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What is the solution(s) to log base 4 of 4x + log base 4 of (x+3) = 2
log4[(4x)(x+3)]=2
convert to exponential form: base(4) raised to log of number(2)=number(4x)(x+3)
4^2=(4x)(x+3)=16
4x^2+12x=16
4x^2+12x-16=0
x^2+3x-4=0
(x+4)(x-1)-0
x=-4 (reject,x>0)
or x=1

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Write an equation in standard form for the circle that statisfies these conditions
center(0,7) tangent to x-axis
**
If you make a rough sketch of given points, you will see that the radius=7
Standard form of equation for a circle: (x-h)^2+(y-k)^2=r^2, (h,k)=(x,y) coordinates of center, r=radius.
Equation of given circle
x^2+(y-7)^2=7^2
x^2+(y-7)^2=49

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write an equation in vertex form then graph it. Label center,verticies,co-verticies and the foci> x^2+4y^2=16
divide by 16
x^2/16+y^2/4=1
This is an equation of an ellipse with horizontal major axis of the standard form:
(x-h)^2/a^2=(y-k)^2/b^2=1,a>b, (h,k)=(x,y) coordinates of center.
For given ellipse:
center:(0,0)
a^2=16
a=√16=4
vertices: (0±a,0)=(0±4,0)=(-4,0) and (4,0)
..
b^2=4
b=√4=2
co-vertices: (0,0±b)= (0,0±2)=(0,-2) and (0,2)
..
c^2=a^2-b^2=16-4=12
c=√12≈3.46
Foci:(0±c,0)=(0±√12,0)=(-3.46,0) and (3.46,0)
see graph below:
y=±(4-x^2/4)^.5

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In which direction does the parabola that is given by the equation below open?
x = 1.5(y - 7)2 + 5
**
Standard form of equation for parabola: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex.
For given equation:
vertex:(7,5)
A=1.5
If A>0, parabola opens upwards.
if A<0, parabola opens downwards.
So given parabola opens upwards

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The vertex of the parabola below is at the point (4, 6.5)
and the point (5, 4.5) is on the parabola.
What is the equation of the parabola?
**
Standard form of equation for a parabola: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex.
y=A(x-4)^2+6.5
Solving for A using coordinates of given point (5,4.5)
4.5=A(5-4)+6.5
4.5=A(1)+6.5
A=-2
Equation: y=-2(x-4)^2+6.5
answers are c and d which are duplicates.

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write an equation in standard form of the hyperpola with foci at (+3, 0) (-3,0) if the difference in the distances from a poin (x,y) on the hyperbola to the foci is 4.
**
Standard form of equation for a hyperbola with horizontal transverse axis:
(x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center
For given hyperbola:
center:(0,0)
Difference in distances from a point (x,y) on the hyperbola to each of the foci is 4=2a
a=2
a^2=4
c=3
c^2=9
c^2=a^2+b^2
b^2=c^2-a^2=9-4=5
Equation of given hyperbola:
x^2/4-y^2/5=1

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Solve the inequality. Write the solution set in interval notation.
2/3y-1/2>_ y+ 11/3
2y/3-1/2≥ y+ 11/3
LCD:6
4y-3≥6y+22
-2y≥25
divide both sides by -2 and reverse inequality sign
y≤-25/2
solution:
(-∞,-25/2)

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5/9(t+2)<= 1/4 (2t-6)
5(t+2)/9≤(2t-6)/4
5(t+2)/9-(2t-6)/4≤0
LCD:36
20(t+2)-9(2t-6)=0
20t+40-18t+54=0
2t+94=0
2t=-94
t=-47
number line
<.....-.....-47......+......>
solution: (-∞,-47]

