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# Recent problems solved by 'lwsshak3'

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 Exponential-and-logarithmic-functions/736276: 1/2logx+log4=21 solutions Answer 449758 by lwsshak3(6505)   on 2013-04-10 20:44:11 (Show Source): You can put this solution on YOUR website!1/2logx+log4=2 log[x^(1/2)(4)]=log(100) 4√x=100 square both sides 16x=10^4=10000 x=10000/16 x=625
 logarithm/735534: 9000 = 5000 (1+r/360)^1440 How do I get the r out of the equation?1 solutions Answer 449614 by lwsshak3(6505)   on 2013-04-10 03:14:00 (Show Source): You can put this solution on YOUR website!9000 = 5000 (1+r/360)^1440 divide both sides by 1000 9=5(1+r/360)^1440 9/5=(1+r/360)^1440 1.8=(1+r/360)^1440 raise both sides by 1/1440 1.8^(1/1440)=1+r/360 1.8^(1/1440)=(360+r)/360 360*1.8^(1/1440)=(360+r) r=360*1.8^(1/1440)-360
 logarithm/735703: Write an equivalent exponential or logarithmic function. e^-2=x^61 solutions Answer 449613 by lwsshak3(6505)   on 2013-04-10 03:01:47 (Show Source): You can put this solution on YOUR website!e^-2=x^6 -2lne=6lnx 6ln(x)=-2 ln(x)=-2/6=-1/3 x=e^(-1/3)
 logarithm/735809: Log5x-Log(2x-1)=Log71 solutions Answer 449612 by lwsshak3(6505)   on 2013-04-10 02:56:50 (Show Source): You can put this solution on YOUR website!Log5x-Log(2x-1)=Log7 log[5x/(2x-1)]=log(7) 5x/(2x-1)=7 5x=14x-7 9x=7 x=7/9
 logarithm/735947: e^(2x)-2e^x+1 =0 Solve:1 solutions Answer 449611 by lwsshak3(6505)   on 2013-04-10 02:41:53 (Show Source): You can put this solution on YOUR website!e^(2x)-2e^x+1 =0 Solve: let u=e^x u^2=e^(2x) .. u^2-2u+1=0 (u-1)^2=0 u=1 (mult. 2) u=e^x=1 x=0
 Trigonometry-basics/735889: i need help on this problem. I am stuck mid way through the problem. Find all solutions in the interval [0,2pi] or [0,360degrees]. may pick degrees or radians. give exact value. sin(3x + pi)+ 1/2 = 1 Thank you1 solutions Answer 449592 by lwsshak3(6505)   on 2013-04-09 22:30:30 (Show Source): You can put this solution on YOUR website!Find all solutions in the interval [0,2pi] or [0,360degrees]. may pick degrees or radians. give exact value. sin(3x+π)+ 1/2 = 1 sin(3x+π)=-1/2 3x+π=inverse sin(1/2)=π/6 3x=π/6-π=-5π/6 x=-5π/18 radians in (quadrant IV) check: sin(3x+π)+ 1/2 =sin((-15π/18)+(18π/18))+1/2 =sin(3π/18)+1/2 =sin(π/6)+1/2 =1/2+1/2 =1 .. note: domain of inverse sin: [-π/2, π/2]
 Trigonometry-basics/735714: How do you solve 7=5+sec(4x+7pi/4) 5+csc(2x+pi/3)=(15+2(sqroot3)/3) Thanks a lot!1 solutions Answer 449531 by lwsshak3(6505)   on 2013-04-09 17:33:35 (Show Source): You can put this solution on YOUR website!How do you solve 7=5+sec(4x+7pi/4) 5+csc(2x+pi/3)=(15+2(sqrt(3)/3) *** 7=5+sec(4x+7pi/4) 7-5=sec(4x+7pi/4) 2=1/cos(4x+7pi/4) cos(4x+7pi/4)=1/2 LCD:12 ... 5+csc(2x+pi/3)=(15+2(sqrt(3)/3) csc(2x+pi/3)=15+2(sqrt(3)/3)-5 csc(2x+pi/3)=10+2(sqrt(3)/3)≈11.1547 sin(2x+pi/3)=1/11.1547≈0.0896 2x+pi/3=inverse sin(0.0896)≈0.0897 2x=0.0897-pi/3≈-0.9575 x≈-0.4787 radians
 Exponential-and-logarithmic-functions/735676: Graph the Exponential Function f(x)= 4^x+2 x+2 is the exponent of the 4.1 solutions Answer 449530 by lwsshak3(6505)   on 2013-04-09 16:07:14 (Show Source): You can put this solution on YOUR website!Graph the Exponential Function f(x)= 4^(x+2) x+2 is the exponent of the 4. *** Start with 4^x (red curve) where you have the x-axis as a horizontal asymptote and a y-intercept at (0,1) 4^(x+2) then shifts the curve 2 units to the left.(green curve) See graphs below of these two curves:
 Money_Word_Problems/735670: how long does it take for \$900 to double if it is invested at 7% compound continuously?1 solutions Answer 449526 by lwsshak3(6505)   on 2013-04-09 15:49:33 (Show Source): You can put this solution on YOUR website!how long does it take for \$900 to double if it is invested at 7% compound continuously? Formula for continuous compounding: A=Pe^rt, P=initial amt, r=interest rate, A=amt after t years For given problem: A/P=2 r=.07 e^rt=2 take log of both sides rt*lne=ln2 lne=1 rt=ln2 t=ln2/r=ln2/.07≈9.9≈10 years At 7% compound continuously,\$900 to double in approximately 10 yrs
 Graphs/735647: How many quarts of water must be added to 40 quarts of solution which is 5 percent acid solution to dilute it to a 2 percent solution? Thanks!1 solutions Answer 449516 by lwsshak3(6505)   on 2013-04-09 15:32:29 (Show Source): You can put this solution on YOUR website!How many quarts of water must be added to 40 quarts of solution which is 5 percent acid solution to dilute it to a 2 percent solution? *** let x=amt of water to be added 40+x=amt of final mixture .. 5%*40+0%x=2%(40+x) 2+0=0.8+.02x .02x=1.2 x=60 amt of water to be added=60 quarts
 Graphs/735653: A car travels 400 miles in 7 hours. Before noon the driver averages 60 miles per hour, but after noon he averages only 50 miles per hour. At what time did he leave?1 solutions Answer 449513 by lwsshak3(6505)   on 2013-04-09 15:20:23 (Show Source): You can put this solution on YOUR website!A car travels 400 miles in 7 hours. Before noon the driver averages 60 miles per hour, but after noon he averages only 50 miles per hour. At what time did he leave? *** let x=distance traveled before noon 400-x=distance traveled after noon Travel time=distance/speed .. travel time before noon plus travel time after noon=7 hrs 50x+24000-60x=7*50*60 -10x+24000=21000 10x=3000 x=300 travel time before noon=x/60=300/60=5 hrs At what time did he leave? 5 hrs before noon=7 am
 Quadratic-relations-and-conic-sections/734909: Find an asymptote of this conic section.. 9x^2-36x-4y^2+24y-36=01 solutions Answer 449456 by lwsshak3(6505)   on 2013-04-09 04:33:05 (Show Source): You can put this solution on YOUR website!Find an asymptote of this conic section.. 9x^2-36x-4y^2+24y-36=0 complete the square: 9(x^2-4x+4)-4(y^2-6y+9)=36+36-36 9(x-2)^2-4(y-3)^2=36 This is an equation of a hyperbola with horizontal transverse axis: Its standard form of equation: , (h,k)=(x,y) coordinates of center For given equation: center: (2,3) a^2=4 a=2 b^2=9 b=3 Slopes of asymptotes for hyperbolas with horizontal transverse axis=±b/a=±3/2 Asymptotes are equations of straight lines that go thru the center of the hyperbola: y=mx+b, m=slope, b=y-intercept .. Equation of asymptote with negative slope, m=-3/2 y=-3x/2+b solve for b using coordinates of the center (2,3) 3=-3*2/2+b b=6 equation:y=-3x/2+6 .. Equation of asymptote with positive slope, m=3/2 y=3x/2+b solve for b using coordinates of the center (2,3) 3=3*2/2+b b=0 equation:y=3x/2 .. Equations of asymptotes: y=-3x/2+6 and y=3x/2
