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Richard’s average driving speed is 6 kilometers per hour faster than Thor’s. In the same length of time it takes Richard to drive 640 kilometers, Thor drives only
592 kilometers. What is Richard’s average speed?
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let x=Thor's average driving speed
x+6=Richard's average driving speed
Travel time=distance/speed (same for both)
..
592/x=640/(x+6)
592(x+6)=640x
592x+3552=640x
48x=3552
x=74
x+6=80
Richard's average driving speed=80 km/hr

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The height of a triangle is 2 units more than its base. If the base is increased by 4 units and the height is decreased by 2 units, the area of the resulting triangle will be 8 square units more than the area of the original triangle. Find the base and the height of the original triangle.
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original triangle:
let x=base
x+2=height
Area=(1/2)(x)(x+2)
..
2nd triangle:
x+4=base
(x+2)-2=x=height
Area=(1/2)(x+4)(x)
..
area of 2nd triangle-area of original triangle=8
(1/2)(x+4)(x)-(1/2)(x)(x+2)=8
LCD:2
x(x+4)-x(x+2)=16
x^2+4x-x^2-2x=16
2x=16
x=8
x+2=10
..
base of original triangle=8 units
height of original triangle=10 units

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Find the equation of the locus of the points which are equidistant from the two parallel lines
This is an equation of a straight line parallel and equidistant from given parallel lines
3x - 2y + 4 = 0 and 3x - 2y - 8 = 0
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3x - 2y + 4 = 0
2y=3x+4
y=3x/2+2
..
3x - 2y - 8 = 0
2y=3x-8
y=3x/2-4
..
since equation is parallel to given lines, its slope is the same as that of given lines=3/2
equation: y=3x/2+b
y-intercept, b=mid point of y- intercepts of two given parallel lines=(2+(-4))/2=-2/2=-1
equation of the locus of the points which are equidistant from the two parallel lines: y=3x/2-1

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what is the domain, range, and zeros for y=2tan((x+(3.14/2))
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domain: all real numbers other than 0, ±π, ±2π, ±3π ... (where the asymptotes are)
range: (-∞,∞)
zeros: ±π/2, ±3π/2, ±5π/2...

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Please help me find the amplitude, period and phase shift for this function. y=11 cos[8(x+pi/3)]
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Equation for cos function: y=Acos(Bx-C), A=amplitude, period=2π/B, phase shift=C/B
For given equation: y=11 cos[8(x+pi/3)]=11 cos[8x+8pi/3)]
Amplitude=11
B=8
period=2π/B=2π/8=π/4
C=8π/3
phase shift=C/B=(8π/3)/8=π/3 (shift to the left)

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find the exact values of the other trigonometric functions if tan = 5/3, x in quadrant 3
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tan=5/3=opp side/adj side (in quadrant 3 where sin and cos <0, tan>0)
hypotenuse=√(5^2+3^2)=√(25+9)=√34
..
sin=-5/√34=-(5*√34)/34
cos=-4/√34=-(4*√34)/34
tan=5/3(given)
csc=-√34/5
sec=-√34/4
cot=3/5

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Find all t in the interval [0, 2π] satisfying 2 cos 3t - cot 3t = 0.
I honestly don't know where to even begin on this problem.
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2cos3t-cot3t=0
2cos3t=cot3t
2cos3t=cot3t/sin3t
2sin3tcos3t=cos3t
sin6t=cos3t
6t=π/3
t=π/18
note: at t=π/18, 6t and 3t become complementary angles

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find a equation of y=A sin(b(x)) and y=A cos(b(x)-c) +d with and of the following points on a sin graph with the amplitude of 2: (-3pi, -2), (6pi, 2), and (pi/2, 1)
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I'm not sure exactly what is wanted but I found the given points on the following equations:
2sin(x+π/2) for points (-3π, -2), and (6π, 2)
..
2cos(x-π/2)-1 for point (π/2, 1)