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find the graph of the equation, and the period and phase shift.
y=sin(1/4x+pi/3)
**
Equation used for graphing sin functions: y=Asin(Bx-C), A=amplitude, Period=2π/B, Phase shift=C/B.
For given equation: y=sin(1/4x+pi/3):
A=1
B=1/4
Period: 2π/B=2π/(1/4)=8π
1/4 period=8/4=2π
phase shift: C/B=(π/3)/(1/4)=4π/3 (shift to the left)
..
Graph for one period:
scale of x-axis in radians
With no phase shift, (x,y) coordinates for the given sin function would be as follows:
(0,0), (2π,1), (4π,0), (6π,-1), (8π,0)
..
A phase shift of 4π/3 radians would shift x-coordinates to the left as follows:
(-4π/3,0), (2π/3,1), (8π/3,0), (14π/3,-1), (20π/3,0)
..
y-intercept
set x=0
y=sin(π/3)
y=√3/2
You now have all the points you need including the y-intercept to plot given sin function.

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If log_x^2(9)-log_x(36)=1
What does x equal?
**
log(9x^2)-log(36x)=1
log[(9x^2)/(36x)]=1
convert to exponential form
10^1=(9x^2)/(36x)=10
x/4=10
x=40

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how do you solve
log7 x+2 log7 x - log7 3=log7 72
log7(x)+log7(x^2)-log7(3)=log7(72)
log7(x)+log7(x^2)-log7(3)-log7(72)=0
place under single log
log[x(x^2)/(3*72)]=0
convert to exponential form
10^0=x^3/216=1
x^3=216
x=6

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Solve Equation
ln(3x+1)/ln(2x-1)=2
ln(3x+1)=2ln(2x-1)
ln(3x+1)=ln(2x-1)^2
3x+1=(2x-1)^2
3x+1=4x^2-4x+1
4x^2-7x=0
x(4x-7)=0
x=0 (reject,(2x-1)>0)
x=7/4

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Solve Equation(Do no use mixed numbers in your answer).
ln(5x-6)+ln(x)=ln(e)
ln[(5x-6)(x)]=1
convert to exponential form
e^1=5x^2-6x
5x^2-6x-e=0
solve by quadratic formula:
a=5, b=-6, c=e
x=[6±√(36-4*5*e)]/2*10
x=[6±√(36-20e)]/20
plug in e=2.7183 to solve for x

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log base X(4)+log base X(9)=2
logx(4)+logx(9)=2
logx(4*9)=2
logx(36)=2
base(x) raised to log of number(2)=number(36)
x^2=36
x=-6(reject, base>1)
x=6 (ans)

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Assume that log 2 = .301 and log 3 = .477, Solve
1)log 1/9
=log 1/3^2
=log 3^-2
=-2log3
=-2*.477
=-0.954
..
2)log 30
=log 10*3
=log 10+log 3
=1+.477
=1.477
..
3)log 5
=log(30/6)
=log 30-log 6
=log 10+log3-(log 3+log 2)
=1+.477-.477-.301
=1-.301
=.699
..
4)log 0.006
=log 6*10^-3
=10^-3*log6
=log(.001)+(log3+log2)
=-3+(.301+.477)
=7.78*10^-4
=-2.222

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the center of the hyperbola is at (2,5). One vertex of the hyperbola is at (0,5) where is the other vertex
**
Note that the horizontal transverse axis and one vertex of given hyperbola lie on the line y=5.
Given vertex is 2 units to the left of center at (0,5) so the other vertex must be 2 units to the right of center at (4,5).

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x^2/16+y^2/9=1
What is the foci and vertices of this ellipse equation
**
Given equation is that of an ellipse with horizontal major axis of the standard form:
(x-h)^2/a^2+(y-k)^2/b^2=1,a>b, (h,k)=(x,y) coordinates of center.
center:0,0)
a^2=16
a=√16=4
vertices:(0±a,0)=(0±4,0)=(-4,0) and ((4,0)
b^2=9
b=√9=3
c^2=a^2-b^2=16-9=7
c=√7
Foci: (0±c,0)=(0±√7,0)=(-√7,0) and (√7,0)

