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# Recent problems solved by 'lwsshak3'

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 Quadratic_Equations/716248: I have a word problem. "A picture has a square frame that is 2 inches wide. The area of the picture is one-third of the total area of the picture and frame. What are the dimensions of the picture to the nearest quarter of an inch?" I know that it is a quadratic equation and my teacher said that I have to solve it by completing the square. 1 solutions Answer 439784 by lwsshak3(6460)   on 2013-02-19 15:31:02 (Show Source): You can put this solution on YOUR website!A picture has a square frame that is 2 inches wide. The area of the picture is one-third of the total area of the picture and frame. What are the dimensions of the picture to the nearest quarter of an inch?" ** let x=length of side of square picture x^2=area of picture (x+4)=length of side of square frame (x+4)^2=total area of picture and frame .. x^2=1/3(x+4)^2=(x^2+8x+16)/3 3x^2=x^2+8x+16 2x^2-8x-16=0 x^2-4x-8=0 solve for x by completing the square: (x^2-4x+4)-4-8=0 (x-2)^2=12 take sqrt of both sides x-2=±√12 x=2±√12 x=2±3.46 x≈-1.46 (reject, x>0) or x≈5.5 dimensions of the picture: 5.5 by 5.5 inches
 Travel_Word_Problems/716032: Two cars (a race car and a normal car) each travel 300 miles per day. The race car travels 40 mph faster and takes 2 hours and 40 minutes less time than the normal car. Find the speed of the race car. 1 solutions Answer 439711 by lwsshak3(6460)   on 2013-02-19 03:15:34 (Show Source): You can put this solution on YOUR website!Two cars (a race car and a normal car) each travel 300 miles per day. The race car travels 40 mph faster and takes 2 hours and 40 minutes less time than the normal car. Find the speed of the race car. ** let x=speed of the race car x-40=speed of the normal car 2 hours and 40 minutes=2-2/3 hours=8/3 hrs .. travel time=distance/speed travel time of normal car-travel time of race car=2-2/3 hrs 300/(x-40)-300/x=8/3 900/(x-40)-900/x=8 900x-900(x-40)=8x(x-40) 900x-900x+36000=8x^2-320x 8x^2-320x-36000=0 x^2-40x-4500=0 (x+50)(x-90)=0 x=-50 (reject) or x=90 x-40=50 speed of the race car=90 mph speed of the normal car=50 mph
 Miscellaneous_Word_Problems/716036: Seven times one number decreased by twice another equals 27, and 3 times the first added to 5 times the second equals 35. Find the two numbers.1 solutions Answer 439706 by lwsshak3(6460)   on 2013-02-19 02:41:58 (Show Source): You can put this solution on YOUR website!Seven times one number decreased by twice another equals 27, and 3 times the first added to 5 times the second equals 35. Find the two numbers. ** let x=first number let y=2nd number .. 7x-2y=27 3x+5y=35 .. 35x-10y=135 6x+10y=70 add 41x=205 x=5 2y=7x-27 2y=35-27 2y=8 y=4 .. 1st number=5 2nd number=4
 Miscellaneous_Word_Problems/716037: Find the ages of A and B from this information: 6 times A's age is 100 more than 5 times B's age, and 9 times A's age is 170 more than 7 time's B's age.1 solutions Answer 439704 by lwsshak3(6460)   on 2013-02-19 02:19:03 (Show Source): You can put this solution on YOUR website!Find the ages of A and B from this information: 6 times A's age is 100 more than 5 times B's age, and 9 times A's age is 170 more than 7 time's B's age. ** let A=A's age let B=B's age .. 6A=100+5B 9A=170+7B .. 42A=700+35B 45A=850+35B subtract: 3A=150 A=50 .. 5B=6A-100 5B=300-100 5B=200 B=40 .. A's age=50 B's age=40