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I need to find the vertex, focus, directrix and axis of symmetry of each parabola
6x+y^2=0
y^2=-6x
This is an equation of a parabola that open leftwards.
Its standard form: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given equation: y^2=-6x
vertex:(0,0)
axis of symmetry: y=0
4p=6
p=6/4=3/2
focus: (-3/2,0) (p-distance left of vertex on axis of symmetry
directrix: x=3 (p-distance right of vertex on axis of symmetry)
..
x^2=y+2x
y=x^2-2x
This is an equation of a parabola that open upwards.
Its standard form:y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex, A>0, curve opens upwards, A<0, curve opens downwards.
for given equation: y=x^2-2x
complete the square
y=(x^2-2x+1)-1
y=(x-1)^2-1
rewrite to standard form of first equation above
(x-1)^2=(y+1)
vertex: (1,-1)
axis of symmetry: x=1
4p=1
p=1/4
focus: (1,-3/4) (p-distance above vertex on axis of symmetry)
directrix: y=-5/4 (p-distance below vertex on axis of symmetry)

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For each equation, identify the equation of the axis of symmetry, the coordinates of the vertex, the y-intercept, the zeros, and the maximum or minimum values.
y = x2 + 4x + 3
y = -2(x - 3)2 + 9
y = 3(x - 0.5)2
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All 3 given equations are parabolas and take the standard form: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex. A is a coefficient that affects the width of the curve. If A>0, parabola opens upwards and parabola has a minimum. If A<0, parabola opens downwards and parabola has a maximum.
..
y = x^2 + 4x + 3
complete the square
y = (x^2+4x+4)+3-4
y = (x+2)^2-1
vertex: (-2,-1)
axis of symmetry: x=-2
minimum: -1
..
y-intercept
set x=0
y=3 (from original equation)
..
x-intercepts (zeros)
set y=0
(x+2)^2-1=0
(x+2)^2=1
x+2=±√1=±1
x=-2±1
x=-3 and -1
..
I will let you do the remaining 2 equations

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cos2xcos3x+sin2xsin3x
Identity: cos(s-t)=cos s cos t+sins sin t
cos2xcos3x+sin2xsin3x
=cos(2x-3x)
=cos(-x)
=cosx (cos is an even function)
..
sin5xcosx-cos5xsinx
Identity: sin(s-t)=sin s cos t-cos s sin t
sin5xcosx-cos5xsinx
=sin(5x-x)
=sin (4x)

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Find all solutions of the equation in the interval [0,2pi)
sin (x+7pi) - sinx - sqrt.2 =0
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sin addition formula: sin(a+b)=sin a cos b+cos a sin b
sin(x+7π)=sin x cos 7π+cos x sin 7π)
cos 7π=-1
sin 7π=0
sin x cos 7π+cos x sin 7π)
sin x*(-1)+cos x*(0)
sin(x+7π)=-sin x
..
sin (x+7π) - sinx - sqrt.2 =0
-sin x-sin x-√2=0
-2sin x=√2
sin x=-√2/2
x=5π/4 and 7π/4 (in quadrants III and IV where sin<0)

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Derive the identity for tan(a-b)using tan(a-b)=tan[a+(-b)].
After applying the formula for the tangent of the sum of two angles, use the fact that the tangent is an odd function.
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if function is even, f(-x)=f(x)
if function is odd, f(-x)=-f(x)
tan, being an odd function:
tan(-x)=-tan(x)
..
Identity: tan(a+b)=(tana+tanb)/(1-tana tanb)
tan[(a+(-b)]=(tana+tan(-b))/(1-tana tan(-b))
tan(-b)=-tan b
tan(a-b)=(tana-tanb)/(1+tana tanb)

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Solve for theta using tan^-1. tan(6theta+5)=-14
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use x for theta
for interval [0,π]
tan(6x+5)=-14
using inverse tan key on calculator:
tan^-1(-14)=1.64 radians(in quadrant II where tan<0)
6x+5=1.64
6x=1.64-5
x=-0.56

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find all solutions of the equation sec4(theta)-2=0
find the solutions in the interval [0,2pi]
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sec4x-2=0
1/cos4x=2
cos4x=1/2
4x=π/3 and 5π/3 (in quadrants I and IV where cos>0)
x=π/12 and 5π/12

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find all solutions of the equation sec(theta)tan(theta)-cos(theta)cot(theta)=sin(theta)
find the solutions in the interval [0,2pi]
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secxtanx-cosxcotx=sinx
1/cosx*sinx/cosx-cosx*cosx/sinx=sinx
sinx/cos^2x-cos^2x/sinx=sinx
multiply each term by sinx
sin^2x/cos^2x-cos^2x=sin^2x
sin^2x/cos^2x=sin^2x+cos^2x
sin^2x/cos^2x=1
tan^2x=1
tanx=±√1=±1
x=π/4, 3π/4, 5π/4, and 7π/4

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Given: cosθ= -3/5 and sinθ > 0.
You are working with a 3-4-5 right triangle and reference angle in quadrant II where cos<0 and sin>0
cosx=-3/5=adj side/hypotenuse
opp side=√(5^2-3^2)=√(25-9)=√16=4
...
Find the exact value of cos x/2.
Identity: cos (x/2)=±√[(1+cosx)/2]
=-√[(1-3/5)/2]
=-√[(2/5)/2]
=√(2/10)
=√(1/5)
=1/√5
=√5/5
..
solve cos 2x=(sqrt2)/2 on the interval [0,2pi). Four solutions.
cos 2x=√2/2
2x=π/4, and 7π/4 (in quadrants I and IV where cos>0) (2 solutions)
x=π/8 and 7π/8
..
solve cos^(2)x+2cos+1=0 on the interval [0, 2pi)
cos^(2)x+2cos+1=0
(cosx+1)^2=0
cosx=-1
x=π (multiplicity 2)

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Determine two coterminal angles (one positive and one negative) for each angle. Give your answers in radians.
1.-19pi/8 and 13pi/8
2.-8pi/15 and 22pi/15
3.-11pi/6 and pi/6
4. 7pi/6 and -5pi/6

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cos(x) csc(x)
cos(x)*1/sin(x)
cos(x)/sin(x)=cot(x)

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the wheel of a bicycle has a diameter of 26 inches find the radian angle that the wheel turns when the bicycle turns 15 ft.
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circumference of wheel=π*diameter=26π in
For each revolution wheel turns 26π in
rev/26π in*12 in/ft*2π radians/rev*15 ft
rev, π, in, ft cancel out
leaving: 12*2*15/26 ≈13.8 radians or 13.8/2π≈2.2 turns
angle that the wheel turns when the bicycle turns 15 ft=13.8 radians

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Find all t in the interval [0, 2π] satisfying (cos t)2 + 4 cos t + 3 = 0.
cos ^2 t+4cos t+3=0
(cos t+3)(cos t+1)=0
cos t+3=0
cos t=-3 (reject, (-1 ≤ cos t ≤1)
(cos t+1)=0
cos t=-1
t=π

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sin(cos^-1(4/7) (this says to find sin of an angle whose cos is 4/7)
cos=adj side/hypotenuse
opp side=√(7^2-4^2)=√(49-16)=√33)
sin(cos^-1(4/7))=√33/7

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Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)
cos 2θ = cos^2 θ − 1/2
cos^2x-sin^2x=cos^2x-1/2
sin^2x=1/2
sinx=1/√2=√2/2
x=π/4+2πk, 3π/4+2πk, k=integer

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what is the point (x,y) on the unit circle that corresponds to a real number 23pi/6 ?
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Standard position of 23π/6 is in quadrant IV (24π/6-π/6)
The reference angle becomes π/6 in quadrant IV where cos>0 and sin<0
The x-coordinate represents the cos function and y-coordinate represents the sin function
On the unit circle, y/1=-1/2 at π/6, so y=-1/2
Similarly, x/1=√3/2/1 at π/6, so x=√3/2
The point (x,y) on the unit circle that corresponds to a real number 23pi/6 is: (√3/2,-1/2)
Check:
let A=reference angle π/6 in quadrant IV
sin A=opp side/hypotenuse=y/1=-1/2/1=-1/2
cosA=adj side/hypotenuse=x/1=(√3/2)/1=√3/2

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a)f(x)=x2-8x+12. b) f(x)=4x-x2
i) find where the parabola cuts the y-axis
ii) find the quadratic equations roots
iii) find the axis of symmetry. and the turning point
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a)f(x)=x2-8x+12
complete the square.
y=(x^2-8x+16)+12-16
y=(x-4)^2-4
This is a parabola that opens upwards with vertex at (4,-4)
..
i) find where the parabola cuts the y-axis
set x=0
y=16-4=12
parabola cuts the y-axis at 12 (y-intercept)
..
ii) find the quadratic equations roots
set y=0
(x-4)^2-4=0
(x-4)^2=4
x-4=±√4=±2
x=4±2
roots are:
x=6 and 2 (x-intercepts)
..
iii) find the axis of symmetry. and the turning point.
axis of symmetry: x=4
turning point: (4,-4) (vertex)
I will let you do b)

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what is the equation for this ellipse with a foci at (1,-4) and (-3,-4) and a length major axis at 12?
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Foci information show this is an ellipse with horizontal major axis.
Its standard form of equation: , a>b, (h,k)=(x,y) coordinates of center.
For given ellipse:
center: (-1,-4) (midpoint of x-coordinate and y-coordinate of foci)
given length of major axis=12=2a
a=6
a^2=36
c=2 (distance from center to foci on horizontal major axis)
c^2=4
c^2=a^2-b^2
b^2=a^2-c^2=36-4=32
Equation of given ellipse:

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Write the standard equation for each Hyperbola.
vertices ( 0, -12) ( 0, 12) and co-vertices (-11, 0) (11, 0)
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This is a hyperbola with vertical transverse axis.
Its standard form of equation:, (h,k)=(x,y) coordinates of center.
For given hyperbola:
center: (0,0) (midpoints of horizontal conjugate axis and vertical transverse axis)
length of vertical transverse axis=24 (-12 to 12)=2a
a=12
a^2=144
length of horizontal conjugate axis=22 (-11 to 11)=2b
b=11
b^2=121
Equation of given hyperbola:

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what is an equation of an elipse with major axis endpoints (2,8) and (2,-4) and minor axis of length 10
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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: (2,2)
length of vertical major axis=12 (-4 to 8)=2a
a=6
a^2=36
given length of minor axis=10=2b
b=5
b^2=25
Equation of given ellipse:

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Find an equation of the line with slope 3 and containing the point of intersection of 2x+3y-5=0 and 3x-7y+5=0.
**
3x-7y+5=0
2x+3y-5=0
..
6x-14y+10=0 (multiply by 2)
6x+9y-15=0 (multiply by 3)
subtract
-23y+25=0
y=25/23≈1.087
2x=5-3y=5-75/23
x=(5-75/23)/2≈0.870
point of intersection: (0.870,1.087)
equation of line: y=mx+b
y=3x+b
solve for b using coordinates of point of intersection
1.087=3*.870+b
b=-1.523
y=3x-1.523
comment: what horrible answers!

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FOR THE GIVEN EQUATION, LIST THE INTERCEPTS AND TEST FOR SYMMETRY.
y=-x^7/x^6-8
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y-intercepts, set x=0
y=-x^7/(x^6-8)
y=0/(0-8)=0
y-intercept=0
..
x-intercepts, set y=0
0=-x^7/(x^6-8)
-x^7=0
x=0
x-intercept=0
..
Test for symmetry:
-x gives y
x gives -y
when we get opposite values of y at x or -x, graph is symmetrical about the origin

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How would I test for intercepts. I have to show the steps to test the intercepts. If anyone could help me, that would be fantastic!!!!
X^2+y^2+12x+10y+61=16
X^2+12x+36+y^2+10y+25=-45+36+25
(x+6)^2+(y+5)^2=16
C (-6, -5) and r=4
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x-intercepts, set y=0
(x+6)^2+(y+5)^2=16
(x+6)^2+(0+5)^2=16
(x+6)^2+25=16
(x+6)^2=-9
(x+6)=±√-9=±3i
no real roots
x-intercepts: none
..
y-intercepts, set x=0
(0+6)^2+(y+5)^2=16
36+(y+5)^2=16
(y+5)^2=-20
(y+5)^2=-20
(y+5)=±√-20=±√20 i
no real roots
y-intercepts: none