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What is the center of the ellipse whose equation is (x2)/25 + (y2)/4 = 1?
**
Standard form of equation for an ellipse with horizontal major axis (a^2 under x^2):
(x-h)^2/a^2+(y-k)^2/b^2=1,a>b, (h,k)=(x,y) coordinates of the center.
..
For given equation, h and k do not appear in the equation, so it is assumed they are=0, that is, the coordinates of the center are (0,0)

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FIND THE EQUATION OF THE PARABOLA WITH FOCUS (2,5) AND DIRECTRIX Y = 1
**
Standard form of equation for a parabola that opens upwards: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex.
For given parabola:
axis of symmetry: x=2
x-coordinate of vertex=2
y-coordinate of vertex=3 (halfway between focus(5) and directrix(1) on the axis of symmetry
vertex: (2,3)
P=2 (distance from focus to vertex)
4p=8
Equation of given parabola:
(x-2)^2=8(y-3)

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a one-way road passes under an overpass in the shape of half and ellipse, 15ft high at the center and 20ft wide. Assuming a truck is 16ft wide, what is the tallest truck that can pass under the overpass?
**
This problem can be represented by an ellipse with a vertical major axis of the standard form:
(x-h)^2/b^2+(y-k)^2/a^2=1, a>b, (h,k)=(x,y) coordinates of center.
..
center: (0,0) (center line of road)
a=15 ft (height of overpass at center)
a^2=225
b=10 (1/2 the width of the road)
b^2=100
equation of ellipse:
x^2/100+y^2/225=1
The point at which a truck 16 ft wide would touch the overpass would have and x-coordinates ±8 ft from center. Its y-coordinate is the answer to given problem.
Plug in x=8 ft in equation to solve for y.
8^2/100+y^2/225=1
y^2/225=1-64/100
y^2=225-225*64/100=81
y=9 ft
ans:
The tallest truck that can pass under the overpass would be 9 feet in height

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write an equation for the following conic. if it is a parabola, it has a vertex at the origin, and if it is an ellipse or a hyperbola, it is centered at the origin. focus at (4,0) and e= 1/2
**
if 0 (x-h)^2/a^2+(y-k)^2/b^2=1, a>b, (h,k)=(x,y) coordinates of center
center: (0,0) (given)
c=4
c^2=16
e=1/2=c/a
a=2c=8
a^2=64
c^2=a^2-b^2
b^2=a^2-c^2=64-16=48
Equation of ellipse:
x^2/64+y^2/48=1

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find the vertex, the focus, and directrix. y^2= 1/2x
**
This is an equation of a parabola that opens rightwards of the standard form:
(y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex.
For given parabola:
vertex:(0,0)
axis of symmetry: x-axis or y=0
4p=1/2
p=1/8
focus: (1/8,0) (p distance from right of vertex on axis of symmetry)
directrix: x=-1/8 (p distance from left of vertex on axis of symmetry)

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If cos x = -(2/3) and pi/2 < x < pi, what is the exact value of sin2x, cos2x, and tan2x
**
If cos x = -(2/3) (adjacent/hypotenuse in quadrant II where cos<0)
sinx=√5/3 (opposite/hypotenuse in quadrant II where sin>0)
tanx=-√5/2(opposite/adjacent in quadrant II where tan<0)
..
sin2x=2sinxcosx=2*√5/3*-2/3=-4√5/9
cos 2x=cos^2x-sin^2x=(-2/3)^2-(√5/3)^2=4/9-5/9=-1/9
tan2x=2tanx/(1-tan^2x)=2*-√5/2/(1-5/4)=-√5/-1/4=4√5

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wrirte and equation for the following conic. If it is a parabola, it has a vertex at the orgin, and if it is an ellipse or a hyperbola, it is centered at the orgin. Focus at (0,5) and e=1
**
In accordance with the focus-directrix property of conics,
if e=1, the conic is a parabola:
Standard form of equation for parabola opening upwards: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex.
vertex:(0,0) (given)
axis of symmetry: y-axis or x=0
p=5 (distance from focus to vertex on the axis of symmetry)
4p=20
Equation of parabola:
x^2=20y

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If (-3z+6)/3 + z/5 = 12 then 4z^2+2z+50 is what?
**
(-3z+6)/3 + z/5 = 12
-z+2 + z/5 = 12
-5z+10 + z = 60
-4z = 50
z=-50/4=-25/2
..
4z^2+2z+50
=4(-25/2)^2+2(-25/2)+50
=4(625/4)-25+50
=625-25+50=650

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Solve the following exponential equation:
2^(2x) + 2^(x+2) - 12 = 0
2^(2x) + 2^x*2^2 - 12 = 0
2^(2x) + 2^x*4 - 12 = 0
2^2x + 4*2^x - 12 = 0
.
let u=2^x
u^2=2^2x
..
u^2+4u-12=0
(u+6)(u-2)=0
u=-6=2^x
no real solution, (2^x>0)
or
u=2=2^x
x=1

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find exact value of [tan(5pi/4)-tan(pi/12)]/1+tan(5pi/4)tan(pi/12)
**
use tan addition identity: tan (s-t)=(tan s-tan t)/(1+tan s tan t)
For given problem:
[tan(5pi/4)-tan(pi/12)]/1+tan(5pi/4)tan(pi/12)=tan(5π/4-π/12)
=tan(15π/12-π/12)=tan(14π/12)
=tan(7π/6) (ans)

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Ok. So i know the answer is going to be an imaginary number. The problem is this: 6 +or- the square root of (-6)^2-4(2)(9) all divided by 4. PLease help. I can not understand how to break down the problem into imaginary numbers.
**
{6±√[(-6)^2-4*2*9]}/4
{6±√[36-72]}/4
(6±√-36)/4
(6±6i)/4
6(1±i)/4
3(1±i)/2=3(1+i)/2 and 3(1-i)/2
..
note: √-36=√36*√-1=6i

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A tank has supply pipe A and an exhaust pipe B. The pipe A takes 4 minutes less to fill a tank than the pipe B takes to empty it. When both are open, the tank is filled in 24 minutes. Find the time needed for pipe A to fill the tank when pipe B is closed.
**
let x=minutes pipe B takes to empty tank
1/x=pipe B work rate
x-4=minutes pipe A takes to fill tank
1/(x-4)=Pipe A work rate
..
In 24 minutes:
24/(x-4)-24/x=100%
24x-24x+96=x(x-4)=x^2-4x
x^2-4x-96=0
(x-12)(x+8)=0
x=-8 (reject, x>0)
x=12
x-4=8
ans:
Pipe A alone fills tank in 8 minutes (When Pipe B is closed)

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1. Show that log8x=1/3 log2x hence solve the equation for x>0
Log2(3x+1) +log8(x-1)3=0
**
log8x=1/3 log2x
change to base2
log8x=log2x/log2(8)=log2x/3
..
log2(3x+1) +log8(x-1)*3=0
change to base 2
log2(3x+1) +log2(x-1)*3/log2(8)=0
log2(3x+1) +log2(x-1)*3/3=0
log2(3x+1) +log2(x-1)=0
place under single log
log2[(3x+1)(x-1)]=0
convert to exponential form:
2^0=(3x+1)(x-1)=1
3x^2-2x-2=0
solve by quadratic formula:
x≈-.549 (reject,x>0)
or
x=1.215

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In 6 years, Sylvia will be 6/5 Hue's age. Four years ago, Sylvia was 10 years less than twice Hue's age. How old are Sylvia and Hue now?
**
let x=Hue's present age
let y=Sylvia's present age
..
In 6 years:
Hue's age=(x+6)
Sylvia's age=(y+6)
(y+6)=(6/5)(x+6)
y=(6/5)(x+6)-6
y=(6x+36)/5-6
..
4 years ago:
Hue's age=(x-4)
sylvia's age=(y-4)
y-4=2(x-4)-10
y=2x-8-10+4
y=2x-14
..
(6x+36)/5-6=2x-14
LCD+5
6x+36-30=10x-70
4x=76
x=19
y=2x-14=38-14=24
..
ans:
Hue's present age=19
Sylvia's present age=24