 Equations/716056: I cannot figure out how to do the equation 1/2(2y+3)=-2/3(y-4) I get different answers every time, can you help me with what steps to take?1 solutions Answer 439700 by lwsshak3(6460)   on 2013-02-19 02:01:40 (Show Source): You can put this solution on YOUR website!1/2(2y+3)=-2/3(y-4) (2y+3)2=-2(y-4)/3 6y+9=-4y+16 10y=7 y=7/10
 Quadratic_Equations/716026: can you help me find the roots of the quadratic equation x2+6x-41=0 and walk me through step by step 1 solutions Answer 439699 by lwsshak3(6460)   on 2013-02-19 01:49:16 (Show Source): You can put this solution on YOUR website!find the roots of the quadratic equation x^2+6x-41=0 complete the square: (x^2+6x+9)-9-41=0 (x+3)^2=50 x+3=±√50 x=-3±√50 x=-3-√50 or x=-3+√50
 Quadratic_Equations/716053: Hello i need help in finding a solution. Form a quadratic equation and solve also a =/= 01 solutions Answer 439697 by lwsshak3(6460)   on 2013-02-19 01:38:35 (Show Source): You can put this solution on YOUR website!Form a quadratic equation and solve: 2/a^2+3/a=-1 LCD:a^2 2+3a=-a^2 a^2+3a+2=0 (a+2)(a+1)=0 a=-2 or a=-1
 logarithm/714700: I need help asap!!! someone help! Use the given function f to answer parts a) through f)below. f(x)=1+log5(x+3) a)the domain of f b)graph f c) from the graph, determine the range and any asymptotes of f. d)find f^-1, the inverse of f e)find the domain and range of f^-1 f)graph f^-1 1 solutions Answer 439669 by lwsshak3(6460)   on 2013-02-18 20:26:07 (Show Source): You can put this solution on YOUR website!f(x)=1+log5(x+3) a)the domain of f: (-3,∞) .. b)graph f I don't have the means to graph it, but I can try to show you how to do it. Start with the basic log5(x) basic curve which has an asymptote at x=0 or the y-axis and an x-intercept at(1,0). Then move this curve 3 units left, placing the asymptote and x-intercept at x=-3 and (-2,0), respectively. Lastly, bump the curve up one unit. .. c) from the graph, determine the range and any asymptotes of f. Range: (-∞,∞); asymptote:x=-3 .. d)find f^-1, the inverse of f y=1+log5(x+3) interchange x and y, then solve for y x=1+log5(y+3) x-1=log5(y+3) 5^(x+1))=y+3 f^-1=5^(x+1)-3 .. e)find the domain and range of f^-1: Range:(-3,∞); Domain:(-∞,∞) .. f)graph f^-1 curve has a horizontal asymptote at y=-3 and a y-intercept at (0,2)
 logarithm/715667: Solve the equation: log4 30-log4(x-1)-log4(x+2)=log4 31 solutions Answer 439659 by lwsshak3(6460)   on 2013-02-18 19:20:55 (Show Source): You can put this solution on YOUR website!Solve the equation: log4 (30)-log4(x-1)-log4(x+2)=log4 (3) log4 (30)-(log4(x-1)+log4(x+2))=log4 (3) log4 [(30/(x-1)(x-2)]=log(3) (30/(x-1)(x-2)=3 30/3=(x-1)(x+2) 10=x^2+x-2 x^2+x-12=0 (x+4)(x-3)=0 x=-4 (reject, x≥0) x=3
 logarithm/715832: What is the exponential form of log[7] 90 = y? 1 solutions Answer 439655 by lwsshak3(6460)   on 2013-02-18 19:00:39 (Show Source): You can put this solution on YOUR website!What is the exponential form of log[7] 90 = y Exponential Form: Base(7) raised to log of number(y)=number(90) 7^y=90
 Trigonometry-basics/715619: what is the smallest positive number for t? sin(t)+cos(2t)=01 solutions Answer 439586 by lwsshak3(6460)   on 2013-02-18 15:23:24 (Show Source): You can put this solution on YOUR website!what is the smallest positive number for t? sin(t)+cos(2t)=0 sint+cos^2t-sin^2t=0 sint+1-sin^2t-sin^2t=0 2sin^2t-sint-1=0 (2sint+1)(sint-1)=0 sint-1=0 sint=1 t=π/2 or 2sint+1=0 sint=-1/2 t=7π/4 .. smallest positive number for t=π/2
 Trigonometry-basics/715423: Find the exact trigonometric ratios for the angle x whose radian measure is given. (If an answer is undefined, enter UNDEFINED.) −9π1 solutions Answer 439583 by lwsshak3(6460)   on 2013-02-18 15:09:31 (Show Source): You can put this solution on YOUR website!Find the exact trigonometric ratios for the angle x whose radian measure is given. (If an answer is undefined, enter UNDEFINED.) −9π ** angle -9π is like rotating clockwise 4-1/2 times around a unit circle terminating at a reference angle of π: trigonometric ratios at this reference angle: sinx=0 cosx=-1 tanx=0 cscx=undefined secx=-1 cotx=undefined
 Trigonometry-basics/715621: If sinx = -3/5 and x is in quadrant III, then what is sin2x, cos2x, and tan2x? 1 solutions Answer 439580 by lwsshak3(6460)   on 2013-02-18 14:50:43 (Show Source): You can put this solution on YOUR website!If sinx = -3/5 and x is in quadrant III, then what is sin2x, cos2x, and tan2x? let O=opposite side let A=adjacent side H=hypotenuse .. sin x=-3/5=O/H O=-3, H=5 A=√(H^2-O^2) =√(5^2-3^2) =√(25-9) =√16 A=-4 (In quadrant III where cos<0) cos x=A/H=-4/5 tanx=O/A=-3/-4=3/4 .. Identity:sin2x=2sinxcosx =2*-3/5*-4/5 sin2x=24/25 .. Identity:cos2x=cos2^x-sin^2x =(-4/5)^2-(-3/5)^2 =16/25-9/25 cos 2x=7/25 .. Identity: tan2x=(2tanx)/(1-tan^2x) =(2*3/4)/(1-(3/4)^2) =(6/4)/(1-9/16) =(3/2)/(7/16) tan2x=24/7
 Quadratic-relations-and-conic-sections/714369: Hello, i need some guidance on this hyperbola problem. On my test, it says to find the vertices, center and the foci of y^2 over 5, minus x^2 over 11=1. Or, if this is another way to write out my problem-> y^2/5-x^2/11=1. Thank you1 solutions Answer 439320 by lwsshak3(6460)   on 2013-02-16 21:41:17 (Show Source): You can put this solution on YOUR website!find the vertices, center and the foci of y^2 over 5, minus x^2 over 11=1. y^2/5-x^2/11=1 This is an equation of a hyperbola with vertical transverse axis: Its standard form: , (h,k)=)x,y) coordinates of center. .. For given equation: center:(0,0) .. a^2=5 a=√5 vertices: (0,0±a)=(0,0±√5)=(0,-√5), (0,+√5) .. b^2=11 .. c^2=a^2+b^2=5+11=16 c=√16=4 Foci: (0,0±c)=(0,0±4)=(0,-4), (0,+4)
 Quadratic-relations-and-conic-sections/715112: use the eccentricity of each hyperbola to find its equation in standard form eccentricity 4 , vertices (-1,3) and (-1,7)1 solutions Answer 439315 by lwsshak3(6460)   on 2013-02-16 21:22:51 (Show Source): You can put this solution on YOUR website!use the eccentricity of each hyperbola to find its equation in standard form eccentricity 4 , vertices (-1,3) and (-1,7) ** Standard form of equation for a hyperbola with vertical transverse axis: , (h,k)=(x,y) coordinates of center .. x-coordinate of center=-1 y-coordinate of center= (3+7)/2=5 (midpoint formula) center:(-1,5) .. length of vertical transverse axis=4 (3 to 7)=2a a=2 a^2=4 .. c=distance from center to foci Eccentricity=4=c/a c=4a .. c^2=a^2+b^2 16a^2=a^2+b^2 b^2=16a^2-a^2=15a^2 .. equation of given hyperbola:
 Quadratic_Equations/715160: What is the standard formula for the quadratic function that has a vertex of (0.5,4) and passes through points (-1.5,0) and (2.5,0)?1 solutions Answer 439250 by lwsshak3(6460)   on 2013-02-16 17:00:43 (Show Source): You can put this solution on YOUR website!What is the standard formula for the quadratic function that has a vertex of (0.5,4) and passes through points (-1.5,0) and (2.5,0)? ** Standard form of equation for a parabola: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of vertex. For given function: y=A(x-0.5)^2+4 solve for A using one of given points (-1.5,0) 0=A(-1.5-0.5)^2+4 0=A(-2)^2+4 4A=-4 A=-1 Equation: y=-(x-0.5)^2+4
 absolute-value/715161: 3 tothe power x equals 80, whats the value of x1 solutions Answer 439241 by lwsshak3(6460)   on 2013-02-16 16:33:46 (Show Source): You can put this solution on YOUR website!3 tothe power x equals 80, whats the value of x 3^x=80 xlog3=log80 x=log80/log3 x≈3.9887..
