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Recent problems solved by 'lwsshak3'

lwsshak3 answered: 6517 problems
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logarithm/619762: please help me solve this equation 2 log base 5 radical 5 1 solutions Answer 391499 by lwsshak3(6519)   on 2012-06-21 03:14:24 (Show Source): You can put this solution on YOUR website! solve this equation 2 log base 5 radical 5 2log5(√5) =2log5(5^1/2) =log5(5^1/2)^2 =log5(5) =1

test/622651: |2x^2 + 3| = 7x 1 solutions Answer 391498 by lwsshak3(6519)   on 2012-06-21 02:33:19 (Show Source): You can put this solution on YOUR website! |2x^2 + 3| = 7x solve for two conditions: 2x^2+3=7x 2x^2-7x+3=0 (2x-1)(2x-3)=0 x=1/2 or x=3/2 (reject, does not check) .. 2x^2+3=-7x 2x^2+7x+3=0 (2x+1)(2x+3)=0 x=-1/2 (reject, does not check) or x=-3/2 (reject, does not check) solution: x=1/2

test/622650: Two number differ by 5 and 3/5 of the greater number is equal to 3/4 to the smaller. Find the number. 1 solutions Answer 391496 by lwsshak3(6519)   on 2012-06-21 02:15:07 (Show Source): You can put this solution on YOUR website! Two number differ by 5 and 3/5 of the greater number is equal to 3/4 to the smaller. Find the number ** let x=larger number x-5=smaller number 3x/5=3(x-5)/4 3x/5=(3x-15)/4 LCD=20 12x=15x-75 3x=75 x=25 x-5=20 larger number=25 smaller number=20

Coordinate-system/622592: Find the x and y intercepts and graph the equation plotting the intercepts. Please show all of your work. -4x+9y=-36 1 solutions Answer 391494 by lwsshak3(6519)   on 2012-06-21 02:02:59 (Show Source): You can put this solution on YOUR website! Find the x and y intercepts and graph the equation plotting the intercepts. Please show all of your work. -4x+9y=-36 y-intercept set x=0 9y=-36 y=-4 .. x-intercept set y=0 -4x=-36 x=4 see graph below:

Polynomials-and-rational-expressions/622644: Please help! Find the LCD. 1 solutions Answer 391493 by lwsshak3(6519)   on 2012-06-21 01:52:56 (Show Source): You can put this solution on YOUR website! Find the LCD ((5)/(7ab^5)) and ((2)/(7a^3b^3)) select the highest power of each variable in the denominator LCD: 7a^3b^5

Polynomials-and-rational-expressions/622640: Please help! Multiply. 1 solutions Answer 391492 by lwsshak3(6519)   on 2012-06-21 01:47:46 (Show Source): You can put this solution on YOUR website! ((8x^2)/(x^2-9))((x^2+6x+9)/(16x^3)) =(8x^2)(x^2+6x+9)/(x^2-9)(16x^3) =(x+3)^2/(x^2-9)2x =(x+3)^2/2x(x+3)(x-3) =(x+3)/2x(x-3)

Polynomials-and-rational-expressions/622641: Please help! Divide. 1 solutions Answer 391489 by lwsshak3(6519)   on 2012-06-21 01:37:38 (Show Source): You can put this solution on YOUR website! ((3y+12)/(8y^3))/((9y+36)/(16y^3)) =2(3y+12)/(9y+36) =2*3(y+4)/9(y+4) =6/9 =2/3

Polynomials-and-rational-expressions/622638: Please help! Simplify completely. 1 solutions Answer 391486 by lwsshak3(6519)   on 2012-06-21 01:29:07 (Show Source): You can put this solution on YOUR website! (x-2)/(8x^2-32) =(x-2)/8(x^2-4) =(x-2/8(x+2)(x-2) =1/[8(x+2]

test/622626: solve using long division, then check. (x^3+2x^2+3x-6)/(x-1)=? 1 solutions Answer 391485 by lwsshak3(6519)   on 2012-06-21 01:22:50 (Show Source): You can put this solution on YOUR website! solve using long division, then check. (x^3+2x^2+3x-6)/(x-1)=? ** ..............x^2......3x.......6....... ...x-1|....x^3....2x^2....3x...-6 ........................3x^2....3x...... ........................3x^2..-3x...... ....................................6x..-6.. ....................................6x..-6.. ..........................................0.. Check: quotient*divisor+remainder=dividend (x^2+3x+6)(x-1)+0 =x^3+3x^2+6x-x^2-3x-6+0 =x^3+2x^2+3x-6

logarithm/622632: which among log (base 13)169 and log (base 23)521 is greater ?(with out using calculator) thanks 1 solutions Answer 391483 by lwsshak3(6519)   on 2012-06-21 01:03:35 (Show Source): You can put this solution on YOUR website! which among log (base 13)169 and log (base 23)521 is greater Using exponential form of logs: Base raised to log of number=number for first case 13^x=169 x=2 which is the log of 169 base 13 .. for 2nd case 23^x=521 x=<2 which is the log of 529 base 23 (if x=2, the number=23^2=529, so x<2) The log of the first case>log of 2nd case

Percentage-and-ratio-word-problems/622571: A 45 gallon barrel contains a mixture with a concentration of 30%. How much of this mixture must be withdrawn and replaced by 100% concentrate to bring the mixture up to 50% concentration? 1 solutions Answer 391450 by lwsshak3(6519)   on 2012-06-20 20:26:30 (Show Source): You can put this solution on YOUR website! A 45 gallon barrel contains a mixture with a concentration of 30%. How much of this mixture must be withdrawn and replaced by 100% concentrate to bring the mixture up to 50% concentration? ** let x=mixture to be withdrawn and replaced 45-x=remaining 30% mixture 30%(45-x)+100%x=50%*45 13.5-.3x+x=22.5 .7x=9 x≈12.86 gallons of mixture to be withdrawn and replaced≈12.86

Linear-equations/622573: Find the slope Intercept form for the lie satisfying the following condition for x-intercept 2, y-intercept 2/3 1 solutions Answer 391449 by lwsshak3(6519)   on 2012-06-20 20:11:41 (Show Source): You can put this solution on YOUR website! Find the slope Intercept form for the lie satisfying the following condition for x-intercept 2, y-intercept 2/3 Standard form of equation for a straight line: y=mx+b, m=slope, b=y-intercept ** 2 given points: (2,0) and (0,2/3) slope=∆y/∆x=(2/3-0)/(0-2)=(2/3)/-2=-2/6=-1/3 equation: y=-x/3+b y=-x/3+2/3

Trigonometry-basics/622016: please help me solve for all 2cos(x/2)+1 =0 1 solutions Answer 391448 by lwsshak3(6519)   on 2012-06-20 19:51:48 (Show Source): You can put this solution on YOUR website! solve for all 2cos(x/2)+1 =0 2cos(x/2)=-1 cos(x/2)=-1/2 x/2=2π/3 x=π/3+2πn, n=integer and x/2=4π/3 x=2π/3+2πn, n=integer

Trigonometry-basics/622507: please help me verify: cos^3sin^2= (sin^2-sin^4)cosx 1 solutions Answer 391440 by lwsshak3(6519)   on 2012-06-20 19:29:33 (Show Source): You can put this solution on YOUR website! please help me verify: cos^3sin^2= (sin^2-sin^4)cosx starting with right side: (sin^2-sin^4)cosx =sin^2(1-sin^2)cos =sin^2*cos^2*cos =cos^3sin^2 verified: right side=left side

Trigonometry-basics/622497: please help me find the exact values of: sin2u given: sinu= -4/5 and pi < u 3pi/2 1 solutions Answer 391409 by lwsshak3(6519)   on 2012-06-20 16:28:25 (Show Source): You can put this solution on YOUR website! find the exact values of: sin2u given: sinu= -4/5 and pi < u 3pi/2 sinu=-4/5 in quadrants III and IV where sin<0 since quadrant IV is not in the domain,(π,3π/2), reference angle is in quadrant III sinu=-4/5=opposite side/hypotenuse adjacent side=3 since you are working with a 3-4-5 right triangle cosu=-3/5 in quadrants III where cos<0 sin 2u=2sinucosu =2*-4/5*-3/5 =24/25

Trigonometry-basics/622024: please help me find the exact value of tan2u given: sinu= -4/5 & pi < u < 3pi/2 1 solutions Answer 391408 by lwsshak3(6519)   on 2012-06-20 16:10:52 (Show Source): You can put this solution on YOUR website! find the exact value of tan2u given: sinu= -4/5 & pi < u < 3pi/2 ** sinu=-4/5 in quadrants III and IV where sin<0 since quadrant IV is not in the domain,(π,3π/2), reference angle is in quadrant III sinu=-4/5=opposite side/hypotenuse adjacent side=3 since you are working with a 3-4-5 right triangle tanu=opposite side/adjacent side=4/3 in quadrant III where tan>0 tan2u=2tanu/(1-tan^2u)=2*4/3/(1-(4/3)^2)=(8/3)/(1-16/9)=(8/3)/(-7/9)=-72/21

Trigonometry-basics/622255: Find all values of θ, if θ is in the interval [0°, 360°) and has the following function value sinθ = - sqrt3/2 1 solutions Answer 391401 by lwsshak3(6519)   on 2012-06-20 15:49:55 (Show Source): You can put this solution on YOUR website! Find all values of θ, if θ is in the interval [0°, 360°) and has the following function value sinθ = - sqrt3/2 ** sinθ = -√3/2 in quadrants III and IV where sin<0 reference angle=60º θ=240º and 300º

Trigonometry-basics/622075: please helpme solve: add & simplify: (1/csc(beta)+1)-(1/csc(beta) -1) 1 solutions Answer 391396 by lwsshak3(6519)   on 2012-06-20 15:37:39 (Show Source): You can put this solution on YOUR website! please helpme solve: add & simplify: (1/csc(beta)+1)-(1/csc(beta) -1) use x for beta 1/cscx=sinx (1/cscx+1)-(1/cscx -1) =(sinx+1)-(sinx-1) =six+1-sinx+1 =2

Trigonometry-basics/622310: (Sin)sqx+(cos)sqx=1? Why 1 solutions Answer 391288 by lwsshak3(6519)   on 2012-06-20 04:14:09 (Show Source): You can put this solution on YOUR website! (Sin)sqx+(cos)sqx=1? Why Let me try using a right triangle let o=opposite side let a=adjacent side let h=hypotenuse .. sin=o/h cos=a/h sin^2=o^2/h^2 cos^2=a^2/h^2 sin^2+cos^2 =o^2/h^2+a^2/h^2 =(o^2+a^2)/h^2 =1 By Pythagorean Theorem: The sum of the squares of the two sides=the square of the hypotenuse. Therefore, (o^2+a^2)/h^2=1

Travel_Word_Problems/622308: Find the quotient and remainder for the division problem (2t3 + 9t2 − 2t − 3) ÷ (2t + 1) 1 solutions Answer 391287 by lwsshak3(6519)   on 2012-06-20 03:57:00 (Show Source): You can put this solution on YOUR website! Find the quotient and remainder for the division problem (2t3 + 9t2 − 2t − 3) ÷ (2t + 1) ** by long division: .................t^2.....4t........................... ..2t+1|....2t^3....9t^2....-2t.....-3.............................. ...............2t^3......t^2....................... ..........................8t^2...-2t................... ..........................8t^2.....4t............. .................................. ..-6t.....-3...... Quotient: t^2+4t Remainder: -6t-3

Polynomials-and-rational-expressions/622295: Find the quotient y^4-3y+4/ y^2-5 1 solutions Answer 391285 by lwsshak3(6519)   on 2012-06-20 03:38:59 (Show Source): You can put this solution on YOUR website! Find the quotient y^4-3y+4/ y^2-5 By long division: ................ y^2..........5................... .y^2-5..|...y^4... 0. ..0.. ..3y....4..................... .................y^4..... ..-5y^2.................. .............................. 5y^2......-25...... .........................................3y..29.. dividend : y^4-3y+4 divisor: y^2-5 quotient: y^2+5+Remainder Remainder: 3y+29

Travel_Word_Problems/622298: Person A and person B start from the same location, at the same time. They ride their bikes in opposite directions for 3 hours, at which time they are 102 miles apart. person A rides 4 miles per hour faster than person B. Find the rate of each rider. I am having trouble figuring out the equation. I made a table that looks like this: D R T A r+4 3 B r 3 1 solutions Answer 391283 by lwsshak3(6519)   on 2012-06-20 03:01:53 (Show Source): You can put this solution on YOUR website! Person A and person B start from the same location, at the same time. They ride their bikes in opposite directions for 3 hours, at which time they are 102 miles apart. person A rides 4 miles per hour faster than person B. Find the rate of each rider. ** let x=rate of rider B x+4=rate of rider A distance=travel time*rate .. 3x+3(x+4)=102 3x+3x+12=102 6x=90 x=15 x+4=19 ans: rate of rider B=15 mph rate of rider A=19 mph

Age_Word_Problems/622299: f(x)=(1)/(√(4x^2-17x+4)) Find the domain and express in interval notation info: its 1 over the square root of the numbers listed above 1 solutions Answer 391282 by lwsshak3(6519)   on 2012-06-20 02:46:27 (Show Source): You can put this solution on YOUR website! f(x)=(1)/(√(4x^2-17x+4)) Find the domain and express in interval notation info: its 1 over the square root of the numbers listed above ** f(x)=(1)/(√(4x^2-17x+4)) =1/√(4x^2-17x+4) =1/[√(4x-1)(x-4)] check for values of x which makes function undefined: 4x-1=0 x≠1/4 or x-4=0 x≠4 .. check for radican≥0 (4x-1)(x-4)≥0 number line <..+....1/4....-......4.....+.........> Domain: (-∞,1/4) U (4,∞) In other words, all real numbers except x=1/4 and 4 and between x=1/4 and 4 (not including 1/4 and 4)

Graphs/622148: solve the inequality 1 solutions Answer 391155 by lwsshak3(6519)   on 2012-06-19 17:16:31 (Show Source): You can put this solution on YOUR website! solve the inequality 2x^3+5x^2-17x-20<0 To solve this problem, the equation must be in factored form. This seems like a difficult expression to factor and probably it could be done with the rational roots theorem, but I'm just going to use a graphing calculator to find the following factors of this problem: 2x^3+5x^2-17x-20<0 (x+4)(x+1)(2x-5)<0 number line: <..-...-4....+.....-1....-....5/2.....+.......> solution: (-∞,-4) U (-1,5/2)

Money_Word_Problems/622147: how much money would be in an account after 5 years if an original investment of $6500 was compounded quarterly at 4.5%?. compare this amount to the same investment that was compounded daily. round to the nearest cent. 1 solutions Answer 391152 by lwsshak3(6519)   on 2012-06-19 16:47:33 (Show Source): You can put this solution on YOUR website! how much money would be in an account after 5 years if an original investment of $6500 was compounded quarterly at 4.5%?. compare this amount to the same investment that was compounded daily. round to the nearest cent. ** Compound Interest formula: A=P(1+i)^n, P=initial investment, i=interest rate per period, A=amount after n periods For given problem: compounding quarterly P=6500 i=.045/4 (annual interest/number of compounding periods per year) n=5*4=20 periods A=6500(1+.045/4)^20 using calculator A=8129.88 .. compounding daily P=6500 i=.045/365 (annual interest/number of compounding periods per year) n=5*365=1825 periods A=6500(1+.045/365)^1825 using calculator A=8139.98 .. compounding daily gives about $10 more over a 5-year period

Exponents/622138: (1/3)^x=(81)(9^(x-1)) when solving for x is -2/3. I do not know how. 1 solutions Answer 391146 by lwsshak3(6519)   on 2012-06-19 16:17:25 (Show Source): You can put this solution on YOUR website! (1/3)^x=(81)(9^(x-1)) when solving for x is -2/3. I do not know how. ** (1/3)^x=(81)(9^(x-1)) 1/(3^x)=(81)(3^2)^(x-1) 3^-x/(81)(3^(2x-2)) 3^-x/(3^(2x-2))=81 3^(-3x+2)=3^4 -3x+2=4 -3x=2 x=-2/3 note: when moving from denominator to numerator or vise-versa, sign of exponents change.

absolute-value/620927: how would you solve this problem |x/5 - 1| = | 1 - x/3| 1 solutions Answer 391041 by lwsshak3(6519)   on 2012-06-19 03:59:21 (Show Source): You can put this solution on YOUR website! how would you solve this problem |x/5 - 1| = | 1 - x/3| solve for two conditions (x/5-1)=(1-x/3) x/5-1=1-x/3 LCD:15 3x-15=15-5x 8x=30 x=30/8 x=15/4 .. and (x/5-1)=-(1-x/3) (x/5-1)=-1+x/3) LCD:15 3x-15=-15+5x 2x=-30 x=-15

absolute-value/621866: how would you solve this problem |2x - 1| => 1/8 1 solutions Answer 391040 by lwsshak3(6519)   on 2012-06-19 03:45:44 (Show Source): You can put this solution on YOUR website! how would you solve this problem |2x - 1| => 1/8 solve for two conditions: 2x-1≥1/8 multiply by 8 16x-8≥1 16x≥9 x≥9/16 and -2x+1≥1/8 -16x+8≥1 -16x≥-7 x≤7/16 .. number line <====7/16]..........[9/16====> solution: (-∞,7/16] U [9/16,∞)

absolute-value/621872: how would you solve this problem 4<6+2/3x < 8 1 solutions Answer 391039 by lwsshak3(6519)   on 2012-06-19 03:33:01 (Show Source): You can put this solution on YOUR website! how would you solve this problem 4<6+2/3x < 8 4<6+2/3x < 8 add-6 -2<2/3x<2 multiply by 3 -6<2x<6 divide by 2 -3 < x <3 number line: <.......(-3<===>3)........> solution: (-3,3)

absolute-value/621873: |2x+6| <2 1 solutions Answer 391037 by lwsshak3(6519)   on 2012-06-19 03:21:31 (Show Source): You can put this solution on YOUR website! |2x+6| <2 solve for two conditions: 2x+6<2 2x<-4 x<-2 and -2x-6<2 -2x<8 x>-4 .. number line: <,......(-4<====>-2)............> solution: (-4,-2)

absolute-value/621875: how would you solve this problem |6x - 1| - 4 =< 2 1 solutions Answer 391036 by lwsshak3(6519)   on 2012-06-19 03:11:09 (Show Source): You can put this solution on YOUR website! how would you solve this problem |6x - 1| - 4 =< 2 solve for two conditions: (6x-1) and -(6x-1) (6x - 1) - 4 ≤ 2 6x - 1 - 4≤ 2 6x ≤ 7 x≤7/6 .. and -(6x - 1) - 4 ≤ 2 -6x +1 - 4 ≤ 2 -6x ≤ 5 x≥-5/6 .. number line: <.........[-5/6<====>7/6]............> solution: [-5/6,7/6]