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lwsshak3 answered: 6458 problems
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1 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
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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
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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
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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
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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
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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
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�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
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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
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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
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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
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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
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Rational-functions/604007: find the rational zeros of f(x)=-x^3-4x^2+11x+30
1 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
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Radicals/604380: how do you rationalize the denominator and simplify cube root of 4y^2/cube root of 5x^11
1 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
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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
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Trigonometry-basics/604377: what are the six trigonometric functions of theta, where the terminal arm of the angle passes through the point (-12,5). thanks
1 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
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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 = 1
1 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
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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 degrees
1 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
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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
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Trigonometry-basics/604209: given sec(t)=2.345 sin(t)<0, find (t)=(0,2pi), list answers in a set
1 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}
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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
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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
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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
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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
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