 Trigonometry-basics/714571: Will someone please help me with this question. I have reviewed my notes and an old textbook but am completely flummoxed although I will bet it is really an easy question, just not for me or my daughter. Which of the following has the same slope? 29 degrees and 1/2 2.2 and 65 degrees 7/11 and 65% 9.8 and 112.5% Any help would be much appreciated.1 solutions Answer 438974 by lwsshak3(6460)   on 2013-02-15 02:46:40 (Show Source): You can put this solution on YOUR website!Which of the following has the same slope? 29 degrees and 1/2 2.2 and 65 degrees 7/11 and 65% 9.8 and 112.5% ** using a calculator: 29 degrees≠1/2 2.2≠65 degrees 7/11≈0.636≠65% 9.8≠112.5% none of the above are equal to each other. unless you wrote 9.8 when you meant 9/8 9/8=1.125=112.5%
 Trigonometry-basics/714587: Prove the identity: (1/1-cosx)-(cosx/1+cosx)=2csc^2(x)-11 solutions Answer 438971 by lwsshak3(6460)   on 2013-02-15 02:31:56 (Show Source): You can put this solution on YOUR website!Prove the identity: (1/1-cosx)-(cosx/1+cosx)=2csc^2(x)-1 start with left side: (1/1-cosx)-(cosx/1+cosx) =1+cosx-cosx(1-cosx)/(1-cosx)(1-cosx) =1+cosx-cosx+cos^2x/(1-cos^2x) =1+cos^2x/(1-cos^2x) =(1+1-sin^2x)/1-(1-sin^2x) =(2-sin^2x)/sin^2x =(2/sin^2x)/(-(sin^2x)/(sin^2x)) =2csc^2x-1 verified: left side=right side
 Trigonometry-basics/714759: I need to prove this identity tan^2x-sin^2x = tan^2xsin^2x1 solutions Answer 438968 by lwsshak3(6460)   on 2013-02-15 02:01:42 (Show Source): You can put this solution on YOUR website!I need to prove this identity tan^2x-sin^2x = tan^2xsin^2x start with left side: tan^2x-sin^2x =(sin^2x/cos^2x)-sin^2x =(sin^2x-sin^2xcos^2x)/cos^2x =sin^2x(1-cos^2x)/cos^2x =sin^2x*sin^2x/cos^2x =tan^2xsin^2x verified: left side=right side
 Exponential-and-logarithmic-functions/714178: please show steps. Answer the following for the given quadric function. f(x)=x^2-4x+4 a.) what is the vertex (h,k) of f? b.) what is the axis of symmetry? c.) what are the intercepts? d.) graph it. e.) on which interval is f increasing? f.) on what interval is f decreasing? thanks so much this is all one problem!1 solutions Answer 438872 by lwsshak3(6460)   on 2013-02-14 14:56:51 (Show Source): You can put this solution on YOUR website!Answer the following for the given quadric function. f(x)=x^2-4x+4 a.) what is the vertex (h,k) of f? b.) what is the axis of symmetry? c.) what are the intercepts? d.) graph it. e.) on which interval is f increasing? f.) on what interval is f decreasing? ** f(x)=x^2-4x+4 complete the square: y=(x^2-4x+4)-4+4 y=(x-2)^2 .. vertex: (2,0) axis of symmetry: x=2 .. y-intercept set x=0, solve for y y-intercept=4 .. x-intercept set y=0, solve for x x-intercept=2 .. see graph below: .. f increasing: (2,∞) .. f decreasing:(-∞,2)
 Exponential-and-logarithmic-functions/714176: Show work please. find the slant asymptote of the graph of the rational function and Use the slant asymptote to graph the rational function. f(x)= x^2-x-20 ------------------------- x-6 b.) first determine the symmetry of the graph of f. c.) find the x and y-intercepts d.) find the asymptotes. thanks!1 solutions Answer 438776 by lwsshak3(6460)   on 2013-02-13 20:41:41 (Show Source): You can put this solution on YOUR website!find the slant asymptote of the graph of the