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# Recent problems solved by 'lwsshak3'

lwsshak3 answered: 6458 problems
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 Angles/604641: if angle b is 30 degrees more than twice its supplement then what is angle b1 solutions Answer 381322 by lwsshak3(6460)   on 2012-04-28 15:43:36 (Show Source): You can put this solution on YOUR website!if angle b is 30 degrees more than twice its supplement then what is angle b ** given angle=b Its supplement:(181-b) b=2(180-b)+30 b=360-2b+30 3b=390 b=130º
 Miscellaneous_Word_Problems/604644: One number is 1 less than a second number. Twice the second number is 31 less than 5 times the first. Find the smaller of two numbers.1 solutions Answer 381321 by lwsshak3(6460)   on 2012-04-28 15:37:06 (Show Source): You can put this solution on YOUR website!One number is 1 less than a second number. Twice the second number is 31 less than 5 times the first. Find the smaller of two numbers. ** let x=second no. x-1 =first no. .. 2x=5(x-1)-31 2x=5x-5-31 3x=36 x=12 x-1=11 ans: smaller no.=11
 Money_Word_Problems/604645: adult tickets for the senior class play were \$4 each and student tickets were 2 each a total of 1250 tickets were sold and \$3400 was earned how many student tickets were sold? please help ive tried to divide 3400/1250 and then got lost... :(1 solutions Answer 381320 by lwsshak3(6460)   on 2012-04-28 15:29:43 (Show Source): You can put this solution on YOUR website!adult tickets for the senior class play were \$4 each and student tickets were 2 each a total of 1250 tickets were sold and \$3400 was earned how many student tickets were sold? ** let x=no. of \$4(adult) tickets sold 1250-x=no. of \$2(student) tickets sold .. 4x+2(1250-x)=3400 4x+2500-2x=3400 2x=900 x=450 1250-x=800 ans: no. of \$2(student) tickets sold=800
 Triangles/604642: The hypotenuse of an isosceles right triangle has a length of 9 meters. Find the leg lengths to the nearest hundredth of a meter? 1 solutions Answer 381319 by lwsshak3(6460)   on 2012-04-28 15:21:11 (Show Source): You can put this solution on YOUR website!The hypotenuse of an isosceles right triangle has a length of 9 meters. Find the leg lengths to the nearest hundredth of a meter ** let x=length of legs by pythagorean theorem x^2+x^2=9^2=81 2x^2=81 x^2=81/2 x=√(81/2) x≈6.36
 Linear-equations/604633: I am lost, how do I graph the following equation; calculating the slope, x-intercept, and y-intercept, and label the intercepts on the graph. 3x = -2 1 solutions Answer 381317 by lwsshak3(6460)   on 2012-04-28 15:10:11 (Show Source): You can put this solution on YOUR website!I am lost, how do I graph the following equation; calculating the slope, x-intercept, and y-intercept, and label the intercepts on the graph. 3x = -2 x=-2/3 This is an equation of a vertical line which crosses the x-axis at -2/3 Its slope is undefined. There is no y-intercept. x-intercept at -2/3
 Rate-of-work-word-problems/604600: Left on together the hot and cold water fill up the tub in 4minutes. Cold water takes 14 minutes by itself. How many minutes does it the hot water to fill up by itself?1 solutions Answer 381316 by lwsshak3(6460)   on 2012-04-28 15:05:00 (Show Source): You can put this solution on YOUR website!Left on together the hot and cold water fill up the tub in 4minutes. Cold water takes 14 minutes by itself. How many minutes does it the hot water to fill up by itself? ** 1/14=cold water work rate 1/x=hot water work rate 1/4=work rate with both hot and cold water left on sum of individual work rates=work rate working together 1/14+1/x=1/4 1/x=1/4-1/14 =14/56-4/56 =10/56 =5/28 x=28/5 ans: minutes it takes the hot water to fill up by itself=28/5=5.6
 �t�e�s�t/604634: Please help me with this problem: A plane flies 900 miles with a tail wind in 3 hours. It takes the same plane 5 hours to fly the 900 miles flying against the wind. What is the plane's speed in still air?1 solutions Answer 381314 by lwsshak3(6460)   on 2012-04-28 14:46:15 (Show Source): You can put this solution on YOUR website!A plane flies 900 miles with a tail wind in 3 hours. It takes the same plane 5 hours to fly the 900 miles flying against the wind. What is the plane's speed in still air? ** let x=speed of plane in still air let c=speed of wind (x+c)=speed of plane downwind (x-c)=speed of plane against wind distance=travel time*speed .. 5(x-c)=900 3(x+c)=900 .. 5x-5c=900 3x+3c=900 .. 15x-15c=2700 (multi. by 3) 15x+15c=4500 (multi. by 5) add 30x=7200 x=240 5c=5x-900=300 c=60 ans: speed of plane in still air=240 mph speed of wind=60 mph
 Exponential-and-logarithmic-functions/604632: I'm seriously going to fail this: Need help please :) The questions is find the solution set for e^(3x+2)=551 solutions Answer 381313 by lwsshak3(6460)   on 2012-04-28 14:25:04 (Show Source): You can put this solution on YOUR website!find the solution set for e^(3x+2)=55 take log of both sides using calculator (3x+2)*lne=ln55 lne=1 (log of base=1) 3x+2=ln55 3x=ln55-2 x=(ln55-2)/3 x≈.669
 Exponential-and-logarithmic-functions/604631: Please help!!!! Thanks!! Write ￼ in exponential form (do not solve for x). log3 49= 2x1 solutions Answer 381309 by lwsshak3(6460)   on 2012-04-28 14:14:59 (Show Source): You can put this solution on YOUR website!Write ￼ in exponential form (do not solve for x). log3 49= 2x exponential form: base(3) raised to log of number(2x)=number(49) 3^2x=49
 Polynomials-and-rational-expressions/604629: Solve the equation: 4x^3+3x^2=64x+48 *Please show work :) Thank you!1 solutions Answer 381308 by lwsshak3(6460)   on 2012-04-28 14:11:51 (Show Source): You can put this solution on YOUR website!Solve the equation: 4x^3+3x^2=64x+48 Try grouping x^2(4x+3)=16(4x+3) set to zero x^2(4x+3)-16(4x+3)=0 factor out (4x+3) (4x+3)(x^2-16)=0 solve for x 4x+3=0 4x=-3 x=-3/4 or x^2-16=0 (x+4)(x-4)=0 (difference of squares) x=-4 and x=4 roots or zeros are: -3/4, -4 and 4
 Linear_Equations_And_Systems_Word_Problems/604618: Two cyclists leave towns 90 miles apart at the same time, and travel towards each other. One cyclist travels 5 mi/h slower than the other. If they meet in 2 hours, what is the rate of each cyclist? I'm having a hard time figuring out the proper formula for this problem. Thank you in advance for your help!1 solutions Answer 381306 by lwsshak3(6460)   on 2012-04-28 13:59:31 (Show Source): You can put this solution on YOUR website!Two cyclists leave towns 90 miles apart at the same time, and travel towards each other. One cyclist travels 5 mi/h slower than the other. If they meet in 2 hours, what is the rate of each cyclist? ** let x=rate of speed of faster cyclist let (x-5)=rate of speed of slower cyclist distance=rate*travel time (2 hrs) .. 2x+2(x-5)=90 2x+2x-10=90 4x=100 x=25 x-5=20 ans: rate of speed of faster cyclist=25 mph rate of speed of slower cyclist=20 mph
 Age_Word_Problems/604512: in 15 years dorothy will be 3 times as old as stan. 5 years ago the difference in their ages was 50 how old are dorothy and stan?1 solutions Answer 381248 by lwsshak3(6460)   on 2012-04-28 03:40:43 (Show Source): You can put this solution on YOUR website!in 15 years dorothy will be 3 times as old as stan. 5 years ago the difference in their ages was 50 how old are dorothy and stan? ** let x=dorothy's present age let y=stan's present age .. 15 years hence: (x+15)=3(y+15) x+15=3y+45 .. 5 years ago: (x-5)-(y-5)=50 x-5-y+5=50 x-y=50 x=50+y .. x+15=3y+45 sub (50+y) for x 50+y+15=3y+45 2y=20 y=10 x=50+y=60 ans: dorothy's present age=60 let y=stan's present age=10
 Travel_Word_Problems/604515: A driver took a day trip around the California coastline driving at two different speeds. He drove 70 miles at a slower speed and 300 miles at a speed 40 miles per hour faster. If the time spent during the faster speed was twice that spent at a slower speed, find the two speeds during the trip. 1 solutions Answer 381245 by lwsshak3(6460)   on 2012-04-28 02:49:08 (Show Source): You can put this solution on YOUR website!A driver took a day trip around the California coastline driving at two different speeds. He drove 70 miles at a slower speed and 300 miles at a speed 40 miles per hour faster. If the time spent during the faster speed was twice that spent at a slower speed, find the two speeds during the trip ** let x=slower speed x+40=faster speed travel time=distance/speed ... 300/(x+40)=2*70/x 300x=140x+5600 160x=5600 x=35 x+40=75 ans: slower speed=35 mph faster speed=75 mph
 Quadratic-relations-and-conic-sections/604372: Write an equation of the hyperbola with vertices at (–7, 0) and (7, 0) and foci at (–9, 0) and (9, 0)1 solutions Answer 381189 by lwsshak3(6460)   on 2012-04-27 20:44:34 (Show Source): You can put this solution on YOUR website!Write an equation of the hyperbola with vertices at (–7, 0) and (7, 0) and foci at (–9, 0) and (9, 0) This is a hyperbola with horizontal transverse axis. Its standard form of equation: (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=14=2a a=7 a^2=49 .. foci c=9 c^2=81 .. c^2=a^2+b^2 b^2=c^2-a^2=81-49=32 b=√32≈5.6 .. equation: x^2/49-y^2/32=1
 Rational-functions/604007: find the rational zeros of f(x)=-x^3-4x^2+11x+301 solutions Answer 381185 by lwsshak3(6460)   on 2012-04-27 20:29:24 (Show Source): You can put this solution on YOUR website!find the rational zeros of f(x)=-x^3-4x^2+11x+30 f(x)=x^3+4x^2-11x-30 use rational roots theorem: ...0...|.....1.......4.......-11.......-30.... ...1...|.....1.......5........-6........-24.... ...2...|.....1.......6.........1.........-28.... ...3...|.....1.......7........10..........0...( 3 is a root) f(x)=(x-3)(x^2+7x+10)=(x-3)(x+5)(x+2) zeros are: -5, -2 and 3
 Radicals/604380: how do you rationalize the denominator and simplify cube root of 4y^2/cube root of 5x^111 solutions Answer 381139 by lwsshak3(6460)   on 2012-04-27 15:43:51 (Show Source): You can put this solution on YOUR website!how do you rationalize the denominator and simplify cube root of 4y^2/cube root of 5x^11 use √ as cube root symbol for this problem 4y^2/√(5x^11) multiply top and bottom by √(5^2x) to make the inside (radican) of denominator a perfect cube 4y^2*√(5^2x)/√(5x^11)*√(5^2x) =4y^2*√(5^2x)/√(5^3*x^12) =4y^2*√(5^2x)/5x^4 =4y^2*cube root of 25x/5x^4
 Quadratic-relations-and-conic-sections/604373: The path that a satellite travels around Earth is an ellipse with Earth at one focus. The length of the major axis is about 16,000 km, and the length of the minor axis is about 12,000 km. Write an equation for the satellite’s orbit.1 solutions Answer 381131 by lwsshak3(6460)   on 2012-04-27 15:08:12 (Show Source): You can put this solution on YOUR website!The path that a satellite travels around Earth is an ellipse with Earth at one focus. The length of the major axis is about 16,000 km, and the length of the minor axis is about 12,000 km. Write an equation for the satellite’s orbit. ** Ellipse is worked as one with vertical major axis: Its standard form of equation: (x-h)^2/b^2+(y-k)^2/a^2=1, a>b, (h,k)=(x,y) coordinates of center For given problem: center: (0,0) given length of major axis=16000=2a a=8000 a^2=64000000=64*10^6 .. given length of minor axis=12000=2b b=6000 b^2=36000000=36*10^6 .. equation: x^2/36*10^6+y^2/64*10^6=1
 Trigonometry-basics/604377: what are the six trigonometric functions of theta, where the terminal arm of the angle passes through the point (-12,5). thanks1 solutions Answer 381130 by lwsshak3(6460)   on 2012-04-27 14:47:01 (Show Source): You can put this solution on YOUR website!what are the six trigonometric functions of theta, where the terminal arm of the angle passes through the point (-12,5). thanks ** You are working with a reference angle in quadrant II: Opposite side=5 Adjacent side=-12 Hypotenuse=√(12^2+5^2)=√169=13 .. 6 trig functions: sin theta=5/13 cos theta=-12/13 tan theta=-5/12 csc theta=13/5 sec theta=-13/12 cot theta=-12/5
 Trigonometry-basics/603634: solve algebraically for all values of theta in the interval 0-360 that satisfy the equation sin^2 theta/1+cos theta = 11 solutions Answer 381021 by lwsshak3(6460)   on 2012-04-27 03:00:18 (Show Source): You can put this solution on YOUR website!solve algebraically for all values of theta in the interval 0-360 that satisfy the equation sin^2 theta/1+cos theta = 1 ** sin^2x/(1+cosx)=1 sin^2x=1+cosx 1-cos^2x=1+cosx cos^2x+cosx=0 cosx(cosx+1)=0 cosx=0 x=π/2 and 3π/2 or cosx+1=0 cosx=-1 x=π .. ans: π/2, π, and 3π/2
 Trigonometry-basics/603891: The expression cot(-200) degrees is equivalent to 1) -tan 20 degrees 2) tan 70 degrees 3) -cot 20 degrees 4)cot 70 degrees1 solutions Answer 381018 by lwsshak3(6460)   on 2012-04-27 02:50:01 (Show Source): You can put this solution on YOUR website!The expression cot(-200) degrees is equivalent to 1) -tan 20 degrees 2) tan 70 degrees 3) -cot 20 degrees 4)cot 70 degrees ** Cot (-200º) makes a reference angle of 20º in quadrant II where cot<0 So Cot (-200º)=-cot 20º
 Trigonometry-basics/603947: What is the exact value of- tan(sin^-1 3/2)1 solutions Answer 381017 by lwsshak3(6460)   on 2012-04-27 02:41:04 (Show Source): You can put this solution on YOUR website!What is the exact value of- tan(sin^-1 3/2) (sin^-1 3/2) This states that the sin of an angle is 3/2>1 -1 < sin < 1 This problem is not written properly
 Trigonometry-basics/603948: What is the exact value of- cos^-1 (cos 2/7)1 solutions Answer 381016 by lwsshak3(6460)   on 2012-04-27 02:34:51 (Show Source): You can put this solution on YOUR website!What is the exact value of- cos^-1 (cos 2/7) ** This problem is not written properly. The inverse of a trig function requires a value not an angle as it appears in this case. You probably meant cos=2/7.
 Trigonometry-basics/603951: What is the exact value of- csc (sin^-1 0.2)1 solutions Answer 381014 by lwsshak3(6460)   on 2012-04-27 02:15:42 (Show Source): You can put this solution on YOUR website!What is the exact value of- csc (sin^-1 0.2) sinx=0.2=2/10=1/5 cscx=1/sinx=5 csc (sin^-1 0.2)=5
 Trigonometry-basics/604236: Can you please help me with this problem? Cosθ (tan θ+cot θ)=csc θ 1 solutions Answer 381012 by lwsshak3(6460)   on 2012-04-27 02:08:32 (Show Source): You can put this solution on YOUR website!Cosθ (tan θ+cot θ)=csc θ starting with left side cos(tan+cot) =cos[(sin/cos)+(cos/sin)] =cos[(sin^2+cos^2)/(sincos)] =cos(1/sincos) =1/sin =csc verified: left side=right side
 Trigonometry-basics/604206: verify (tanx)/(1-cosx)=cscx(1+secx)1 solutions Answer 381011 by lwsshak3(6460)   on 2012-04-27 02:00:26 (Show Source): You can put this solution on YOUR website!verify (tanx)/(1-cosx)=cscx(1+secx) starting with right side csc(1+sec) =(1/sin)(1+(1/cos)) expand =(1/sin)+(1/sincos) =(cos+1)/(sincos) multiply top and bottom by (cos-1) =(cos+1)(cos-1)/(sincos)(cos-1) =(cos^2-1)/(sincos)(cos-1) =[(cos^2-1)/sincos]*(1/(cos-1)) =(-sin^2)/sincos/(cos-1) =(-sin/cos)/(cos-1) =-tan/(cos-1) =tan/(1-cos) verified: right side=left side
 Trigonometry-basics/604209: given sec(t)=2.345 sin(t)<0, find (t)=(0,2pi), list answers in a set1 solutions Answer 381008 by lwsshak3(6460)   on 2012-04-27 01:07:30 (Show Source): You can put this solution on YOUR website!given sec(t)=2.345 sin(t)<0, find (t)=(0,2pi), list answers in a set ** sec(t)=2.345 cos(t)=1/2.345 reference angle≈1.13 sin<0 in quadrants III and IV cos and sec>0 in quadrants I and IV So, reference angle must be in quadrant IV t≈5.15 {t|t=5.15}
 Trigonometry-basics/604035: prove trigonometric identity for sin(x)+cot(x)cos(x)=csc(x)1 solutions Answer 380944 by lwsshak3(6460)   on 2012-04-26 20:39:18 (Show Source): You can put this solution on YOUR website!prove trigonometric identity for sin(x)+cot(x)cos(x)=csc(x) start with left side sin+cotcos =sin+(cos/sin)*cos =sin+(cos^2/sin) =(sin^2+cos^2)/sin =1/sin =csc verified: left side=right side
 Trigonometry-basics/604037: prove trigonometric identity for tan^2(x)-tan^2(x)sin^2(x)=sin^2(x)1 solutions Answer 380935 by lwsshak3(6460)   on 2012-04-26 20:28:56 (Show Source): You can put this solution on YOUR website!prove trigonometric identity for tan^2(x)-tan^2(x)sin^2(x)=sin^2(x) tan^2-tan^2sin^2=sin^2 starting with left side tan^2-tan^2sin^2 =tan^2(1-sin^2) =tan^2*cos^2 =(sin^2/cos^2)*cos^2 =sin^2 verified: left side=right side
 Trigonometry-basics/604042: prove trigonometric identity (csc^2(x)-1)(sin(x))= cot(x)cos(x)1 solutions Answer 380929 by lwsshak3(6460)   on 2012-04-26 20:11:16 (Show Source): You can put this solution on YOUR website!prove trigonometric identity (csc^2(x)-1)(sin(x))= cot(x)cos(x) Starting with left side: (csc^2-1)(sin) =((1/sin^2)-1)sin =(1/sin)-sin =(1-sin^2)/sin =cos^2/sin =cos/sin*cos =cotcos verified: left side=right side
 Trigonometry-basics/604043: prove trigonometric identity sin(x)/1-sin(x) + sin(x)/1+sin(x)= 2tan(x)sec(x)1 solutions Answer 380927 by lwsshak3(6460)   on 2012-04-26 19:53:23 (Show Source): You can put this solution on YOUR website!prove trigonometric identity sin(x)/1-sin(x) + sin(x)/1+sin(x)= 2tan(x)sec(x) ** Starting with left side: sin/1-sin + sin/1+sin =sin(1+sin+1-sin)/(1-sin)(1+sin) =2sin/(1-sin^2) =2sin/(cos^2) =2sin/cos*1/cos =2tansec verified: left side=right side
 logarithm/602034: Expand the expression using the logarithm identities and write answer without exponents. log10 ((x-4)^2/(3x^2+4))1 solutions Answer 380810 by lwsshak3(6460)   on 2012-04-26 04:07:22 (Show Source): You can put this solution on YOUR website!Expand the expression using the logarithm identities and write answer without exponents. log10 ((x-4)^2/(3x^2+4)) 2log(x-4)-log(3x^2+4) Note: Exponent in second term cannot be eliminated because it is associated with +4