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# Recent problems solved by 'lwsshak3'

lwsshak3 answered: 6520 problems
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 logarithm/683836: how to solve log₂(x²)=(log₂x)²1 solutions Answer 423764 by lwsshak3(6522)   on 2012-11-22 01:10:18 (Show Source): You can put this solution on YOUR website!how to solve: log₂(x²)=(log₂x)² 2log2(x)=[log2(x)]^2 [log2(x)]^2-2log2(x)=0 log2(x)(log2(x)-2)=0 log2(x)=0 (reject, log2(x)>0) log2(x)-2=0 log2(x)=2 x=4
 Quadratic-relations-and-conic-sections/683687: Write an equation for the ellipse Eccentricity 3/4; foci (0,-2) (0,2)1 solutions Answer 423762 by lwsshak3(6522)   on 2012-11-22 00:15:13 (Show Source): You can put this solution on YOUR website!Write an equation for the ellipse Eccentricity 3/4; foci (0,-2) (0,2) This is an ellipse with vertical major axis. its standard form of equation: , a>b, (h,k)=(x,y) coordinates of center For given ellipse: center: (0,0) c=2 (center to focus) c^2=4 eccentricity:=3/4=c/a a=4c/3=16/3 a^2=256/9 c^2=a^2-b^2 b^2=a^2-c^2=256/9-4=256/9-36/9=220/9 Equation:
 Quadratic-relations-and-conic-sections/683786: graph the parabola with the following equation.. Give the focus coordinates, vertex coordinates, equation of directrix and one additional point on the curve.. The equation is: x= 1/16y^2 1 solutions Answer 423718 by lwsshak3(6522)   on 2012-11-21 20:12:36 (Show Source): You can put this solution on YOUR website!graph the parabola with the following equation.. Give the focus coordinates, vertex coordinates, equation of directrix and one additional point on the curve.. The equation is: x= 1/16y^2 ** rewrite equation: y^2=16x This is a parabola that opens rightwards. Its standard form: (y-k)^2=4p(x-h), (h,k)=coordinates of the vertex For given equation: y^2=16x vertex: (0,0) axis of symmetry: y=0 4p=16 p=4 focus=(4,0) (p-distance right of vertex on axis of symmetry) directrix: x=4 ( (p-distance left of vertex on axis of symmetry) 2 points on curve: (1,4) and (1,-4) see graph below: y=(16x)^.5
 Quadratic-relations-and-conic-sections/683282: Solve the equation. 6/m-5 - 2/m+5 = 8/m^2-251 solutions Answer 423711 by lwsshak3(6522)   on 2012-11-21 19:51:21 (Show Source): You can put this solution on YOUR website!Solve the equation. 6/m-5 - 2/m+5 = 8/m^2-25 m^2-25=(m+5)(m-5) LCD:(m+5)(m-5) 6(m+5)-2(m-5)=8 6m+30-2m+10=8 4m=-32 m=-8
 Trigonometry-basics/682705: Use a half-angle formula to find the exact value of sin ((3pi)/8) Thanks!!1 solutions Answer 423611 by lwsshak3(6522)   on 2012-11-21 02:54:01 (Show Source): You can put this solution on YOUR website!Use a half-angle formula to find the exact value of sin ((3pi)/8) Identity:sin(s/2)=±√[(1-cos s)/2] sin(3π/8)=sin[(3π/4)/2]=√[1-cos(3π/4)/2]=√[1-(-√2/2)/2]=√[(1+√2/2)/2] sin(3π/8)=√[(1+√2/2)/2] check with calculator: sin(3π/8)≈.9238.. √[(1+√2/2)/2]≈.9238..
 Trigonometry-basics/683000: Solve: csc^2x-2cot^2x=01 solutions Answer 423610 by lwsshak3(6522)   on 2012-11-21 02:30:15 (Show Source): You can put this solution on YOUR website!For interval [0,π] Solve: csc^2x-2cot^2x=0 (1/sin^2x)-(2cos^2x/sin^2x)=0 divide by sin^2x 1-2cos^2x=0 cos^2x=1/2 cosx=1/√2=√2/2 x=π/4
 Trigonometry-basics/683009: if cos theta = 3/5 and sin theta < 0, find sin theta1 solutions Answer 423609 by lwsshak3(6522)   on 2012-11-21 01:50:47 (Show Source): You can put this solution on YOUR website!if cos theta = 3/5 and sin theta < 0, find sin theta cos>0 and sin<0 only in quadrant IV reference angle is that of a 3-4-5 right triangle therefore, sin theta=-4/5
 Trigonometry-basics/683216: IF (-4,6) is on the terminal side of theta what are the 6 trig functions.1 solutions Answer 423608 by lwsshak3(6522)   on 2012-11-21 01:40:58 (Show Source): You can put this solution on YOUR website!IF (-4,6) is on the terminal side of theta what are the 6 trig functions. visualize a circle with center at (0,0) where a point on the circle is at (-4,6) This gives a right triangle where the x-coordinate(-4) is the adjacent side and y-coordinate(6) is the opposite sidee. The hypotenuse then=√(4^2+6^2)=√(16+36)=√52 6 trig functions as follows: sin=6/√52 cos=-4/√52 tan=-6/4 csc=√52/6 sec=-√52/4 cot=-4/6
 Trigonometry-basics/683542: Find all solutions of the equation cos2x + cosx = 2 in the interval [0, 4pi]1 solutions Answer 423607 by lwsshak3(6522)   on 2012-11-21 01:07:51 (Show Source): You can put this solution on YOUR website!Find all solutions of the equation cos2x + cosx = 2 in the interval [0, 4pi] cos2x + cosx = 2 2cos^2x-1+cosx=2 2cos^2x+cosx-3=0 (2cosx+3)(cosx-1)=0 2cosx+3=0 cosx=-3/2 (reject, -1≤cosx≤1) or cosx-1=0 cosx=1 x=0, 2π, 4π
 Trigonometry-basics/683395: Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to two decimal places where appropriate.) sin theta = -0.4 1 solutions Answer 423606 by lwsshak3(6522)   on 2012-11-21 00:36:57 (Show Source): You can put this solution on YOUR website!Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to two decimal places where appropriate.) sin theta = -0.4 sinx=0.4 x≈0.41+2πk, 2.73+2πk, k=integer
 Trigonometry-basics/683396: Solve the equation for (x) over the interval [0(degrees),360(degrees)). What is the solution set? 1 solutions Answer 423605 by lwsshak3(6522)   on 2012-11-21 00:32:15 (Show Source): You can put this solution on YOUR website!Solve the equation for (x) over the interval [0(degrees),360(degrees)). What is the solution set?2sin^2(x) = sqrt(3)sin(x) 2sin^2(x) =√3)sin(x) 2sin^2(x)-√3)sin(x=0 sinx(2sinx-√3)=0 sinx=0 x=0, 180º or 2sinx-√3=0 sinx=√3/2 x=60º, 120º (in quadrants I and II where sin>0)
 Trigonometry-basics/683403: Solve the equation for (theta) over the interval [0(degrees),360(degrees)). If necessary, round your answers to the nearest tenth. What is the solution set?1 solutions Answer 423604 by lwsshak3(6522)   on 2012-11-21 00:17:26 (Show Source): You can put this solution on YOUR website!Solve the equation for (theta) over the interval [0(degrees),360(degrees)). If necessary, round your answers to the nearest tenth. 3secx-2√3) = 0 secx=2√3/3 cosx=1/secx=3/(2√3)=√3/2 x=30º, 300º (in quadrants I and IV where cos>0)
 Trigonometry-basics/683501: Find the general solutions to the equation . List all solutions on the interval [0,4pi]1 solutions Answer 423603 by lwsshak3(6522)   on 2012-11-21 00:08:42 (Show Source): You can put this solution on YOUR website!List all solutions on the interval [0,4pi] csc^2theta+6csc(theta)+5=0 (cscx+5)(cscx+1)=0 cscx+5=0 cscx=-5 sinx=1/cscx=-1/5 x≈3.343,6.081,9.626,12.635 (in quadrants III and IV where sin<0) or cscx+1=0 cscx=-1 sinx=1/cscx=-1 x=3π/2 and 7π/2
 Trigonometry-basics/683333: find all values of α; 0°<α<360° that satisfy 2cos2x=11 solutions Answer 423553 by lwsshak3(6522)   on 2012-11-20 17:51:15 (Show Source): You can put this solution on YOUR website!find all values of α; 0°<α<360° that satisfy 2cos(2x)=1 2(1-2sin^2x)=1 1-2sin^2x=1/2 2sin^2x=1-1/2=1/2 sin^2x=1/4 sinx=1/2 x=30º and 150º
 Trigonometry-basics/683350: Solve the following, finding all solutions in radians in [0,2) 2cos^2(x)-√(3)cos(x)=01 solutions Answer 423550 by lwsshak3(6522)   on 2012-11-20 17:35:35 (Show Source): You can put this solution on YOUR website!Solve the following, finding all solutions in radians in [0,2) 2cos^2(x)-√(3)cos(x)=0 cosx(2cosx-√3)=0 cosx=0 x=π/2+2πk and 3π/2+2πk, k=integer and 2cosx-√3=0 cosx=√3/2 x=π/6+2πk and 11π/6+2πk, k=integer (in quadrants I and IV where cos>0)
 Trigonometry-basics/683117: Please solve equation without using a calculator. Show all work {{Cos(2sin^-1(1/4)}}1 solutions Answer 423477 by lwsshak3(6522)   on 2012-11-20 02:50:58 (Show Source): You can put this solution on YOUR website!Please solve equation without using a calculator. Show all work {{Cos(2sin^-1(1/4)}} sin^-1(1/4)=an angle x whose sin=1/4 Cos(2sin^-1(1/4)=cos2x =1-2sin^2x=1-2*(1/4)^2 =1-(2/16) =1-(1/8) =7/8 check with calculator: 2sin^-1(1/4)≈.2527*2≈.5054 cos(.5054)≈.875=7/8
 Trigonometry-basics/683118: Please solve equation for x. Show all work. {{Arctanx=arccos5/13}} 1 solutions Answer 423476 by lwsshak3(6522)   on 2012-11-20 01:40:33 (Show Source): You can put this solution on YOUR website!Please solve equation for x. Show all work. {{Arctanx=arccos5/13}} arccos5/13=an angle whose cos=5/13=adjacent side/hypotenuse opposite side=√(13^2-5^2)=√(169-25)=√144=12 tan=opposite side/adjacent side=12/5 Arctanx=to the same angle whose tan=x x=12/5
 Trigonometry-basics/683120: Please solve equation in radians for all exact solutions. Show all work {{2√3cos(x/2)=-3}}1 solutions Answer 423475 by lwsshak3(6522)   on 2012-11-20 01:30:05 (Show Source): You can put this solution on YOUR website!Please solve equation in radians for all exact solutions. Show all work {{2√3cos(x/2)=-3}} 2√3cos(x/2)=-3 cos(x/2)=-3/(2√3)=-3√3/2*3=-√3/2 cos(x/2)=-√3/2 x/2=5π/6+2πk and 7π/6+2πk, k=integer (in quadrants II and III where cos<0) x=5π/3+2πk and 7π/3+2πk, k=integer
 Trigonometry-basics/682943: I need help slove this identity, please show the steps becasue I subed in 2 for B and both sides were the same so I know it is an identity but I do not how to prove it without subing in a number (1+2inBcosB)/(sinB+cosB)=sinB+cosB Thanks :)1 solutions Answer 423473 by lwsshak3(6522)   on 2012-11-20 01:07:07 (Show Source): You can put this solution on YOUR website!prove following identity: (1+2sinBcosB)/(sinB+cosB)=sinB+cosB start with left side (1+2sinBcosB)/(sinB+cosB) multiply top and bottom by (sinB+cosB) =(1+2inBcosB)/(sinB+cosB)* (sinB+cosB)/ (sinB+cosB) =(1+2inBcosB)(sinB+cosB/ (sinB+cosB)/ (sinB+cosB) =(1+2inBcosB)(sinB+cosB/ (sin^2B+2sinBcosB+cos^2B) =(1+2inBcosB)(sinB+cosB/ (1+2sinBcosB) =sinB+cosB verified: left side=right side
 Polynomials-and-rational-expressions/682962: How do I solve the quadratic inequality x^2-8x+12<0 and answer it in interval notation?1 solutions Answer 423403 by lwsshak3(6522)   on 2012-11-19 19:13:57 (Show Source): You can put this solution on YOUR website!How do I solve the quadratic inequality x^2-8x+12<0 and answer it in interval notation? x^2-8x+12<0 factor: (x-6)(x-2)<0 number line <...+..2....-....6....+......> solution: (2,6)
 Trigonometry-basics/682945: I need help sloving this identity, please show the steps becasue I subed in 2 for B and both sides were the same so I know it is an identity but I do not how to prove it without subing in a number (1-cosB)/(sinB)=(sinB)/(1+cosB) Thanks :)1 solutions Answer 423400 by lwsshak3(6522)   on 2012-11-19 18:59:18 (Show Source): You can put this solution on YOUR website!prove following identity: (1-cosB)/(sinB)=(sinB)/(1+cosB) stare with left side (1-cosB)/(sinB) multiply both top and bottom by 1+cosB, making the denominator a difference of 2 squares (1-cosB)/(sinB)*1+cosB/1+cosB =(1-cos^2B)/sinB(1+cosB) =sin^2B/(sinB(1+cosB)) =sinB/1+cosB verified: left side=right side
 Trigonometry-basics/682757: Find the following exactly in both radians and degrees from [0,2) and [0,360). tan^-1(-1) 1 solutions Answer 423351 by lwsshak3(6522)   on 2012-11-19 16:16:19 (Show Source): You can put this solution on YOUR website!Find the following exactly in both radians and degrees from [0,2π) and [0,360). tan^-1(-1) tan^-1(-1)=3π/4 and 7π/4 or 135º and 315º( in quadrants II and IV where tan<0)
 Trigonometry-basics/682764: please help me solve this equation. Solve the equation for x over the interval [0,2pi) {{4cos2x=squareroot3}}1 solutions Answer 423350 by lwsshak3(6522)   on 2012-11-19 16:09:30 (Show Source): You can put this solution on YOUR website!please help me solve this equation. Solve the equation for x over the interval [0,2pi) {{4cos2x=squareroot3}} 4cos(2x)=√3 cos(2x)=√3/4 Identity: cos(2x)=1-2sin^2x 1-2sin^2x=√3/4 2sin^2x=1-√3/4=(4-√3)/4 sin^2x=(4-√3)/8≈.2835 sinx=√.2835=≈.5324 using calculator set to radians x=0.5685 and 2.5801 (in Q1 and Q2 where sin>0)
 Trigonometry-basics/682779: solve the equation on the interval [0,2pi) tan x = -3.5618 What is the solution set answered in radians and rounded to 4 decimal places or There is no solution1 solutions Answer 423348 by lwsshak3(6522)   on 2012-11-19 15:48:07 (Show Source): You can put this solution on YOUR website!solve the equation on the interval [0,2pi) tan x = -3.5618 What is the solution set answered in radians and rounded to 4 decimal places ** arctan(-3.5618)=1.8445 and 4.9681(in Q2 and Q4 where tan<0) x=1.8445 and 4.9681
 Exponential-and-logarithmic-functions/682812: Can you show me step by step how to solve 4-ln(5-x)=5/2 1 solutions Answer 423345 by lwsshak3(6522)   on 2012-11-19 15:28:53 (Show Source): You can put this solution on YOUR website!Can you show me step by step how to solve 4-ln(5-x)=5/2 ln(5-x)=4-5/2=8/2-5/2=3/2 ln(5-x)=3/2 convert to exponential form: e^(3/2)=5-x x=5-e^(3/2) x≈0.5183
 Trigonometry-basics/682804: Find Approximate Values of theta between 0 degrees and 360 degrees that make the equation a true statement Sec(theta)= -1.058 1 solutions Answer 423344 by lwsshak3(6522)   on 2012-11-19 15:19:10 (Show Source): You can put this solution on YOUR website!Find Approximate Values of theta between 0 degrees and 360 degrees that make the equation a true statement Sec(theta)= -1.058 cos(theta)=1/sec(theta)=-1/1.058≈-.945 using calculator arccos(-.945) theta≈161º and 199º (in Q2 and Q3 where cos<0)
 Trigonometry-basics/682406: Cot x + tan x = 2 csc2x Prove the following is an identity. Having a difficult time fitting the left side to equal the right. I know that cot x= cos x/ sin x and tan x = sin x / cos x1 solutions Answer 423255 by lwsshak3(6522)   on 2012-11-19 02:41:25 (Show Source): You can put this solution on YOUR website!Cot x + tan x = 2 csc2x Prove the following is an identity. ** start with left side: Cot x + tan x=cosx/sinx+sinx/cosx=(cos^2x+sin^2x)/sinx cosx =1/sinx cosx=2/2sinx cosx=2/sin(2x)=2csc(2x) verified: left side=right side
 Trigonometry-basics/682513: Use a sum or difference formula to find the exact value of the trigonometric function cos 75degree simplyfy the answer including any radicals use integers or fractions for any number in the expression rationalize all denominators1 solutions Answer 423254 by lwsshak3(6522)   on 2012-11-19 02:26:41 (Show Source): You can put this solution on YOUR website!Use a sum or difference formula to find the exact value of the trigonometric function cos 75degree simplyfy the answer including any radicals use integers or fractions for any number in the expression rationalize all denominators ** Identity: (s+t)=cos s cos t-sin s sin t cos(75)=cos(45+30)=cos 45 cos 30-sin 45 sin 30 =√2/2*√3/2-√2/2*1/2 =√6/4-√2/4 =(√6-√2)/4
 Trigonometry-basics/682524: I don't understand how to solve the equation for (x) over the interval of [0,360(degrees)]. 4cos(2x)= sqrt( 3 )1 solutions Answer 423253 by lwsshak3(6522)   on 2012-11-19 02:13:24 (Show Source): You can put this solution on YOUR website!solve the equation for (x) over the interval of [0,360(degrees)]. 4cos(2x)= sqrt( 3 ) cos(2x)=√3/4 2x=arccos√3/4 using calculator(cos^-1 key) 2x≈64.34º, 295.66º (in Q1 and Q4 where cos>0) x≈32.17º, 147.83º
 Trigonometry-basics/682511: solve the equation in radians for all exact solutions. 2sqrt( 3 )cos(x/2)=-31 solutions Answer 423252 by lwsshak3(6522)   on 2012-11-19 01:54:46 (Show Source): You can put this solution on YOUR website!solve the equation in radians for all exact solutions. 2sqrt( 3 )cos(x/2)=-3 ** 2√3cos(x/2)=-3 cos(x/2)=-3/(2√3) Identity: cos(x/2)=±√[(1+cos x)/2] ±√[(1+cos x)/2]=-3/(2√3) square both sides (1+cos x)/2=9/12 1+cos x=18/12=3/2 cos x=3/2-1=1/2 x=π/3+2πk, 5π/3+2πk, k=integer
 Trigonometry-basics/682514: solve the equation for x. arctan x = arccos(5/13)1 solutions Answer 423251 by lwsshak3(6522)   on 2012-11-19 01:26:22 (Show Source): You can put this solution on YOUR website!solve the equation for x. arctan x = arccos(5/13) ** arccos(5/13) means an angle whose cos=5/13=adjacent side/hypotenuse opposite side=√(13^2-5^2)=√(169-25)=√144=12 (by Pythagorean Theorem) tan=opposite side/adjacent side=12/5 arctan x=the same angle whose cos=5/13 arctan(12/5) = arccos(5/13) x=12/5