rational function and Use the slant asymptote to graph the rational function. f(x)= (x^2-x-20)/(x-6) f(x)=(x-5)(x+4)/(x-6) .. b.) first determine the symmetry of the graph of f. Graph has no symmetry as can be seen by the graph below. .. c.) find the x and y-intercepts y-intercept: set x=0 y-intercept=-20/-6=10/3 .. x-intercept: set y=0 x^2-x-20 (x-5)(x+4)=0 x-intercepts at 5 and-4 .. d.) find the asymptotes. vertical asymptote: set denominator=0, then solve for x x-6=0 vertical asymptote: x=6 .. slant asymptote: divide numerator by denominator: (x^2-x-20)/(x-6)=(x+5)+remainder slant asymptote: y=x+5 .. number line <...-....-4....+.....5....-.....6....+.....>
 Exponential-and-logarithmic-functions/714277: Simplify without a calculator. Show all steps 4 log base 3 of 1/3 + 2log base 27 of 91 solutions Answer 438752 by lwsshak3(6460)   on 2013-02-13 19:53:38 (Show Source): You can put this solution on YOUR website!Simplify without a calculator. Show all steps 4 log base 3 of 1/3 + 2log base 27 of 9 .. 4 log base 3 of 1/3 =log3(1/3)^4 =log3(1/81) =log3(1/3)^4) =log3(1/3)^3)*(1/3) =log3[(3^-3)*(3^-1)] =log3(3^-3)+log3(3^-1) log of base raised to a power=power =-3-1=-4 .. 2log base 27 of 9 =log27(9^2) =log27(81) =log27(3^3)+log27(3^1) =log27(27)+log27(3) =1+1/3 =4/3 .. 4 log base 3 of 1/3 + 2log base 27 of 9=-4+4/3=-12/3+4/3=-8/3
 Quadratic-relations-and-conic-sections/713184: What is y=x^(2)+8x-1 in vertex form?1 solutions Answer 438702 by lwsshak3(6460)   on 2013-02-13 14:58:31 (Show Source): You can put this solution on YOUR website!What is y=x^(2)+8x-1 in vertex form? ** y=x^(2)+8x-1 complete the square y=(x^2+8x+16)-1-16 y=(x+4)^2-17 This is an equation of a parabola that opens upwards with vertex at (-4,-17) Its standard form of equation:y= (x+h)^2+k, (h,k) =(x,y) coordinates of the vertex.
 Quadratic-relations-and-conic-sections/713388: Write the standard form equation of an ellipse: Vertices: (5,-3) (-3,-3) Co-vertices: (1,0), (1, -6) I know how to find the standard form with only two endpoints rather than four. Thanks for your help!1 solutions Answer 438698 by lwsshak3(6460)   on 2013-02-13 14:41:12 (Show Source): You can put this solution on YOUR website!Write the standard form equation of an ellipse: Vertices: (5,-3) (-3,-3) Co-vertices: (1,0), (1, -6) I know how to find the standard form with only two endpoints rather than four. Thanks for your help! ** Given data shows ellipse has a horizontal major axis. Its standard form of equation: , a>b, (h,k)=(x,y) coordinates of center. x-coordinate of center= (5+(-3))/2=2/2=1 (midpoint formula) y-coordinate of center=-3 center:(1,-3) length of horizontal major axis=8 (-3 to 5)=2a a=4 a^2=16 length of minor axis (co-vertices)=6 (0 to 6)=2b b=3 b^2=9 .. Equation of given ellipse:
 Trigonometry-basics/713946: prove that (sinx-1)(tanx+secx)= -cosx1 solutions Answer 438643 by lwsshak3(6460)   on 2013-02-13 02:54:42 (Show Source): You can put this solution on YOUR website!prove that (sinx-1)(tanx+secx)= -cosx start with left side: (sinx-1)(tanx+secx)=(sinx-1)[(sinx/cosx)+(1/cosx)] =(sinx-1)(sinx+1)/cosx) =(sin^2x-1)/cosx =-(1-sin^2x)/cosx =-cos^2x/cosx =-cosx verified: left side=right side
