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A picture has a square frame that is 2 inches wide. The area of the picture is one-third of the total area of the picture and frame. What are the dimensions of the picture to the nearest quarter of an inch?"
**
let x=length of side of square picture
x^2=area of picture
(x+4)=length of side of square frame
(x+4)^2=total area of picture and frame
..
x^2=1/3(x+4)^2=(x^2+8x+16)/3
3x^2=x^2+8x+16
2x^2-8x-16=0
x^2-4x-8=0
solve for x by completing the square:
(x^2-4x+4)-4-8=0
(x-2)^2=12
take sqrt of both sides
x-2=±√12
x=2±√12
x=2±3.46
x≈-1.46 (reject, x>0)
or
x≈5.5
dimensions of the picture: 5.5 by 5.5 inches

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Two cars (a race car and a normal car) each travel 300 miles per day. The race car travels 40 mph faster and takes 2 hours and 40 minutes less time than the normal car. Find the speed of the race car.
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let x=speed of the race car
x-40=speed of the normal car
2 hours and 40 minutes=2-2/3 hours=8/3 hrs
..
travel time=distance/speed
travel time of normal car-travel time of race car=2-2/3 hrs
300/(x-40)-300/x=8/3
900/(x-40)-900/x=8
900x-900(x-40)=8x(x-40)
900x-900x+36000=8x^2-320x
8x^2-320x-36000=0
x^2-40x-4500=0
(x+50)(x-90)=0
x=-50 (reject)
or
x=90
x-40=50
speed of the race car=90 mph
speed of the normal car=50 mph

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Seven times one number decreased by twice another equals 27, and 3 times the first added to 5 times the second equals 35. Find the two numbers.
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let x=first number
let y=2nd number
..
7x-2y=27
3x+5y=35
..
35x-10y=135
6x+10y=70
add
41x=205
x=5
2y=7x-27
2y=35-27
2y=8
y=4
..
1st number=5
2nd number=4

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Find the ages of A and B from this information: 6 times A's age is 100 more than 5 times B's age, and 9 times A's age is 170 more than 7 time's B's age.
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let A=A's age
let B=B's age
..
6A=100+5B
9A=170+7B
..
42A=700+35B
45A=850+35B
subtract:
3A=150
A=50
..
5B=6A-100
5B=300-100
5B=200
B=40
..
A's age=50
B's age=40

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1/2(2y+3)=-2/3(y-4)
(2y+3)2=-2(y-4)/3
6y+9=-4y+16
10y=7
y=7/10

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find the roots of the quadratic equation
x^2+6x-41=0
complete the square:
(x^2+6x+9)-9-41=0
(x+3)^2=50
x+3=±√50
x=-3±√50
x=-3-√50
or
x=-3+√50

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Form a quadratic equation and solve:
2/a^2+3/a=-1
LCD:a^2
2+3a=-a^2
a^2+3a+2=0
(a+2)(a+1)=0
a=-2
or
a=-1

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f(x)=1+log5(x+3)
a)the domain of f: (-3,∞)
..
b)graph f
I don't have the means to graph it, but I can try to show you how to do it.
Start with the basic log5(x) basic curve which has an asymptote at x=0 or the y-axis and an
x-intercept at(1,0). Then move this curve 3 units left, placing the asymptote and x-intercept at
x=-3 and (-2,0), respectively. Lastly, bump the curve up one unit.
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c) from the graph, determine the range and any asymptotes of f. Range: (-∞,∞); asymptote:x=-3
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d)find f^-1, the inverse of f
y=1+log5(x+3)
interchange x and y, then solve for y
x=1+log5(y+3)
x-1=log5(y+3)
5^(x+1))=y+3
f^-1=5^(x+1)-3
..
e)find the domain and range of f^-1: Range:(-3,∞); Domain:(-∞,∞)
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f)graph f^-1
curve has a horizontal asymptote at y=-3 and a y-intercept at (0,2)

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Solve the equation:
log4 (30)-log4(x-1)-log4(x+2)=log4 (3)
log4 (30)-(log4(x-1)+log4(x+2))=log4 (3)
log4 [(30/(x-1)(x-2)]=log(3)
(30/(x-1)(x-2)=3
30/3=(x-1)(x+2)
10=x^2+x-2
x^2+x-12=0
(x+4)(x-3)=0
x=-4 (reject, x≥0)
x=3

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What is the exponential form of log[7] 90 = y
Exponential Form: Base(7) raised to log of number(y)=number(90)
7^y=90

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what is the smallest positive number for t?
sin(t)+cos(2t)=0
sint+cos^2t-sin^2t=0
sint+1-sin^2t-sin^2t=0
2sin^2t-sint-1=0
(2sint+1)(sint-1)=0
sint-1=0
sint=1
t=π/2
or
2sint+1=0
sint=-1/2
t=7π/4
..
smallest positive number for t=π/2

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Find the exact trigonometric ratios for the angle x whose radian measure is given. (If an answer is undefined, enter UNDEFINED.)
−9π
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angle -9π is like rotating clockwise 4-1/2 times around a unit circle terminating at a reference angle of π:
trigonometric ratios at this reference angle:
sinx=0
cosx=-1
tanx=0
cscx=undefined
secx=-1
cotx=undefined

