relations-and-conic-sections/590855: FIND THE EQUATION OF THE PARABOLA WITH FOCUS (2,5) AND DIRECTRIX Y = 11 solutions Answer 375463 by lwsshak3(6463)   on 2012-03-24 16:40:51 (Show Source): You can put this solution on YOUR website!FIND THE EQUATION OF THE PARABOLA WITH FOCUS (2,5) AND DIRECTRIX Y = 1 ** Standard form of equation for a parabola that opens upwards: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex. For given parabola: axis of symmetry: x=2 x-coordinate of vertex=2 y-coordinate of vertex=3 (halfway between focus(5) and directrix(1) on the axis of symmetry vertex: (2,3) P=2 (distance from focus to vertex) 4p=8 Equation of given parabola: (x-2)^2=8(y-3)
 Quadratic-relations-and-conic-sections/590887: a one-way road passes under an overpass in the shape of half and ellipse, 15ft high at the center and 20ft wide. Assuming a truck is 16ft wide, what is the tallest truck that can pass under the overpass?1 solutions Answer 375460 by lwsshak3(6463)   on 2012-03-24 16:12:22 (Show Source): You can put this solution on YOUR website!a one-way road passes under an overpass in the shape of half and ellipse, 15ft high at the center and 20ft wide. Assuming a truck is 16ft wide, what is the tallest truck that can pass under the overpass? ** This problem can be represented by an ellipse with a vertical major axis of the standard form: (x-h)^2/b^2+(y-k)^2/a^2=1, a>b, (h,k)=(x,y) coordinates of center. .. center: (0,0) (center line of road) a=15 ft (height of overpass at center) a^2=225 b=10 (1/2 the width of the road) b^2=100 equation of ellipse: x^2/100+y^2/225=1 The point at which a truck 16 ft wide would touch the overpass would have and x-coordinates ±8 ft from center. Its y-coordinate is the answer to given problem. Plug in x=8 ft in equation to solve for y. 8^2/100+y^2/225=1 y^2/225=1-64/100 y^2=225-225*64/100=81 y=9 ft ans: The tallest truck that can pass under the overpass would be 9 feet in height
 Quadratic-relations-and-conic-sections/590886: write an equation for the following conic. if it is a parabola, it has a vertex at the origin, and if it is an ellipse or a hyperbola, it is centered at the origin. focus at (4,0) and e= 1/21 solutions Answer 375376 by lwsshak3(6463)   on 2012-03-24 04:10:59 (Show Source): You can put this solution on YOUR website!write an equation for the following conic. if it is a parabola, it has a vertex at the origin, and if it is an ellipse or a hyperbola, it is centered at the origin. focus at (4,0) and e= 1/2 ** if 0 (x-h)^2/a^2+(y-k)^2/b^2=1, a>b, (h,k)=(x,y) coordinates of center center: (0,0) (given) c=4 c^2=16 e=1/2=c/a a=2c=8 a^2=64 c^2=a^2-b^2 b^2=a^2-c^2=64-16=48 Equation of ellipse: x^2/64+y^2/48=1
 Quadratic-relations-and-conic-sections/590891: find the vertex, the focus, and directrix. y^2= 1/2x1 solutions Answer 375375 by lwsshak3(6463)   on 2012-03-24 03:58:19 (Show Source): You can put this solution on YOUR website!find the vertex, the focus, and directrix. y^2= 1/2x ** This is an equation of a parabola that opens rightwards of the standard form: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex. For given parabola: vertex:(0,0) axis of symmetry: x-axis or y=0 4p=1/2 p=1/8 focus: (1/8,0) (p distance from right of vertex on axis of symmetry) directrix: x=-1/8 (p distance from left of vertex on axis of symmetry)