 Trigonometry-basics/714018: Use a table of trigonometric values to find the angle θ in the right triangle in the following problem. Round to the nearest degree, if necessary. sin θ = ? O = 8 H = 16 Please explain how to do this, this is one of a series of questions1 solutions Answer 438642 by lwsshak3(6460)   on 2013-02-13 02:41:25 (Show Source): You can put this solution on YOUR website!Use a table of trigonometric values to find the angle θ in the right triangle in the following problem. Round to the nearest degree, if necessary. sin θ = ? O = 8 H = 16 O=opposite side H=hypotenuse sin θ=O/H=8/16=1/2 From table of trig values: θ=30º
 Trigonometry-basics/713908: Evaluate the cot[arccos(5-x)]1 solutions Answer 438641 by lwsshak3(6460)   on 2013-02-13 02:35:16 (Show Source): You can put this solution on YOUR website!Evaluate the cot[arccos(5-x)] let A=adjacent side let O=opposite side H=hypotenuse .. cot[arccos(5-x)] This reads: cot of an angle whose cos=(5-x) cos(5-x)=A/H=(5-x)/1 A=(5-x) H=1 O=√(H^2-A^2) (Pythagorean Theorem) =√[1^2-(5-x)^2] =√[1-(25-10x+x^2)] O=√(-24+10x-x^2) cot[arccos(5-x)]=A/O=(5-x)/√(-24+10x-x^2)
 logarithm/712208: Use the definition of logarithm to simplify each expression. (a) log10^10 (b)log10 1,000 (c) log 10 ^10 -21 solutions Answer 438281 by lwsshak3(6460)   on 2013-02-11 01:51:26 (Show Source): You can put this solution on YOUR website!Use the definition of logarithm to simplify each expression. (a) log10^10=1 (b)log10 1,000=3 (c) log 10 ^10 -2=-2 .. Definition: The base raised to the log of the number=to the number
 Quadratic-relations-and-conic-sections/712313: What are the foci of the ellipse given by the equation 100x^2+64y^2=64,0001 solutions Answer 438278 by lwsshak3(6460)   on 2013-02-11 01:33:23 (Show Source): You can put this solution on YOUR website!What are the foci of the ellipse given by the equation 100x^2+64y^2=64,000 x^2/640+y^2/1000=1 This is an equation of an ellipse with vertical major axis. Its standard form: , a>b, (h,k)=(x,y) coordinates of center center:(0,0) a^2=1000 b^2=640 c^2=a^2-b^2=1000-640=360 c=√360≈18.97 Foci: (0, 0±c)=(0,0±18.97)=(0,-18.97) and (0,18.97)
 Quadratic-relations-and-conic-sections/712593: A hyperbolic mirror can be used to take panoramic photos, if the camera is pointed toward the mirror with the lens at one focus of the hyperbola. write the equation of the hyperbola that can be used to model a mirror that has a vertex 4 inches from the center of the hyperbola and a focus 1 inch in front of the surface of the mirror. assume the mirror has a horizontal trasverse axis and the hyperbola is centered at (0,0).1 solutions Answer 438275 by lwsshak3(6460)   on 2013-02-11 01:10:51 (Show Source): You can put this solution on YOUR website!A hyperbolic mirror can be used to take panoramic photos, if the camera is pointed toward the mirror with the lens at one focus of the hyperbola. write the equation of the hyperbola that can be used to model a mirror that has a vertex 4 inches from the center of the hyperbola and a focus 1 inch in front of the surface of the mirror. assume the mirror has a horizontal trasverse axis and the hyperbola is centered at (0,0). ** Standard form of equation for a hyperbola with horizontal transverse axis: , (h,k)=(x,y) coordinates of center For given problem: center: (0,0) vertices a=4 a^2=16 .. Foci c=5 c^2=25 .. c^2=a^2+b^2 b^2=c^2-a^2=25/16=9 .. Equation of hyperbola: