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Quadratic-relations-and-conic-sections/745114: Which is the equation with focus of (6,0)
A) Y=-1/6^2
B) y=1/12x^2
C) x=1/24y^2
D) x=-1/6y^2
1 solutions
Answer 453796 by lwsshak3(6505)   on 2013-05-05 19:51:19 (Show Source):
You can put this solution on YOUR website!Which is the equation with focus of (6,0)
A) Y=-1/6^2
B) y=1/12x^2
C) x=1/24y^2
D) x=-1/6y^2
***
Parabola opens rightward:
Its basic equation: (y-k)^2=4p(x-h), (hk)=(x,y) coordinates of the vertex
For given equation:
vertex: (0,0)
axis of symmetry: y=0
p=6 (distance from vertex to focus on the axis of symmetry)
4p=24
equation: y^2=4px=24x
x=y^2/24 ans. C)
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logarithm/744933: For the functions ƒ(x) = log_6(x) and g(x)=log_1/4(x), for what values of x is g(x) > ƒ(x)?
1 solutions
Answer 453673 by lwsshak3(6505) on 2013-05-05 03:06:56 (Show Source):
You can put this solution on YOUR website!For the functions ƒ(x) = log_6(x) and g(x)=log_1/4(x), for what values of x is g(x) > ƒ(x)?
..
..
At x=1, g(x)=f(x)
At x<1, g(x)>f(x)
Between (0 and 1), f becomes more negative as x becomes more negative; whereas, g becomes more positive as x becomes more negative.
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logarithm/744937: Identify the transformations of the function ƒ(x) = log_2(2(x + 6)) from the base graph g(x) = log_2(x).
1 solutions
Answer 453669 by lwsshak3(6505) on 2013-05-05 01:50:52 (Show Source):
You can put this solution on YOUR website!Identify the transformations of the function ƒ(x) = log_2(2(x + 6)) from the base graph g(x) = log_2(x).
***
transformations: moves x-intercept 6 units left, from (0,1) to (0,-5) and stretches basic curve up vertically. Asymptote moves from y-axis to x=-6, 6 units to the left.
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Trigonometry-basics/744248: Find solutions in radians for 0 less then or equal to x less then 2pi:
sin^2(3x)-3sin(3x)-4=0
1 solutions
Answer 453569 by lwsshak3(6505) on 2013-05-04 04:16:01 (Show Source):
You can put this solution on YOUR website!Find solutions in radians for 0 less then or equal to x less then 2pi:
sin^2(3x)-3sin(3x)-4=0
sin((3x)-4)(sin(3x)+1)=0
..
sin((3x)-4)=0
sin(3x)=4 (reject,(-1 ≤ sin(3x) ≤ 1))
..
sin(3x)+1=0
sin(3x)=-1
3x=3π/2
x=3π/6=π/2
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Trigonometry-basics/744764: Evaluate the given expression without using a calculator. (Assume any variables represent positive numbers)
Sin( arccos 3/5 + arctan 1/2)
1 solutions
Answer 453537 by lwsshak3(6505) on 2013-05-03 21:09:12 (Show Source):
You can put this solution on YOUR website!Evaluate the given expression without using a calculator. (Assume any variables represent positive numbers)
Sin( arccos 3/5 + arctan 1/2)
***
let x=arccos 3/5
let y=arctan 1/2
..
..
..
Sin( arccos 3/5 + arctan 1/2)=sin(x+y)
=sin(x) cos(y)+cos(x) sin(y)
=4/5*2/√5+3/5*1/√5
=8/5√5+3/5√5=11/(5√5)
=(11√5)/25
..
Check: (with calculator)
cos(x)=3/5
x≈53.13º
tan(y)=1/2
y≈26.57º
x+y≈79.7º
sin(x+y)=sin79.7º≈0.9839
exact answer=(11√5)/25≈0.9839
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Trigonometry-basics/744730: If sec x = -sqrt10 with 180 degrees less than or equal to x less than or equal to 270 degrees, find the following and write your answer in an exact form.
A. Sin(2x)
B. Cos(x/2)
I appreciate all the help. :)
1 solutions
Answer 453530 by lwsshak3(6505) on 2013-05-03 20:43:37 (Show Source):
You can put this solution on YOUR website!If sec x = -sqrt10 with 180 degrees less than or equal to x less than or equal to 270 degrees, find the following and write your answer in an exact form.
A. Sin(2x)
B. Cos(x/2)
***
 (180º ≤ x ≤ 270º)
..
A. 
..
B. 
= 
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Trigonometry-basics/744624: Given that sin A= -4/5 where A is in quadrant 3 and sin B= 12/13 where B is in quadrant 2. Find:
A. cosA
B. cosB
C. sec(A-B)
D. tan(2A)
E. cos(B/2)
Help!
1 solutions
Answer 453509 by lwsshak3(6505) on 2013-05-03 16:54:18 (Show Source):
You can put this solution on YOUR website!Given that sin A= -4/5 where A is in quadrant 3 and sin B= 12/13 where B is in quadrant 2. Find:
A. cosA= 
..
B. cosB = 
..
C. sec(A-B)=1/cos(A-B)
cos(A-B)=cosA cosB+SinA sinB=-3/5*-5/13+(-4/5)*12/13=15/65-48/65=-33/65
sec(A-B)=-65/33
..
D. tan(2A)=(2tanA)/(1-tan^2A)
tanA=sinA/cosA=(-4/5)/(-3/5)=4/3
tan(2A)=(8/3)/(1-16/9)=(8/3)/(-7/9)=-72/21
..
E. cos(B/2)=  =√((8/13)/2))=√(8/26)
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logarithm/744486: 1. Solve the following equations
a) log(base 2) x+1 + log(base 2) x-1 = 3
b) log(base 2) x^2-1 =3
c) explain why the equations log(base 2) x+1 + log(base 2) x-1 = 3 and log(base 2) x^2-1 =3 are not equal.
Can you please help me out? Thanks so much in advance:)
Can you please show all the steps it would really help me understand:)
1 solutions
Answer 453356 by lwsshak3(6505) on 2013-05-03 03:10:24 (Show Source):
You can put this solution on YOUR website!1. Solve the following equations
a) log(base 2) x+1 + log(base 2) x-1 = 3
(x+1)(x-1)=8
x^2-1=8
x^2=9
x=±3
..
b) log(base 2) x^2-1 =3
x^2-1=8
x^2=9
x=±3
..
c) explain why the equations log(base 2) x+1 + log(base 2) x-1 = 3 and log(base 2) x^2-1 =3 are not equal.
The equations of a) and b) are equal as shown above.
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Quadratic-relations-and-conic-sections/743371: Graph the ellipse and identify the center,vertrices,foci and endpoints of 25x^2+9y^2=225
1 solutions
Answer 453355 by lwsshak3(6505) on 2013-05-03 02:47:54 (Show Source):
You can put this solution on YOUR website!Graph the ellipse and identify the center,vertrices,foci and endpoints of
This ellipse has a vertical major axis.
Its standard form of equation:  , a>b, (h,k)=(x,y) coordinates of center
center: (0,0)
a^2=25
a=5
vertices: (0,0±a)=(0,0±5)=(0,-5) and (0,5)
..
b^2=9
b=3
end points of minor axis: (0±b,0)=(0±3,0)=(-3,0) and (3,0)
..
c^2=a^2-b^2=25-9=16
c=4
foci: (0,0±c)=(0,0±4)=(0,-4) and (0,4)
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Quadratic-relations-and-conic-sections/743930: consider the ellipse 9x^2+49y^2=9
Its vertices are (0,+or-A)
Its Foci are (0,+or-B)
Its eccentricity is?
The length of its major and minor axis are?
1 solutions
Answer 453354 by lwsshak3(6505) on 2013-05-03 02:27:13 (Show Source):
You can put this solution on YOUR website!consider the ellipse 9x^2+49y^2=9
Its vertices are (0,+or-A)
Its Foci are (0,+or-B)
Its eccentricity is?
The length of its major and minor axis are?
***
I will use standard notation which you can convert to your own notation:
Given ellipse has a horizontal major axis.
Its standard form of equation:  ,(a>b), (h,k)=(x,y) coordinates of center.
a^2=1
a=1
length of major axis=2a=2
b^2=9/49
b=3/7
length of minor axis=2b=6/7≈0.8571
c^2=a^2-b^2=1-9/49=49/49-9/49=40/49
c=√(40/49)≈0.9035
eccentricity=c/a=0.9035/1≈0.9035
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Quadratic-relations-and-conic-sections/744434: focus (2,5) and directrix y=3
write standard form of the equation of the parabola with the given criteria.
1 solutions
Answer 453351 by lwsshak3(6505) on 2013-05-03 01:28:43 (Show Source):
You can put this solution on YOUR website!focus (2,5) and directrix y=3
write standard form of the equation of the parabola with the given criteria.
***
Given parabola opens upward. (directrix below focus)
Its basic form of equation: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex.
y-coordinate of vertex=midpoint between focus and directrix on the axis of symmetry=(5+3)/2=4
x-coordinate of vertex=2
vertex: (2,4)
axis of symmetry: x=2
p=1 (distance from vertex to focus or directrix on the axis of symmetry)
4p=4
Equation of given parabola: (x-2)^2=4(y-4)
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Quadratic-relations-and-conic-sections/743933: Identify the vertex, focus, and directrix of the graph of y=1/12(x+5)^2-3
1 solutions
Answer 453312 by lwsshak3(6505) on 2013-05-02 19:56:29 (Show Source):
You can put this solution on YOUR website!Identify the vertex, focus, and directrix of the graph of
y=1/12(x+5)^2-3
rewrite equation to basic form: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex
1/12(x+5)^2=y+3
(x+5)^2=12(y+3)
vertex:(-5,-3)
axis of symmetry: x=-5
4p=12
p=3
focus: (-5,0)
directrix:(-5,-6)
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Quadratic-relations-and-conic-sections/744040: Classify the conic section and write its ezuation in standard form. Then graph the equation.
4x^2+9y^2+40x+72y+208=0
It is an ellipse. I Completed the square and all that but I can't factor it after that..
1 solutions
Answer 453306 by lwsshak3(6505) on 2013-05-02 19:30:24 (Show Source):
You can put this solution on YOUR website!Classify the conic section and write its equation in standard form. Then graph the equation.
4x^2+9y^2+40x+72y+208=0
complete the square:
4x^2+40x+9y^2+72y+208=0
4(x^2+10x+25)+9(y^2+8y+16)=-208+100+144
4(x+5)^2+9(y+4)^2=36
Given conic section is an ellipse with horizontal major axis with center at (-5,-4)
y=±(4-4(x+5)^2/9)^.5-4
see graph below as a visual check:
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Quadratic-relations-and-conic-sections/744051: A satellite dish has a parabolic cross section and is 6 ft deep. The focus is 4 ft from the vertex. Find the width of the satellite dish at the opening. Round your answer to the nearest foot.
1 solutions
Answer 453301 by lwsshak3(6505) on 2013-05-02 19:06:30 (Show Source):
You can put this solution on YOUR website!A satellite dish has a parabolic cross section and is 6 ft deep. The focus is 4 ft from the vertex. Find the width of the satellite dish at the opening. Round your answer to the nearest foot.
***
place vertex at (0,0)
basic equation: x^2=4py
p=4 (distance from vertex to focus)
4p=16
equation: x^2=16y
y=6 at ends of parabolic cross section
x^2=16*6=96
x=±√96
width of the satellite dish at the opening=2*√96≈20 ft
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Quadratic-relations-and-conic-sections/744406: write an equation for the conic section
parabola with vertex at(0,0) and directrix x = 5
1 solutions
Answer 453298 by lwsshak3(6505) on 2013-05-02 18:49:34 (Show Source):
You can put this solution on YOUR website!write an equation for the conic section
parabola with vertex at(0,0) and directrix x = 5
***
This is a parabola that opens rightward.
Its basic form of equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given parabola:
given vertex: (0,0)
axis of symmetry: y=0 or x-axis
p=5 (distance from vertex to directrix on the axis of symmetry)
4p=20
Equation: y^2=20x
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Quadratic-relations-and-conic-sections/744390: parabola with vertex at (0,0) and directrix at x=-3
1 solutions
Answer 453290 by lwsshak3(6505) on 2013-05-02 18:04:11 (Show Source):
You can put this solution on YOUR website!parabola with vertex at (0,0) and directrix at x=-3
Given parabola opens rightward.
Its basic form of equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
given vertex: (0,0)
axis of symmetry: y=0
p=3 (distance from vertex to directrix on the axis of symmetry
4p=12
Equation of given parabola: y^2=12x
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Quadratic-relations-and-conic-sections/744237: How do i find the vertices and foci of; (x^2/64)-(9y^2/4)=1
1 solutions
Answer 453265 by lwsshak3(6505) on 2013-05-02 15:30:26 (Show Source):
You can put this solution on YOUR website!How do i find the vertices and foci of; (x^2/64)-(9y^2/4)=1
Given ellipse has a horizontal major axis.
Its standard form of equation:  ,a>b, (h,k)=(x,y) coordinates of center
(x^2/64)-(9y^2/4)=1
(x^2/64)-y^2/(4/9)=1
center: (0,0)
a^2=64
a=8
vertices: (0±a,0)=(0±8,0)=(-8,0) and (8,0)
b^2=4/9
b=2/3
c^2=a^2-b^2=64-4/9=(576/9)-(4/9)=572/9
c=√(572/9)≈7.97
foci: (0±c,0)=(0±7.97,0)=(-7.97,0) and (7.97,0)
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Quadratic-relations-and-conic-sections/744231: I need to write an equation for an ellipse with the endpoints of the major axis (7,3) (7,9) , endpoints of the minor axis at (5,6) , (9,6).
1 solutions
Answer 453256 by lwsshak3(6505) on 2013-05-02 15:02:06 (Show Source):
You can put this solution on YOUR website!I need to write an equation for an ellipse with the endpoints of the major axis (7,3) (7,9) , endpoints of the minor axis at (5,6) , (9,6).
***
Given ellipse has a vertical major axis. (y-coordinates of major axis change but x-coordinates do not)
Its standard form of equation:  , a>b, (h,k)=(x,y) coordinates of center
y-coordinate of center=6 (midpoint of vertical major axis, (9+3)/2=12/2=6
x-coordinate of center=7
center:(7,6)
a=6 (distance from center to end points of major axis)
a^2=36
b=2 (distance from center to end points of minor axis)
b^2=4
Equation for given ellipse:
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Quadratic-relations-and-conic-sections/744294: Find the equation for the hyperbola with foci (4,6) and (-8,6) and a transverse axis length of 8
1 solutions
Answer 453247 by lwsshak3(6505) on 2013-05-02 14:23:36 (Show Source):
You can put this solution on YOUR website!Find the equation for the hyperbola with foci (4,6) and (-8,6) and a transverse axis length of 8
***
Given hyperbola has a horizontal transverse axis.
Its standard form of equation:  , (h,k)=(x,y) coordinates of center.
For given hyperbola:
center: (-2, 6)
length of transverse axis=8=2a
a=4
a^2=16
c=6 (distance from center to foci on the transverse axis)
c^2=36
c^2=a^2+b^2
b^2=c^2-a^2=36-16=20
Equation of given hyperbola:
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Trigonometry-basics/744146: Please help me determine two other angles that have the same trig ratio as sin 2.3. I dont know where to begin; can i switch the radian to a degree? Please help!!
1 solutions
Answer 453220 by lwsshak3(6505) on 2013-05-02 03:22:45 (Show Source):
You can put this solution on YOUR website!Please help me determine two other angles that have the same trig ratio as sin 2.3. I dont know where to begin; can i switch the radian to a degree?
***
using calculator set to radians:
sin(2.3)≈0.7457 (in quadrant II where sin>0)
sin(π-2.3)=sin(.8416)≈0.7457 (in quadrant I where sin>0)
sin (π+2.3)=sin(5.4416)≈-0.7457 (in quadrant III where sin<0)
sin(2π-2.3)=sin(3.9832)≈-0.7457 (in quadrant IV where sin<0)
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Trigonometry-basics/743932: Find all (theta) in the interval [0,2pi] satisfying
a) 2sin^2(theta) + sin(theta) = 1
b) 5cos)theta) + 2cos(theta + pi) + 3sin(pi/2 - (theta)) = 3
Please show work.
1 solutions
Answer 453204 by lwsshak3(6505) on 2013-05-02 01:45:01 (Show Source):
You can put this solution on YOUR website!Find all (theta) in the interval [0,2pi] satisfying
a) 2sin^2(theta) + sin(theta) = 1
2sin(x)-1=0
sin(x)=1/2
x=π/6, 5π/6
or
sin(x)+1=0
sin(x)=-1
x=3π/2
..
b) 5cos)theta) + 2cos(theta + pi) + 3sin(pi/2 - (theta)) = 3
6cos(x)=3
cos(x)=3/6=1/2
x=π/3, 5π/3
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Trigonometry-basics/743938: please help me solve this problem!!! What is the value of csc (-19pi/3)?
1 solutions
Answer 453203 by lwsshak3(6505) on 2013-05-02 01:19:46 (Show Source):
You can put this solution on YOUR website!What is the value of csc (-19pi/3)?
(-19π/3)=(-18π/3-π/3)=(-6π-π/3)=3(2π) revolutions +π/3 clockwise.
you are working with a reference angle, π/3 in quadrant IV where csc and sin<0
csc (-19pi/3)=-csc(π/3)=1/-sin(π/3)=1/-(√3/2)=-2/√3=-2√3/3
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