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if cos(a) is 2/3 and sin(b) is 4/5 what is cos(a+b)
..
cos a=2/3=adj side/hypotenuse
opp side=√(3^2-2^2)=√(9-4)=√5
sin a=√5/3
..
sin b=4/5 (3-4-5 right triangle)
cos b=3/5
..
cos(a+b)=cos a cos b -sin a sin b
=2/3*3/5 -√5/3*4/5
=6/15-4√5/15
=(6-4√5)/15
..
calculator check:
a=cos^1(2/3)≈.8411
b=sin^-1(4/5)=.9273
cos(a+b)=cos(1.7684)=-0.1962..
(6-4√5)/15=-0.1962..

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Find the equation of the ellipse with the given conditions.
Center (1,1), vertex (1,3), passing through (0,0).
**
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: (1,1)
length of major axis=4=2a
a=2
a^2=4
plug in coordinates of given point on ellipse: (0,0)
(x-1)^2/b^2+(y-1)^2/4=1
(0-1)^2/b^2+(0-1)^2/4=1
1/b^2+1/4=1
1/b^2=3/4
b^2=4/3
Equation of given ellipse:

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Find an equation in standard form of the parabola passing through the following points (-1,1) (0,5) (2,7)
y=Ax^2+Bx+C
(1) 1=A-B+C (-1,1)
(2) 5=C (0,5)
(3) 7=4A+2B+C (2,7)
..
1=A-B+5
7=4A+2B+5
2=2A-2B+10
add
9=6A+15
6A=-6
A=-1
..
B=A+5-1=3
Equation:
y=Ax^2+Bx+C
y=-x^2+3x+5
complete the square
y=-(x^2-3x+9/4)+9/4+5
y=-(x-3/2)^2+29/4
This is an equation of a parabola that opens downwards with vertex at(3/2,29/4)

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Verify the following equation:
cosx(cosx-secx)=-sin^2x
start with left side
cosx(cosx-secx)
=cosx(cosx-1/cosx)
=cos^2x-1
=-(1-cos^2x)
=-sin^2x
verified:
left side=right side

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given sec theta equals -10 with 180< theta< 270 find and simplify sin theta/2 as an exact value
use x for theta
secx=-10=1/cos
cosx=-1/10=adjacent side/hypotenuse
opposite side=√(10^2-1^2)=√(100-1)=√99
sinx=-√99/10 (in quadrant III where sin<0)
sin(x/2)
=√[(1-cosx)/2]
=√[(1-(-1/10)/2]
=√[(1+(1/10)/2]
=√(1.1/2)
..
check with calculator:
sin^-1(√99/10)=84.26º+180=264.26º(in quadrant III as given)
sin(x/2)=132.13 (in quadrant II where sin>0)
reference angle=132.13-180=47.87º
sin 47.87º≈0.7416..
√(1.1/2)≈0.7416..

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how do I found the exact answer of sec(-5 pi/8)?
use identity for cos half-angle formula: cos (s/2)=±√[(1+cos s)/2]
-sec(-5π/8)=-1/cos(-5π/8)
solve for cos(-5π/8), then take the negative reciprocal.
cos(-5π/8)=cos(-5π/4)/2)=-√[1+cos(-5π/4)/2]
cos(-5π/4)=cos π/4=-√2/2 in quadrant II where cos<0
-√[1+cos(-5π/4)/2]=-√[1-√2/2/2]=-√(2-√2)/2
-sec(-5π/8)=-1/cos(-5π/8)
-sec(-5π/8)=-1/-√(2-√2)/2
-sec(-5π/8)=2/√(2-√2)
check with computer:
cos(-5π/8)=-.3836
-sec(-5π/8)≈-1/-.3836=2.6131..
2/√(2-√2)≈2.6131..

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What is the reference angle for 5pi/8 and -490 degrees?
5π/8=1.9635 radians in quadrant II
reference angle=π-1.9635≈1.1781
..
-490º+360º=-130º in quadrant III
reference angle=-130+180=50º

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find period and phase shift, y= -4 tan(5x + pi/2)
Equation of tan function: y=tan(Bx-C), period=π/B, phase shift=C/B
For given tan function:
B=5
period=π/B=π/5
C=π/2
phase shift=C/B=(π/2)/5=π/10 (shift to the left)

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Find the exact value of the six trig functions. (5,9) is a point on the terminal side of theta
use x for theta
you are working with a reference angle in quadrant I where all 6 functions>0
tanx=9/5=opposite side/adjacent side
hypotenuse=√(9^2+5^2)=√(81+25)=√106
sinx=9/√106
cosx=5/√106
tanx=9/5
cscx=√106/9
secx=√106/5
cotx=5/9

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Write the standard equation for the ellipse with the given characteristics Foci: (5, 0), (-5, 0) Vertices: (9, 0), (-9, 0)
**
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: (0,0)
a=9 (center to vertex on the major axis)
a^2=81
c=5 (center to foci on the major axis)
c^2=25
c^2=a^2-b^2
b^2=a^2-c^2=81-25=56
equation:

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bOB OWN A WATCH REPAIR SHOP. hE HAS found that the cost of operating his shop is given by c(x) = 2x^2-112x+58, where c is cost and x is the number of watches repaired. how many watches must he repair to have the lowest cost?
**
c(x) = 2x^2-112x+58
This is an equation of a parabola that opens upwards, (curve has a minimum)
Its standard form: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex. (y-value is the minimum in this case.
c(x) = 2x^2-112x+58
complete the square
c(x)=2(x^2-56x+28^2)-2*28^2+58
c(x)=2(x^2-56x+784)-1568+58
c(x)=2(x-28)^2-1510
vertex: (28,-1510)
how many watches must he repair to have the lowest cost? 28
see graph below as a visual check

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Solve the following rational inequality.
x>= 3/(x+2)
write your answer in interval notation
x+2=0
x≠-2
number line:
<.....-.....-2......+.......>
solution: (-2,∞)

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Find the coordinates of the vertex and the equation of the axis of symmetry for the parabola given.
Y= -1/4 x squared -1
y=-(1/4)x^2-1
4y=-x^2-4
x^2=-4y-4
x^2=-4(y+1)
This is an equation of a parabola that opens downwards
Its standard form: (x-h)^2=-4p(y-k), (h,k)=(x,y) coordinates of vertex
For given equation:x^2=-4(y+1)
vertex: (0,-1)
axis of symmetry: x=0 or y-axis

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how do you find the vertices, foci, of the ellipse.
56x^2 + 7y^2 = 8
**
56x^2+7y^2 = 8
7x^2+7y^2/8 = 1
This is an equation of an ellipse with vertical major axis.
Its standard form:, a>b, (h,k)=(x,y) coordinates of center
For given equation:7x^2+7y^2/8 = 1
center: (0,0)
a^2=8/7
a=√(8/7)≈1.07
vertices:(0,0±a)=(0,0±1.07)=(0,-1.07) and (0,1.07 )
b^2=1/7
b=√7/7≈.38
c^2=a^2-b^2=8/7-1/7=7/7=1
c=1
foci: (0,0±c)=(0,0±1)=(0,-1) and (0,1)

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How do I write an equation for a parabola whose vertex is (4,1) and passes through the point (2,13)?Standard form of equation for a parabola:
(x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex
plug in given coordinates of the vertex(4,1)
(x-4)^2=4p(y-1)
plug in coordinates of given point (2,13)
(2-4)^2=4p(13-1)
4=4p(12)
4p=4/12=1/3
equation of parabola:
(x-4)^2=(y-1)/3

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what is the solution to this problem ln(x-9)-ln(x+1)=ln(x-6)-ln(x+5) ?
ln(x-9)-ln(x+1)=ln(x-6)-ln(x+5)
ln[(x-9)/(x+1)]=ln[(x-6)/(x+5)]
(x-9)/(x+1)=(x-6)/(x+5)
(x+1)(x-6)=(x-9)(x+5)
x^2-5x-6=x^2-4x-45
x=39

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find the foci and asymptotes to
(y-1)^2/9-(x-3)^2/4=1
This is an equation of a hyperbola with vertical transverse axis (y-term listed first)
Its standard form of equation: , (h,k)=(x,y) coordinates of center
For given equation:
center: (3,1)
a^2=9
a=√9=3
b^2=4
b=√4=2
c^2=a^2+b^2=9+4=13
c=√13≈3.6
foci: (3,1±c)=(3,1±3.6)=(3,-2.6) and (3,4.6)
Asymptotes: Straight line equations that go thru center, y=mx+b, m=slope, b=y-intercept
slopes of asymptotes for hyperbolas with vertical transverse axis=±a/b=±3/2
equation of asymptote with negative slope:
y=-3x/2+b
solve for b using coordinates of center(3,1)
1=-3*3/2+b
b=1+9/2=11/2
equation of asymptote:
y=-3x/2+11/2
..
equation of asymptote with positive slope:
y=3x/2+b
solve for b using coordinates of center(3,1)
1=3*3/2+b
b=1-9/2=-7/2
equation of asymptote:
y=3x/2-7/2

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Can someone please help me find the vertices and foci of the ellipse
3x^2+7y^2=21
x^2/7+y^2/3=1
This is an equation of an ellipse with horizontal major axis.
Its standard form of equation: , a>b, (h,k)=(x,y) coordinates of center.
center: 0,0)
a^2=7
a=√7
vertices: (0±a,0)=(0±√7,0)=(-√7,0) and (√7,0)
b^2=3
c^2=a^2-b^2=7-3=4
c=2
foci: (0±c,0)=(0±2,0)=(-2,0) and (2,0)

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Can someone please help me find the radius and center of the circle
x^2+y^2-3x=39-6y.
complete the square
x^2-3x+y^2+6y=39
(x^2-3x+9/4)+(y^2+6y+9)=39+9/4+9
(x-3/2)^2+(y+3)^2=50.25
center: (3/2,-3)
radius=√50.25≈7.0887..
you got it right!

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Use properties of logarithms to condense the logarithmic expression. Write the expression as a single logarithm whose coefficient is 1. Where possible, evaluate logarithmic expressions.
1/2(log7 (r - 7) - log7 r)
log7[(r-7)^(1/2)/r]

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Given that Tan y = 2/5, find 4/5Cos y - 3/4Sin y
tany=2/5=opposite side/adjacent side
hypotenuse=√(2^2+5^2)=√(4+25)=√29
cosy=5/√29
siny=2/√29
4/5Cos y-3/4Sin y
=(4/5)(5/√29)-(3/4)(2/√29)
=(4√29)–(3/2√29)
=(8-3)/2√29
=5/(2√29)

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what is the reference angle of -95º?
95º

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logx + log(x+6)= log 7
Can you please solve x exactly.
log[x(x+6)]=log7
x(x+6)=7
x^2+6x-7=0
(x+7)(x-1)=0
x=-7
or
x=1

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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.)
cos(theta)=3/4
using calculator set to radians
cos^-1(3/4)=0.72+2πk,5.56+2πk, k=any integer

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ln(x)= 3- ln(x+2) solve for x
ln(x)+ln(x+2)=3
ln[x(x+2)]=3
convert to exponential form
e^3=x(x+2)
e^3=x^2+2x
x^2+2x-e^3=0
x^2+2x-e^3=0
solve by following quadratic formula:

a=1, b=2, c=-e^3
ans:
x=3.5919
or
x=-5.5919 (reject, x>0)

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Find all solutions of the equation in the interval [0,2pi)
cscθ-1=0
cscθ=1
1/sinθ=1
sinθ=1
θ=π/2

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for [0,2π)
2sin^2theta-5sintheta-3=0
2sin^2x-5sinx-3=0
(2sinx+1)(sinx-3)=0
2sinx+1=0
sinx=-1/2
x=5π/6 and 11π/6
or
sinx=3(no solution) -1 < sinx ≤ 1

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express sin2x/1+cos2x in terms of tanx
sin2x/(1+cos2x)
2sinxcosx/(1+2cos^2x-1)
2sinxcosx/(2cos^2x)
2sinxcosx/(2cos^2x)
sinx/cosx=tanx

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For the equation of the parabola find the coordinates for the VERTEX, FOCUS and the equations of the DIRECTRIX AND AXIS OF SYMMETRY
Y^2+6y+9=12-12x
(Y^2+6y+9)=12-12x-9+9
(y+3)^2=12-12x=12(1-x)=-12(x-1)
(y+3)^2=-12(x-1)
This is an equation of a parabola that opens leftwards
Its standard form: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given parabola:
vertex: (1,-3)
axis of symmetry: y=-3
4p=12
p=3
focus: (-2,-3) (p-distance to the left of the vertex on the axis of symmetry)
directrix: x=4 (p-distance to the right of the vertex on the axis of symmetry)

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How do i find the Center, Vertices, Foci and the Equation of the asymptotes of the following equation
(((x-2)^2)/25)-(((y-1)^2)/9) = 1
This is an equation of a hyperbola with horizontal transverse axis. (x-term listed first)
Its standard form of equation: , (h,k)=(x,y) coordinates of center
For given equation: (x-2)^2)/25)-(y-1)^2)/9)=1
center: (2,1)
a^2=25
a=√25=5
vertices:(2±a,1)=(2±5,1)=(-3,1) and (7,1)
b^2=9
b=√9=3
c^2=a^2+b^2=25+9=34
c=√34≈5.8
foci:(2±c,1)=(2±5.8,1)=(-3.8,1) and (7.8,1)
asymptotes: (straight line equations that go thru center, y=mx+b, m=slope, b=y-intercept)
slopes of asymptotes for hyperbolas with horizontal transverse axis=±b/a=±3/5
Equation of asymptote with negative slope:
y=-3x/5+b
solve for b using coordinates of center(2,1)
1=-3*2/5+b
b=1+6/5=11/5
equation: y=-3x/5+11/5
..
Equation of asymptote with positive slope:
y=3x/5+b
solve for b using coordinates of center(2,1)
1=3*2/5+b
b=1-6/5=-1/5
equation: y=-3x/5-1/5

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An arch is in the form of a semiellipse. It's 50 metres wide at the base and has a height of 20 metres. How wide is the arch at the height of 10 metres above the base?
**
Use standard form of equation for an ellipse wit horizontal major axis:
(x-h)^2/a^2+(y-k)^2/b^2=1, a>b, (h,k)=(x,y) coordinates of center.
For given problem:
place center at (0,0)
a=25
a^2=625
b=20
b^2=400
Equation: x^2/625+y^2/400=1
at a height of 20 metres, the coordinates are (x,20)
plug in these coordinates and solve for x
x^2/625+y^2/400=1
x^2/625+10^2/400=1
x^2/625+100/400=1
x^2/625=1-1/4=3/4=.75
x^2=.75*625=468.75
x=√468.75
x≈21.65
2x≈43.3
At the height of 10 metres above the base, the arch is 43.3 metres wide.