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# Recent problems solved by 'lwsshak3'

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 Quadratic-relations-and-conic-sections/721008: write the equation of a hyperbola in standard form whose center is (-2, -4), a focus at (-2, 6), and eccentricity of 5/4.1 solutions Answer 442227 by lwsshak3(6522)   on 2013-03-03 01:47:30 (Show Source): You can put this solution on YOUR website!write the equation of a hyperbola in standard form whose center is (-2, -4), a focus at (-2, 6), and eccentricity of 5/4. *** Given hyperbola has a vertical transverse axis. (From given center and focus data, y-coordinates change but x-coordinates do not.) Its standard form of equation: , (h,k)=(x,y) coordinates of center, For given hyperbola: center: (-2,-4) c=10 (-4 to 6) (distance from center to focus on vertical transverse axis) c^2=100 eccentricity=5/4=c/a a=4c/5=40/5=9 a^2=81 c^2=a^2+b^2 b^2=c^2-a^2=100-81=19 Equation of given hyperbola:
 Trigonometry-basics/721064: Find the values of x in the interval 0° ≤ x ≤ 360° for which sin(x) = cos(x)1 solutions Answer 442224 by lwsshak3(6522)   on 2013-03-03 01:16:06 (Show Source): You can put this solution on YOUR website!Find the values of x in the interval 0° ≤ x ≤ 360° for which sin(x) = cos(x) sin(45º)=cos(45º) (in quadrant I where both sin(x) and cos(x)=√2/2) sin(225º)=cos(225º) (in quadrant III where both sin(x) and cos(x)=-√2/2)
 Travel_Word_Problems/721138: How much time does a train 50m long moving at 65 km/hr take to pass another train 75 m long moving at 50km/hr in the same direction?What is the answer?1 solutions Answer 442223 by lwsshak3(6522)   on 2013-03-03 00:59:10 (Show Source): You can put this solution on YOUR website!How much time does a train 50m long moving at 65 km/hr take to pass another train 75 m long moving at 50km/hr in the same direction? travel time=distance/speed distance=50+75=125 m speed=65-50=15 km/hr travel time=.125/15*60=0.5 min ans:0.5 min
 Trigonometry-basics/721131: How fast would you have to travel on the surface of earth at the equator to keep up with the sun (that is, so that the sun would appear to remain in the same position in the sky)? Assume the radius of the earth at the equator is 3960 miles. Here is what I have so far, I believe that I am trying to solve for linear velocity (v=rw:radius times angular velocity. I think that w=2pi (or 365 degrees). and radius=3960miles. after that point I'm stuck.1 solutions Answer 442221 by lwsshak3(6522)   on 2013-03-03 00:27:28 (Show Source): You can put this solution on YOUR website!How fast would you have to travel on the surface of earth at the equator to keep up with the sun (that is, so that the sun would appear to remain in the same position in the sky)? Assume the radius of the earth at the equator is 3960 miles. *** The earth makes one revolution every 24 hours. Each revolution=2π*radius=7920π miles linear speed of earth=7920π miles/24 hours≈1037 mi/hr In order to make it appear the sun stays in the same position, one must travel at this speed against the rotation of the earth which is west to east. Traveling at this speed east to west at the equator in effect makes one stationary with respect to the sun. I think.