 Trigonometry-basics/590934: If cos x = -(2/3) and pi/2 < x < pi, what is the exact value of sin2x, cos2x, and tan2x1 solutions Answer 375374 by lwsshak3(6463)   on 2012-03-24 03:46:25 (Show Source): You can put this solution on YOUR website!If cos x = -(2/3) and pi/2 < x < pi, what is the exact value of sin2x, cos2x, and tan2x ** If cos x = -(2/3) (adjacent/hypotenuse in quadrant II where cos<0) sinx=√5/3 (opposite/hypotenuse in quadrant II where sin>0) tanx=-√5/2(opposite/adjacent in quadrant II where tan<0) .. sin2x=2sinxcosx=2*√5/3*-2/3=-4√5/9 cos 2x=cos^2x-sin^2x=(-2/3)^2-(√5/3)^2=4/9-5/9=-1/9 tan2x=2tanx/(1-tan^2x)=2*-√5/2/(1-5/4)=-√5/-1/4=4√5
 Quadratic-relations-and-conic-sections/590889: wrirte and equation for the following conic. If it is a parabola, it has a vertex at the orgin, and if it is an ellipse or a hyperbola, it is centered at the orgin. Focus at (0,5) and e=11 solutions Answer 375372 by lwsshak3(6463)   on 2012-03-24 03:21:45 (Show Source): You can put this solution on YOUR website!wrirte and equation for the following conic. If it is a parabola, it has a vertex at the orgin, and if it is an ellipse or a hyperbola, it is centered at the orgin. Focus at (0,5) and e=1 ** In accordance with the focus-directrix property of conics, if e=1, the conic is a parabola: Standard form of equation for parabola opening upwards: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex. vertex:(0,0) (given) axis of symmetry: y-axis or x=0 p=5 (distance from focus to vertex on the axis of symmetry) 4p=20 Equation of parabola: x^2=20y
 Equations/590931: If then is what?1 solutions Answer 375368 by lwsshak3(6463)   on 2012-03-24 02:39:19 (Show Source): You can put this solution on YOUR website!If (-3z+6)/3 + z/5 = 12 then 4z^2+2z+50 is what? ** (-3z+6)/3 + z/5 = 12 -z+2 + z/5 = 12 -5z+10 + z = 60 -4z = 50 z=-50/4=-25/2 .. 4z^2+2z+50 =4(-25/2)^2+2(-25/2)+50 =4(625/4)-25+50 =625-25+50=650
 Exponential-and-logarithmic-functions/590926: Solve the following exponential equation: 1 solutions Answer 375367 by lwsshak3(6463)   on 2012-03-24 02:22:33 (Show Source): You can put this solution on YOUR website!Solve the following exponential equation: 2^(2x) + 2^(x+2) - 12 = 0 2^(2x) + 2^x*2^2 - 12 = 0 2^(2x) + 2^x*4 - 12 = 0 2^2x + 4*2^x - 12 = 0 . let u=2^x u^2=2^2x .. u^2+4u-12=0 (u+6)(u-2)=0 u=-6=2^x no real solution, (2^x>0) or u=2=2^x x=1
 Trigonometry-basics/590645: find exact value of [tan(5pi/4)-tan(pi/12)]/1+tan(5pi/4)tan(pi/12)1 solutions Answer 375222 by lwsshak3(6463)   on 2012-03-23 04:10:58 (Show Source): You can put this solution on YOUR website!find exact value of [tan(5pi/4)-tan(pi/12)]/1+tan(5pi/4)tan(pi/12) ** use tan addition identity: tan (s-t)=(tan s-tan t)/(1+tan s tan t) For given problem: [tan(5pi/4)-tan(pi/12)]/1+tan(5pi/4)tan(pi/12)=tan(5π/4-π/12) =tan(15π/12-π/12)=tan(14π/12) =tan(7π/6) (ans)
 Square-cubic-other-roots/590663: Ok. So i know the answer is going to be an imaginary number. The problem is this: 6 +or- the square root of (-6)^2-4(2)(9) all divided by 4. PLease help. I can not understand how to break down the problem into imaginary numbers.1 solutions Answer 375220 by lwsshak3(6463)   on 2012-03-23 03:51:41 (Show Source): You can put this solution on YOUR website!Ok. So i know the answer is going to be an imaginary number. The problem is this: 6 +or- the square root of (-6)^2-4(2)(9) all divided by 4. PLease help. I can not understand how to break down the problem into imaginary numbers. ** {6±√[(-6)^2-4*2*9]}/4 {6±√[36-72]}/4 (6±√-36)/4 (6±6i)/4 6(1±i)/4 3(1±i)/2=3(1+i)/2 and 3(1-i)/2 .. note: √-36=√36*√-1=6i
 Linear_Equations_And_Systems_Word_Problems/590666: A tank has supply pipe A and an exhaust pipe B. The pipe A takes 4 minutes less to fill a tank than the pipe B takes to empty it. When both are open, the tank is filled in 24 minutes. Find the time needed for pipe A to fill the tank when pipe B is closed.1 solutions Answer 375219 by lwsshak3(6463)   on 2012-03-23 03:28:02 (Show Source): You can put this solution on YOUR website!A tank has supply pipe A and an exhaust pipe B. The pipe A takes 4 minutes less to fill a tank than the pipe B takes to empty it. When both are open, the tank is filled in 24 minutes. Find the time needed for pipe A to fill the tank when pipe B is closed. ** let x=minutes pipe B takes to empty tank 1/x=pipe B work rate x-4=minutes pipe A takes to fill tank 1/(x-4)=Pipe A work rate .. In 24 minutes: 24/(x-4)-24/x=100% 24x-24x+96=x(x-4)=x^2-4x x^2-4x-96=0 (x-12)(x+8)=0 x=-8 (reject, x>0) x=12 x-4=8 ans: Pipe A alone fills tank in 8 minutes (When Pipe B is closed)
 logarithm/590236: 1. Show that log8x=1/3 log2x hence solve the equation for x>0 Log2(3x+1) +log8(x-1)3=0 1 solutions Answer 375083 by lwsshak3(6463)   on 2012-03-22 04:15:30 (Show Source): You can put this solution on YOUR website!1. Show that log8x=1/3 log2x hence solve the equation for x>0 Log2(3x+1) +log8(x-1)3=0 ** log8x=1/3 log2x change to base2 log8x=log2x/log2(8)=log2x/3 .. log2(3x+1) +log8(x-1)*3=0 change to base 2 log2(3x+1) +log2(x-1)*3/log2(8)=0 log2(3x+1) +log2(x-1)*3/3=0 log2(3x+1) +log2(x-1)=0 place under single log log2[(3x+1)(x-1)]=0 convert to exponential form: 2^0=(3x+1)(x-1)=1 3x^2-2x-2=0 solve by quadratic formula: x≈-.549 (reject,x>0) or x=1.215
 Linear-systems/590196: In 6 years, Sylvia will be 6/5 Hue's age. Four years ago, Sylvia was 10 years less than twice Hue's age. How old are Sylvia and Hue now?1 solutions Answer 375082 by lwsshak3(6463)   on 2012-03-22 03:41:02 (Show Source): You can put this solution on YOUR website!In 6 years, Sylvia will be 6/5 Hue's age. Four years ago, Sylvia was 10 years less than twice Hue's age. How old are Sylvia and Hue now? ** let x=Hue's present age let y=Sylvia's present age .. In 6 years: Hue's age=(x+6) Sylvia's age=(y+6) (y+6)=(6/5)(x+6) y=(6/5)(x+6)-6 y=(6x+36)/5-6 .. 4 years ago: Hue's age=(x-4) sylvia's age=(y-4) y-4=2(x-4)-10 y=2x-8-10+4 y=2x-14 .. (6x+36)/5-6=2x-14 LCD+5 6x+36-30=10x-70 4x=76 x=19 y=2x-14=38-14=24 .. ans: Hue's present age=19 Sylvia's present age=24