 Quadratic-relations-and-conic-sections/735390: What can I do to solve for an from just a vertex at a origin and with the directrix of y= -3/4?1 solutions Answer 449455 by lwsshak3(6505)   on 2013-04-09 03:41:46 (Show Source): You can put this solution on YOUR website!solve for an equation from just a vertex at a origin and with the directrix of y= -3/4? *** Given directrix (-3/4) shows this is a parabola that opens downward: Basic form of equation for such a parabola with vertex at (0,0): x^2=-4py axis of symmetry:x=0 p=3/4 (distance from vertex to directrix on the axis of symmetry) 4p=3 equation: x^2=-3y
 Quadratic-relations-and-conic-sections/735422: I have a question on Hyperbolas. This is what I think I know so far, Orientation=Horizontal, Center=(0,0), a=7, b=5,c=8.602 The equation is 1 solutions Answer 449453 by lwsshak3(6505)   on 2013-04-09 03:27:51 (Show Source): You can put this solution on YOUR website! This is an equation of a hyperbola with horizontal transverse axis. Its standard form:,(h,k)=(x,y) coordinates of center. For given equation: center:(0,0) a^2=25 a=5 b^2=49 b=7 c^2=a^2+b^2=25+49=74 c=√74≈8.60 You got everything right except a and b which you had in reverse. For hyperbolas, a and b don't change positions like in ellipses For hyperbolas with vertical transverse axis, the y-term is placed ahead of the x-term, but a and b remain in the same position as that for a hyperbola with horizontal transverse axis
 Trigonometry-basics/735099: If f(x)= cos(2x)-3sin(4x), find the exact value of f'(PIE/6) The first step I thought of was cos(2A)= 1-Sin^2A because you'll need to make the cos into sin for the equation to work. Thanks.1 solutions Answer 449449 by lwsshak3(6505)   on 2013-04-09 03:05:25 (Show Source): You can put this solution on YOUR website!If f(x)= cos(2x)-3sin(4x), find the exact value of f'(PIE/6) *** For[0,2π] f(x)= cos(2x)-3sin(4x) f(π/6)=cos(2π/6)-3sin(4π/6) .. cos(2π/6)=cos(π/3)=1/2 sin(4π/6)=sin(2π/3)=√3/2 (in Q2 where sin>0) .. f(π/6)=cos(2π/6)-3sin(4π/6) =cos(π/3)-3sin(2π/3) =1/2-3√3/2 =(1-3√3)/2
 Trigonometry-basics/735214: Solve the equation on the given interval, expressing the solution for x in terms of inverse trigonometric functions. (Enter your answers as a comma-separated list.) 7(cos x)^2 − cos x − 6 = 0 on [π/2, π] Thank you!1 solutions Answer 449447 by lwsshak3(6505)   on 2013-04-09 02:45:22 (Show Source): You can put this solution on YOUR website!Solve the equation on the given interval, expressing the solution for x in terms of inverse trigonometric functions. (Enter your answers as a comma-separated list.) 7(cos x)^2-cos x−6=0 on [π/2, π] .. 7(cos x)^2-cos x−6=0 (7cosx+6)(cosx-1)=0 .. 7cosx+6=0 7cosx=-6 cosx=-6/7 x=arccos(-6/7) .. cosx-1=0 cosx=1 x=arccos(1) .. solutions: x=arccos(-6/7), arccos(1)
 Trigonometry-basics/735468: verify the identity cos^2 B-sin^2 B=1-2sin^2 B1 solutions Answer 449446 by lwsshak3(6505)   on 2013-04-09 02:32:14 (Show Source): You can put this solution on YOUR website!verify the identity cos^2 B-sin^2 B=1-2sin^2 B start with left side: cos^2 B-sin^2 B 1-sin^2B-sin^2B=1-2sin^2B verified: left side=right side
 Trigonometry-basics/735475: find all solutions to the equations in the interval [0,2pie) sin^2x-1=01 solutions Answer 449445 by lwsshak3(6505)   on 2013-04-09 02:26:20 (Show Source): You can put this solution on YOUR website!sin^2x-1=0 sin^2x=1 take sqrt of both sides sinx=1 x=π/2
 Trigonometry-basics/735476: verify all solutions to the equations in the interval [0,2pie) 2sin^2x=2+cosx1 solutions Answer 449444 by lwsshak3(6505)   on 2013-04-09 02:21:59 (Show Source): You can put this solution on YOUR website!verify all solutions to the equations in the interval [0,2pie) 2sin^2x=2+cosx 2(1-cos^2x)=2+cosx 2-2cos^2x=2+cosx 2cos^2x+cosx=0 cosx(2cosx+1)=0 .. cosx=0 x=1 .. 2cosx+1=0 cosx=-1/2 x=2π/3, 4π/3 (in Q2 and Q3 where cos<0) .. solutions: x=1, 2π/3, 4π/3
 Trigonometry-basics/735477: find all solutions to the equations [0,2pie) 2sin^2x+3sinx+1=01 solutions Answer 449442 by lwsshak3(6505)   on 2013-04-09 02:12:22 (Show Source): You can put this solution on YOUR website!find all solutions to the equations [0,2pie) 2sin^2x+3sinx+1=0 (2sinx+1)(sinx+1)=0 .. 2sinx+1=0 sinx=-1/2 x=7π/6, 11π/6 (in Q3 and Q4 where sin<0) .. sinx+1=0 sinx=-1 x=3π/2 solutions: x=7π/6, 11π/6, 3π/2
 Trigonometry-basics/735272: Use the trigonometric function values to evaluate. cos90° + 3sin270° 1 solutions Answer 449370 by lwsshak3(6505)   on 2013-04-08 17:29:01 (Show Source): You can put this solution on YOUR website!Use the trigonometric function values to evaluate. cos90° + 3sin270° ** cos90º=0 3sin270º=-1*3=-3 cos90°+3sin270°=0+(-3)=-3
 Trigonometry-basics/735275: Determine the signs of the trigonometric functions for 218°. sin___; cos___; tan___; csc___; sec___; cot___ 1 solutions Answer 449369 by lwsshak3(6505)   on 2013-04-08 17:23:24 (Show Source): You can put this solution on YOUR website!Determine the signs of the trigonometric functions for 218° This places the angle 218º in Q3 where: sin<0 cos<0 tan>0 csc<0 sec<0 cot>0
 Trigonometry-basics/735279: sin2theta-costheta=01 solutions Answer 449368 by lwsshak3(6505)   on 2013-04-08 17:17:15 (Show Source): You can put this solution on YOUR website!For: [0,π/2] sin2theta-costheta=0 sin2x-cosx=0 2sinxcosx-cosx=0 cosx(2sinx-1)=0 .. cosx=0 x=1 .. 2sinx-1=0 sinx=1/2 x=π/6
 Trigonometry-basics/735271: Sketch an angle θ in standard position and find the values of the six trig functions for the angle. (-4,3) 1. sin__ cos__ tan__ csc__ sec__ cot__1 solutions Answer 449364 by lwsshak3(6505)   on 2013-04-08 17:10:31 (Show Source): You can put this solution on YOUR website!Sketch an angle θ in standard position and find the values of the six trig functions for the angle. (-4,3) *** Given coordinates places θ in Q2 where sin>0, cos<0, tan<0. We are working with a (3-4-5) reference right triangle. sin θ=3/5 cos θ=-4/5 tan θ=-3/4 csc θ=5/3 sec θ=-5/4 cot θ=-4/3
 Polygons/735266: The diagonal of a square is 56 centimeters long. Find the perimeter of the square to the nearest tenth.1 solutions Answer 449356 by lwsshak3(6505)   on 2013-04-08 16:54:55 (Show Source): You can put this solution on YOUR website!The diagonal of a square is 56 centimeters long. Find the perimeter of the square to the nearest tenth. *** let x=side of the square The diagonal and two of the sides form a right triangle By the pythagorean theorem, perimeter of the square=4x=158.4 cm
 Trigonometry-basics/734534: Angular velocity : a trucker drives 55 miles per hour. His truck's tires have a diameter of 26 inches. What is the angular velocity of the wheels in revolutions per second.1 solutions Answer 449269 by lwsshak3(6505)   on 2013-04-08 04:40:08 (Show Source): You can put this solution on YOUR website!Angular velocity : a trucker drives 55 miles per hour. His truck's tires have a diameter of 26 inches. What is the angular velocity of the wheels in revolutions per second. *** 1 revolution(rev)=circumference of truck's tires=π*diameter=π*26 inches cancel out:mi,hr,ft,inches
 Trigonometry-basics/735057: if sin s = -1/4 and s is in quadrant 4 and if cos t = -2/5 and t is in quadrant 2, find the following: cos(s + t) sin(s - t)1 solutions Answer 449268 by lwsshak3(6505)   on 2013-04-08 03:54:13 (Show Source): You can put this solution on YOUR website!if sin s = -1/4 and s is in quadrant 4 and if cos t = -2/5 and t is in quadrant 2, find the following: cos(s + t) sin(s - t) *** let O=opposite side of reference right triangle let A=adjacent side of reference right triangle let H=hypotenuse of reference right triangle sin s=-1/4 (in Q4)=O/H O=-1, H=4 .. cos t = -2/5(in Q2)=A/H A=-2, H=5 .. cos(s + t)=cos s cos t-sin s sin t =√15/4*-2/5-(-1/4)*√21)/5 =-2√15/20+√21/20 =(√21-2√15)/(20) .. sin(s-t)=sin s cos t-cos s sin t =-1/4*-2/5-√15/4*√21/5 =2/20-√315/20 =(2-√315)/20 .. Check with calculator: sin(s)=-1/4(in Q4) s≈345.52º cos(t)=-2/5 (in Q2) t≈113.58 .. s+t≈459.1 reference angle=80.9º(in Q2 where cos<0) Cos(s+t)=cos(99.1)≈-0.1581.. (√21-2√15)/20≈-0.1581.. .. (s-t)≈231.94 reference angle≈51.94 (in Q3 where sin<0) sin(s-t)=sin(51.94)≈-0.7874... (2-√315)/20≈-0.7874...
 Trigonometry-basics/733798: A pulley with a radius of 8 inches rotates three times every five seconds. Find the angular velocity of the pulley in radians/sec (round to the nearest hundredth). Find the linear velocity to the nearst ft/hr.1 solutions Answer 449081 by lwsshak3(6505)   on 2013-04-07 04:09:09 (Show Source): You can put this solution on YOUR website!A pulley with a radius of 8 inches rotates three times every five seconds. Find the angular velocity of the pulley in radians/sec (round to the nearest hundredth). Find the linear velocity to the nearst ft/hr. *** let C=circumference of 8 inch pulley cross out: rev, sec, inches ..
 Trigonometry-basics/734052: if cotx + cosecx = 1.5, then prove that cosx = 5/131 solutions Answer 449079 by lwsshak3(6505)   on 2013-04-07 03:01:29 (Show Source): You can put this solution on YOUR website!if cotx + cosecx = 1.5, then prove that cosx = 5/13 let O=opposite side let A=adjacent side let H=hypotenuse .. cosx=5/13=A/H A=5, H=13 This proves that if cotx+cscx=1.5, cosx = 5/13, because both equations come from the same 5-12-13 right triangle
 Trigonometry-basics/734621: Find all solutions in the interval {0,2pi)or {0,360 degree). you may us degrees or radians, your choice. 3sec^2(x)tan(x)=4tan(x) 1 solutions Answer 449077 by lwsshak3(6505)   on 2013-04-07 02:22:15 (Show Source): You can put this solution on YOUR website! take sqrt of both sides x=30º, 330º (in Q1 and Q4 where cos>0)