 Trigonometry-basics/720955: A wave is modeled with the function y=1/2 sin3x. Describe the graph of this function, including its period, amplitude, and points of intersection with the x axis. I have the amp. is 1/2 the period is (2 pi)/3 but don't know how to get the points of intersection. I know one is (0,0)1 solutions Answer 442158 by lwsshak3(6522)   on 2013-03-02 18:37:04 (Show Source): You can put this solution on YOUR website!A wave is modeled with the function y=1/2 sin3x. Describe the graph of this function, including its period, amplitude, and points of intersection with the x axis. I have the amp. is 1/2 the period is (2 pi)/3 but don't know how to get the points of intersection. I know one is (0,0) *** Basic equation for sin function: y=ASin(Bx-C), A=amplitude, Period=2π/B, phase shift=C/B. For given equation: y=1/2 sin3x Amplitude=1/2 B=3 Period=2π/B=2π/3 C=0 phase shift: none Because there is no phase shift, the graph is like the basic sin curve with amplitude=1/2 and period=2π/3 As with the basic sin curve, it intersects the x-axis at zero, half the period, and at the end of the period. So, for given function, points of intersection for one period are: (0,0), (π/3,0) and (2π/3,0)
 Trigonometry-basics/720972: Solve all equations for (0, 2pi) sin(x + (pi/4)) + sin(x - (pi/4)) = 1 1 solutions Answer 442149 by lwsshak3(6522)   on 2013-03-02 18:03:59 (Show Source): You can put this solution on YOUR website!Solve all equations for (0, 2pi) sin(x + (pi/4)) + sin(x - (pi/4)) = 1 *** Using sin addition identities: sin(x+π/4)=sinxcos(π/4)+cosxsin(π/4) sin(x-π/4)=sinxcos(π/4)-cosxsin(π/4) sinxcos(π/4)+cosxsin(π/4)+sinxcos(π/4)-cosxsin(π/4)=1 2sinxcos(π/4)=1 2sinx√2/2=1 sinx=1/√2=√2/2 x=π/4, 3π/4 (in quadrants I and II where sin>0)
 Trigonometry-basics/721023: Find the exact value of x between 0 degrees and 180 degrees for which each is true. cos x = sqrt(-3)/2 1 solutions Answer 442144 by lwsshak3(6522)   on 2013-03-02 17:40:03 (Show Source): You can put this solution on YOUR website!Find the exact value of x between 0 degrees and 180 degrees for which each is true. cos x = sqrt(-3)/2 I believe this should be written:cosx=-√3/2 For given domain, [0,180º], reference angle is in quadrant II where cos<0. cosx=-√3/2 x=150º
 Trigonometry-basics/721039: How can do you prove that secx - (tanx)(sinx) = cosx?1 solutions Answer 442140 by lwsshak3(6522)   on 2013-03-02 17:26:39 (Show Source): You can put this solution on YOUR website!How can do you prove that secx - (tanx)(sinx) = cosx? Start with left side: secx-(tanx)(sinx) =(1/cosx)-(sinx/cosx)sinx =(1/cosx)-sin^2x/cosx =(1-sin^2x)/cos) =cos^2x/cosx =cosx verified: left side=right side
 Quadratic-relations-and-conic-sections/719601: Find the coordinates of the vertex,focus,ends of the latus rectum and the equation of the directrix.Draw the parabola of: x^2=12(y+7)1 solutions Answer 442045 by lwsshak3(6522)   on 2013-03-02 02:37:14 (Show Source): You can put this solution on YOUR website!Find the coordinates of the vertex,focus,ends of the latus rectum and the equation of the directrix.Draw the parabola of: x^2=12(y+7) ... This is an equation of a parabola that opens upwards: Its basic form of equation:(x-h)^2=4p(y-k) For given equation: x^2=12(y+7) vertex: (0,-7) axis of symmetry: x=0 4p=12 p=3 focus: (0,-4) (p-units above vertex on the axis of symmetry) Ends of latus rectum(focal width): plugs in y-coordinate of focus(-4) then solve for x x^2=12(y+7) x^2=12(-4+7)=12*3=36 x=±√36=±6 ends of latus rectum: (-6,-4) and (6,-4) see graph below:
 Trigonometry-basics/720476: if the point (-2,-7) is on the terminal arm of N, an angle in standard position, what is the value of cos N?1 solutions Answer 442038 by lwsshak3(6522)   on 2013-03-02 01:47:11 (Show Source): You can put this solution on YOUR website!if the point (-2,-7) is on the terminal arm of N, an angle in standard position, what is the value of cos N? *** The value of angles in standard positions is measured by their reference angles. The reference angle is the angle the terminal arm makes with the horizontal x-axis The reference angle is in quadrant III where tan>0 The x-value(-2) represents the horizontal(cos) leg of the reference right triangle The y-value(-7) represents the vertical(sin) leg of the reference right triangle The hypotenuse=√(2^2+7^2)=√(4+49)=√53 cosN=-2/√53≈-0.2747
 Trigonometry-basics/720575: Not Urgent, Just VERY Curious Hello, I couldn't find the correct section for my question. I am having difficulty with a certian concept in my math 170 class. We were recently taught about building functions, and specifically how to express the area of a rectangle as a function of x. In the example we are shown that a rectangle with its corners on origin, positive x, positive y, and a point in quadrant I on the graph y = 25 -x^2. I get all that, I can crunch the numbers, and come up with the right answers, but I have no idea what this function represents. It later says that the maximum area is 48.11 square units at x=2.89 units. What is x representing here? What is this area related to? How does this information corrospond to a practical application of mathematics? I am very sorry if I'm wasting your time, but I am on the verge of chewing my own fingernails off. I have to know what this actualy means. I've tried using the internet, and I keep getting function problems, not explainations. Thank you for your time, and have a wonderful day.1 solutions Answer 442037 by lwsshak3(6522)   on 2013-03-02 01:15:18 (Show Source): You can put this solution on YOUR website!y = 25 -x^2 I will give it a try: The given equation is that of a parabola that open downward (curve has a maximum). Its standard form: y=A(x-h)^+k, (h,k)=(x,y) coordinates of the vertex. For given equation: y=25-x^2 rewrite: y=-x^2+25 Since you don't see h, the x-coordinate of the vertex is=0 The y-coordinate of the vertex is=k=25 which is also the maximum value for that particular parabolic function. A which is assumed=1, affects the slope of steepness of the curve. ... As for the maximum area is 48.11 square units at x=2.89 units, this must represent a different curve from the one above. If it is a parabola it would be like: y=-(x-2.89)^2+48.11, that is, coordinates of the vertex would be (2.89,48.11) ... In math and physics applications of the parabola is quite common especially in higher math courses like calculus. To get a better understanding of what the different curves look like, get a graphics calculator and punch in the equation to display the curve. Hope this helps!
 Trigonometry-basics/720839: Write the following in terms of sin θ and cos θ; then simplify if possible. (Leave your answer in terms of sin θ and/or cos θ.) sin theta cot theta + 7 cos theta1 solutions Answer 442036 by lwsshak3(6522)   on 2013-03-02 00:33:53 (Show Source): You can put this solution on YOUR website!Write the following in terms of sin θ and cos θ; then simplify if possible. (Leave your answer in terms of sin θ and/or cos θ.) sin theta cot theta+7 cos theta .. sin θ*cot θ+7cos θ sin θ*(cos θ/sin θ)+7cos θ cos θ+7cos θ=8cos θ
 Trigonometry-basics/720868: A pick-up truck is fitted with new tires which have a diameter of 44 inches. How fast will the pick up truck be moving when the wheels are rotating at 285 revolutions per minute?1 solutions Answer 442034 by lwsshak3(6522)   on 2013-03-02 00:26:22 (Show Source): You can put this solution on YOUR website!A pick-up truck is fitted with new tires which have a diameter of 44 inches. How fast will the pick up truck be moving when the wheels are rotating at 285 revolutions per minute? *** circumference=π*diameter=44π inches speed of truck=285 rev/min*44π inches/rev=44π*285=1254π inches/min
 Trigonometry-basics/720869: sec3x = sqrt2 solve for [0,2pi]1 solutions Answer 442033 by lwsshak3(6522)   on 2013-03-02 00:17:51 (Show Source): You can put this solution on YOUR website!sec3x = sqrt2 solve for [0,2pi] sec3x = √2=1/cos3x cos3x=1/√2=√2/2 3x=π/4, 5π/4 x=π/12, 5π/12
 Trigonometry-basics/720870: tan3x = sqrt 3 solve for [0,2pi]1 solutions Answer 442032 by lwsshak3(6522)   on 2013-03-02 00:12:48 (Show Source): You can put this solution on YOUR website!tan3x = sqrt 3 solve for [0,2pi] tan 3x=√3 3x=π/3, 4π/3 x=π/9, 4π/9
 Trigonometry-basics/720867: 2sin^2x-5sinx=-31 solutions Answer 442031 by lwsshak3(6522)   on 2013-03-02 00:06:53 (Show Source): You can put this solution on YOUR website!2sin^2x-5sinx=-3 2sin^2x-5sinx+3=0 (2sinx-3)(sinx-1)=0 2sinx-3=0 sinx=3/2 (reject, -1 ≤ sinx ≤ 1) or sinx-1=0 sinx=1 x=π/2
 Quadratic-relations-and-conic-sections/720102: how can I solve this use eccentricity of each ellipse to find its equation is standrad forom eccentricity 2/5 , major axis on the x-axis and of lenth 10 center (0,0)1 solutions Answer 441984 by lwsshak3(6522)   on 2013-03-01 19:05:41 (Show Source): You can put this solution on YOUR website!how can I solve this use eccentricity of each ellipse to find its equation is standrad forom eccentricity 2/5 , major axis on the x-axis and of lenth 10 center (0,0) *** Standard form of equation for an ellipse with horizontal major axis: , (a>b), (h,k) = (x,y) coordinates of the center For given problem: center: (0,0) length of horizontal major axis=10=2a a=5 a^2=25 eccentricity=c/a=2/5 c=2a/5=10/5=2 c^2=4 c^2=a^2-b^2 b^2=a^2-c^2=25-4=21 Equation of given ellipse:
 logarithm/719738: write x in terms of a,b and c log x=2log a-(3logb(1/2)log c)1 solutions Answer 441983 by lwsshak3(6522)   on 2013-03-01 18:50:10 (Show Source): You can put this solution on YOUR website!write x in terms of a,b and c log x=2log a-(3logb(1/2)log c) log x=log[a^2/b^3*c^(1/2)]
 logarithm/720805: Express as a single logarithm with a coefficient of 1: 3*log(x)-2*log(y)+4*log(z)1 solutions Answer 441981 by lwsshak3(6522)   on 2013-03-01 18:44:32 (Show Source): You can put this solution on YOUR website!Express as a single logarithm with a coefficient of 1: 3*log(x)-2*log(y)+4*log(z) log[(x^3*z^4)/y^2]
 logarithm/720806: Solve: log5x-log5(x-2)=log5(4)1 solutions Answer 441980 by lwsshak3(6522)   on 2013-03-01 18:40:48 (Show Source): You can put this solution on YOUR website!Solve: log5x-log5(x-2)=log5(4) log5[x/(x-2)]=log5(4) x/(x-2)=4 x=4x-8 3x=8 x=8/3
 logarithm/720721: write the expression 2logx-(3logY+logz) as a single log1 solutions Answer 441973 by lwsshak3(6522)   on 2013-03-01 18:09:41 (Show Source): You can put this solution on YOUR website!write the expression 2logx-(3logY+logz) as a single log log[(x/(y^3)z]
 Exponential-and-logarithmic-functions/720760: 2/log(base x)4 + 3/log(base 4)x = 71 solutions Answer 441969 by lwsshak3(6522)   on 2013-03-01 18:06:27 (Show Source): You can put this solution on YOUR website!2/log(base x)4 + 3/log(base 4)x = 7 convert to Base 10 2/(log4/logx)+3/(logx/log4)=7 2logx/log4+3log4/logx=7 LCD:(logx)(log4) 2(logx)^2+3(log4)^2=7log4logx log4≈.60206 2(logx)^2+1.08743=4.21442logx 2(logx)^2-4.21442logx+1.08743=0 solve for logx using quadratic formula: logx=.30102 x=10^.30102 x≈2 or logx=1.80618 x=10^1.80618 x≈64 .. Check: for x=2 2/log(base x)4+3/log(base 4)x =2/log2(4) + 3/log4(2) =2/2+3/(1/2) =1+6=7 .. for x=64 2/log(base x)4+3/log(base 4)x =2/log64(4) + 3/log4(64) =2/(1/3)+3/3 =6+1=7 ans: x=2 or 64
 Trigonometry-basics/720477: Find the exact value of the trigonometric functions. a. 13pi/2 I have 13pi/2*180/pi=1170-1080= 90 degrees= pi/2 b.(-11pi/2) I have -11pi/2*180/pi=990-720= 270 degrees= 3pi/2 Am I on the right track with my answers?1 solutions Answer 441895 by lwsshak3(6522)   on 2013-03-01 03:00:24 (Show Source): You can put this solution on YOUR website!Find the exact value of the trigonometric functions. a. 13pi/2=6π+π/2=π/2 This is like rotating 3 revolutions (6π) counter-clockwise +π/2 b.(-11pi/2)=5π+π/2=4π+π+π/2=π/2 This is like rotating 2 revolutions (4π) clockwise+π+π/2
 Trigonometry-basics/720499: How do you solve 2sinx cosx = -sinx?1 solutions Answer 441894 by lwsshak3(6522)   on 2013-03-01 02:38:15 (Show Source): You can put this solution on YOUR website!How do you solve 2sinx cosx = -sinx? divide both sides by sinx 2cosx=-1 cosx=-1/2 x=2π/3 or 4π/3 (in quadrants II and III where cos<0)
 Trigonometry-basics/720502: Sec^2x=41 solutions Answer 441893 by lwsshak3(6522)   on 2013-03-01 02:30:27 (Show Source): You can put this solution on YOUR website!Sec^2x=4 secx=2=1/cosx cosx=1/2 x=π/3 or 60º
 Trigonometry-basics/720579: sin(8pi)=? 1 solutions Answer 441891 by lwsshak3(6522)   on 2013-03-01 01:59:42 (Show Source): You can put this solution on YOUR website!sin(8pi)=? on a unit circle, 8π is like making 4 revolutions counter-clockwise starting from zero and returning to zero. So, sin(8π)=0
 Trigonometry-basics/719857: Find the exact value of each of the following under the given conditions: 1 solutions Answer 441890 by lwsshak3(6522)   on 2013-03-01 01:55:11 (Show Source): You can put this solution on YOUR website!Find the exact value of each of the following under the given conditions: *** let o=opposite side let a=adjacent side let h=hypotenuse .. π/2 < A < π (quadrant II) cotA=-24/7=a/o a=-24, o=7 h=√(o^2+24^2)=√(49+576)=√625=25 sinA=o/h=7/25 cosA=a/h=-24/25 tanA=o/a=-7/24 ... 0 < B < π/2 (quadrant I) cosB=5/6=a/h a=5, h=6 o=√(h^2-a^2)=√(36-25)=√11 sinB=o/h=√11/6 cosB=a/h=5/6 tanB=o/a=√11/5 ... Identity: sin(A+B)=sinAcosB+cosAsinB =7/25*5/6+(-24/25*√11/6) =35/150-24√11/150=(35-24√11)/150 ... Identity: sin(A-B)=sinAcosB-cosAsinB =7/25*5/6-(-24/25*√11/6) =35/150+24√11/150 =(35+24√11)/150 ... Identity:cos(A-B)=cosAcosB+sinAsinB =-24/25*5/6+7/25*√11/6 =-120/150+7√11/150 =-120+7√11/150 ... Identity:tan(A+B)=(tanA+tanB)/(1-tanAtanB) =(-7/24+√11/5)/(1-(-7/24*√11/5) =(-7/24+√11/5)/(1+7√11/120)
 Trigonometry-basics/720378: find the exact value of the remaining trigonometric functions: 1. cotθ = 12/5, cosθ < 0 2. cosθ = -3/5, θ in quadrant III1 solutions Answer 441831 by lwsshak3(6522)   on 2013-02-28 19:55:41 (Show Source): You can put this solution on YOUR website!find the exact value of the remaining trigonometric functions: 1. cotθ = 12/5, cosθ < 0 let O=opposite side let A=adjacent side H=hypotenuse .. cot>0 and cos<0 only in quadrant III cotθ = 12/5=A/O A=-12, O=-5 H=√(O^2+A^2)=√(144+25)=√169=13 sinθ=-5/13 cosθ=-12/13 tanθ=5/12 cscθ=-13/5 secθ=-13/12 cotθ=12/5 .. 2. cosθ = -3/5, θ in quadrant III sin and cos<0 cosθ=-3/5=A/H A=-3, H=5 O=√(H^2-A^2)=√(25-9)=√16=-4 sinθ=-4/5 cosθ=-3/5 tanθ=4/3 cscθ=-5/4 secθ=-5/3 cotθ=3/4
 test/720433: I must find the horizontal and vertical asymptotes of : f(x) = x/ x^2 - 91 solutions Answer 441822 by lwsshak3(6522)   on 2013-02-28 19:22:25 (Show Source): You can put this solution on YOUR website!I must find the horizontal and vertical asymptotes of : f(x) = x/ x^2 - 9 .. When the degree of the numerator is 1 less than the degree of the denominator, as in this case, the horizontal asymptote is the x-axis or y=0. Vertical asymptotes are found by setting the denominator=0, then solving for x-values which makes the function undefined. x^2-9=0 (x+3)(x-3)=0 x≠3 and x≠-3 horizontal asymptotes: x-axis Vertical asymptotes: x=3, x=-3)
 absolute-value/720340: |x-9|=x^2-9x1 solutions Answer 441769 by lwsshak3(6522)   on 2013-02-28 16:53:56 (Show Source): You can put this solution on YOUR website!|x-9|=x^2-9x solve for 2 possibilities: for (x-9)>0 x-9=x^2-9x x^2-10x+9=0 (x-9)(x-1)=0 x=9 or x=1 .. for (x-9)<0 -x+9=x^2-9x x^2-8x-9=0 (x-9)(x+1)=0 x=9 or x=-1 Check: for x=9 |x-9|=x^2-9x 0=81-81 0=0 (x=9 ok) .. for x=1 |x-9|=x^2-9x |1-9|=1-9 8≠-8 (reject x=1) .. for x=-1 |x-9|=x^2-9x |-1-9|=x^2-9x 10=10 (x=-1 ok) solution: x=9,-1
 Radicals/720317: Here is my problem. [SQRT(x + 7)] - 2[SQRT(x)] =-2 OR √(x + 7) - 2√(x) = -2 ------- Now, the steps I took to solve this problem are to first square both sides: [√(x + 7) - 2√(x)] * [√(x + 7) - 2√(x)] = 4 so then I FOIL the left side, resulting in: (x+7) - [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] + 4x So then I thought to subtract (x+7) and (4x) to both sides - [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] = 4 - 4x - x - 7 (I think I'm supposed to switch the sign, because I've subtracted it and moved it to the opposite side, right?) I think I'm correct up to this point, but now I have to square both sides again. I think this left hand side could be re-written as: -2[√(x + 7) * - 2√(x)] Is this right? I'm subtracting it from itself, a negative, which could simply multiplied by -2. Anyway, now I need to square this again, so I assume the -2 becomes a 4 and I FOIL them separately? FOILING the left side will get: [√(x + 7) * - 2√(x)] * [√(x + 7) * - 2√(x)] Which, when FOILed, looks like (x+7) - [√(x+7) * -2√(x)] - [ - 2√(x) * √(x + 7)] + 4x It looks exactly the same as before!! I'm just really confused by this problem, and I have a couple more like it, so I want to know if figuring this one out could help me solve the other ones. I'm confused about FOILing the different sides, whether or not I can combine two square roots, and quite frankly, a lot of other things. One of the options on the test is 9, and I think this is the answer, because I've inserted it into the original equation and it works, but I'm just confused about how to actually get 9 out of this.. Sorry for the long question. I hope it's not hard to understand. I'm just hoping someone can walk me through all the steps of solving a problem like this so I can do it easily in the future.1 solutions Answer 441762 by lwsshak3(6522)   on 2013-02-28 16:30:49 (Show Source): You can put this solution on YOUR website!√(x + 7) - 2√(x) = -2 √(x+7)=2√x-2 square both sides x+7=4x-8√x+4 -3x+3=-8√x -3(x-1)=-8√x square both sides again 9(x-1)^2=64x 9(x^2-2x+1)=64x 9x^2-18x+9-64x=0 9x^2-82x+9=0 solve for x using quadratic formula: x≈0.1111..(reject, extraneous root) x=9