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If sinx = -3/5 and x is in quadrant III, then what is sin2x, cos2x, and tan2x?
let O=opposite side
let A=adjacent side
H=hypotenuse
..
sin x=-3/5=O/H
O=-3, H=5
A=√(H^2-O^2)
=√(5^2-3^2)
=√(25-9)
=√16
A=-4 (In quadrant III where cos<0)
cos x=A/H=-4/5
tanx=O/A=-3/-4=3/4
..
Identity:sin2x=2sinxcosx
=2*-3/5*-4/5
sin2x=24/25
..
Identity:cos2x=cos2^x-sin^2x
=(-4/5)^2-(-3/5)^2
=16/25-9/25
cos 2x=7/25
..
Identity: tan2x=(2tanx)/(1-tan^2x)
=(2*3/4)/(1-(3/4)^2)
=(6/4)/(1-9/16)
=(3/2)/(7/16)
tan2x=24/7

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find the vertices, center and the foci of y^2 over 5, minus x^2 over 11=1.
y^2/5-x^2/11=1
This is an equation of a hyperbola with vertical transverse axis:
Its standard form: , (h,k)=)x,y) coordinates of center.
..
For given equation:
center:(0,0)
..
a^2=5
a=√5
vertices: (0,0±a)=(0,0±√5)=(0,-√5), (0,+√5)
..
b^2=11
..
c^2=a^2+b^2=5+11=16
c=√16=4
Foci: (0,0±c)=(0,0±4)=(0,-4), (0,+4)

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use the eccentricity of each hyperbola to find its equation in standard form eccentricity 4 , vertices (-1,3) and (-1,7)
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Standard form of equation for a hyperbola with vertical transverse axis:
, (h,k)=(x,y) coordinates of center
..
x-coordinate of center=-1
y-coordinate of center= (3+7)/2=5 (midpoint formula)
center:(-1,5)
..
length of vertical transverse axis=4 (3 to 7)=2a
a=2
a^2=4
..
c=distance from center to foci
Eccentricity=4=c/a
c=4a
..
c^2=a^2+b^2
16a^2=a^2+b^2
b^2=16a^2-a^2=15a^2
..
equation of given hyperbola:

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What is the standard formula for the quadratic function that has a vertex of (0.5,4) and passes through points (-1.5,0) and (2.5,0)?
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Standard form of equation for a parabola: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of vertex.
For given function:
y=A(x-0.5)^2+4
solve for A using one of given points (-1.5,0)
0=A(-1.5-0.5)^2+4
0=A(-2)^2+4
4A=-4
A=-1
Equation: y=-(x-0.5)^2+4

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3 tothe power x equals 80, whats the value of x
3^x=80
xlog3=log80
x=log80/log3
x≈3.9887..

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Which of the following has the same slope?
29 degrees and 1/2
2.2 and 65 degrees
7/11 and 65%
9.8 and 112.5%
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using a calculator:
29 degrees≠1/2
2.2≠65 degrees
7/11≈0.636≠65%
9.8≠112.5%
none of the above are equal to each other.
unless you wrote 9.8 when you meant 9/8
9/8=1.125=112.5%

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Prove the identity:
(1/1-cosx)-(cosx/1+cosx)=2csc^2(x)-1
start with left side:
(1/1-cosx)-(cosx/1+cosx)
=1+cosx-cosx(1-cosx)/(1-cosx)(1-cosx)
=1+cosx-cosx+cos^2x/(1-cos^2x)
=1+cos^2x/(1-cos^2x)
=(1+1-sin^2x)/1-(1-sin^2x)
=(2-sin^2x)/sin^2x
=(2/sin^2x)/(-(sin^2x)/(sin^2x))
=2csc^2x-1
verified: left side=right side

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I need to prove this identity
tan^2x-sin^2x = tan^2xsin^2x
start with left side:
tan^2x-sin^2x
=(sin^2x/cos^2x)-sin^2x
=(sin^2x-sin^2xcos^2x)/cos^2x
=sin^2x(1-cos^2x)/cos^2x
=sin^2x*sin^2x/cos^2x
=tan^2xsin^2x
verified:
left side=right side

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Answer the following for the given quadric function. f(x)=x^2-4x+4
a.) what is the vertex (h,k) of f?
b.) what is the axis of symmetry?
c.) what are the intercepts?
d.) graph it.
e.) on which interval is f increasing?
f.) on what interval is f decreasing?
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f(x)=x^2-4x+4
complete the square:
y=(x^2-4x+4)-4+4
y=(x-2)^2
..
vertex: (2,0)
axis of symmetry: x=2
..
y-intercept
set x=0, solve for y
y-intercept=4
..
x-intercept
set y=0, solve for x
x-intercept=2
..
see graph below:
..
f increasing: (2,∞)
..
f decreasing:(-∞,2)

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find the slant asymptote of the graph of the rational function and Use the slant asymptote to graph the rational function.
f(x)= (x^2-x-20)/(x-6)
f(x)=(x-5)(x+4)/(x-6)
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b.) first determine the symmetry of the graph of f.
Graph has no symmetry as can be seen by the graph below.
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c.) find the x and y-intercepts
y-intercept:
set x=0
y-intercept=-20/-6=10/3
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x-intercept:
set y=0
x^2-x-20
(x-5)(x+4)=0
x-intercepts at 5 and-4
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d.) find the asymptotes.
vertical asymptote:
set denominator=0, then solve for x
x-6=0
vertical asymptote: x=6
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slant asymptote:
divide numerator by denominator:
(x^2-x-20)/(x-6)=(x+5)+remainder
slant asymptote: y=x+5
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number line
<...-....-4....+.....5....-.....6....+.....>

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Simplify without a calculator. Show all steps
4 log base 3 of 1/3 + 2log base 27 of 9
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4 log base 3 of 1/3
=log3(1/3)^4
=log3(1/81)
=log3(1/3)^4)
=log3(1/3)^3)*(1/3)
=log3[(3^-3)*(3^-1)]
=log3(3^-3)+log3(3^-1)
log of base raised to a power=power
=-3-1=-4
..
2log base 27 of 9
=log27(9^2)
=log27(81)
=log27(3^3)+log27(3^1)
=log27(27)+log27(3)
=1+1/3
=4/3
..
4 log base 3 of 1/3 + 2log base 27 of 9=-4+4/3=-12/3+4/3=-8/3

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What is y=x^(2)+8x-1 in vertex form?
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y=x^(2)+8x-1
complete the square
y=(x^2+8x+16)-1-16
y=(x+4)^2-17
This is an equation of a parabola that opens upwards with vertex at (-4,-17)
Its standard form of equation:y= (x+h)^2+k, (h,k) =(x,y) coordinates of the vertex.

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Write the standard form equation of an ellipse:
Vertices: (5,-3) (-3,-3)
Co-vertices: (1,0), (1, -6)
I know how to find the standard form with only two endpoints rather than four.
Thanks for your help!
**
Given data shows ellipse has a horizontal major axis.
Its standard form of equation: , a>b, (h,k)=(x,y) coordinates of center.
x-coordinate of center= (5+(-3))/2=2/2=1 (midpoint formula)
y-coordinate of center=-3
center:(1,-3)
length of horizontal major axis=8 (-3 to 5)=2a
a=4
a^2=16
length of minor axis (co-vertices)=6 (0 to 6)=2b
b=3
b^2=9
..
Equation of given ellipse:

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prove that (sinx-1)(tanx+secx)= -cosx
start with left side:
(sinx-1)(tanx+secx)=(sinx-1)[(sinx/cosx)+(1/cosx)]
=(sinx-1)(sinx+1)/cosx)
=(sin^2x-1)/cosx
=-(1-sin^2x)/cosx
=-cos^2x/cosx
=-cosx
verified: left side=right side

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Use a table of trigonometric values to find the angle θ in the right triangle in the following problem. Round to the nearest degree, if necessary.
sin θ = ? O = 8 H = 16
O=opposite side
H=hypotenuse
sin θ=O/H=8/16=1/2
From table of trig values:
θ=30º

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Evaluate the cot[arccos(5-x)]
let A=adjacent side
let O=opposite side
H=hypotenuse
..
cot[arccos(5-x)]
This reads: cot of an angle whose cos=(5-x)
cos(5-x)=A/H=(5-x)/1
A=(5-x)
H=1
O=√(H^2-A^2) (Pythagorean Theorem)
=√[1^2-(5-x)^2]
=√[1-(25-10x+x^2)]
O=√(-24+10x-x^2)
cot[arccos(5-x)]=A/O=(5-x)/√(-24+10x-x^2)

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Use the definition of logarithm to simplify each expression.
(a) log10^10=1
(b)log10 1,000=3
(c) log 10 ^10 -2=-2
..
Definition: The base raised to the log of the number=to the number

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What are the foci of the ellipse given by the equation
100x^2+64y^2=64,000
x^2/640+y^2/1000=1
This is an equation of an ellipse with vertical major axis.
Its standard form: , a>b, (h,k)=(x,y) coordinates of center
center:(0,0)
a^2=1000
b^2=640
c^2=a^2-b^2=1000-640=360
c=√360≈18.97
Foci: (0, 0±c)=(0,0±18.97)=(0,-18.97) and (0,18.97)

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A hyperbolic mirror can be used to take panoramic photos, if the camera is pointed toward the mirror with the lens at one focus of the hyperbola. write the equation of the hyperbola that can be used to model a mirror that has a vertex 4 inches from the center of the hyperbola and a focus 1 inch in front of the surface of the mirror. assume the mirror has a horizontal trasverse axis and the hyperbola is centered at (0,0).
**
Standard form of equation for a hyperbola with horizontal transverse axis:
, (h,k)=(x,y) coordinates of center
For given problem:
center: (0,0)
vertices
a=4
a^2=16
..
Foci
c=5
c^2=25
..
c^2=a^2+b^2
b^2=c^2-a^2=25/16=9
..
Equation of hyperbola: