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(1/3)^(5x+2)=(1/9)^3
solving for x
3^-(5x+2)=1/9^3
3^-(5x+2=1/(3^2)^3=1/3^6=3^-6
-(5x+2=-6
5x+2=6
5x=4
x=4/5

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A rectangle is 5 cm longer than it is wide. If the length and breadth are increased by 2 cm each, the area increases by 50 cm squared. Find the dimensions of the original rectangle.
**
let x=width of original rectangle
x+5=length of original rectangle
area of original rectangle=x(x+5)=x^2+5x
..
x+2=width of larger rectangle
x+5+2=length of larger rectangle
area of larger rectangle=(x+2)(x+7)=x^2+9x+14
..
Area of larger rectangle-area of original rectangle=50
x^2+9x+14-x^2-5x=50
4x=36
x=9
x+5=14
..
width of original rectangle=9 cm
length of original rectangle=14 cm

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a farmer decides to enclose a rectangular garden using the side of a barn as one side of the rectangle. What is the maximum area that the farmer can enclose with 60 foot of fence? What should the dimensions of the garden be to get this area?
**
let x=length of each of the two sides perpendicular to the side of the barn
60-2x=length of third side parallel to side of the barn
Area=x(60-2x)
=60x-2x^2
=-2x^2+60x
complete the square
=-2(x^2-30x+225)+450
=-2(x-15)^2+450
maximum area=450 sqft
dimensions of garden: 15 ft by 30 ft

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(x^2-3)^2-43(x^2-3)+42=0
let u=(x^2-3)
u^2-43u+42=0
(u-42)(u-1)=0
u=1=x^2-3
x^2=4
x=±√4=±2
or
x=42=x^2-3
x^2=45
x=±√45
solution: x=-√45, -2, 2, √45

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solve and simplify.
3log of 27(x-1)=2
log27(x-1)=2/3
convert to exponential form
10^(2/3)=27(x-1)=27x-27
27x=10^(2/3)+27≈31.64
x≈31.64/27≈1.17

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Solve for x:
log (x - 6) base of 2 = 3 - log 4 base of 2
log2(x-6)=3-log2(4)
log2(x-6)+log2(4)=3
log2[(x-6)*4]=3
convert to exponential form
2^3=4(x-6)=8
4x-24=8
4x=32
x=8

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Write the equation of a hyperbola with vertices at (-3, 0) and (3, 0) and co-vertices (0, 5) and (0, -5).
Given hyperbola has a horizontal transverse axis.
Its standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center
For given hyperbola:
center:(0,0)
length of horizontal transverse axis=6=2a
a=3
a^2=9
..
length of co-vertices or conjugate axis=10=2b
b=5
b^2=25
..
Equation:
x^2/9-y^2/25=1

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What is the radius of the circle given by the equation below?
Standard form of equation for a circle: (x-h)^2+(y-k)^2=r^2, (h,k)=(x,y) coordinates of center, r=radius
x2 - 14x + y2 + 12y = -69
complete the square.
(x^2-14x+49)+(y^2+12y+36) = -69+49+36
(x-7)^2+(y+6)^2=16
r^2=16
radius=√16=4

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The sum of two numbers is 45. The larger number, y, is 6 less than twice the smaller number, x. What is the smaller number x? Also what is the equation?
**
let x=smaller number
y=2x-6=larger number
x+y=45
x+2x-6=45 (equation)
3x=39
x=13=smaller number

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What is the center and the radius of a circle whose equation is
x^2+y^2 -4x +14y +28=0 ?
complete the square
x^2 -4x+y^2+14y +28=0
(x^2-4x+4)+(y^2+14y+49)=-28+4+49
(x-2)^2+(y+7)^2=25
Standard form of equation for a circle:
(x-h)^2+(y-k)^2=r^2, (h,k)=(x,y) coordinates of center, r=radius
For given equation:
center:(2,-7)
r^2=251
r=√25=5

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Find all rational zeros:
y=2x^3+2x^2-50x-50
y=2x^2(x+1)-50(x+1)
(x+1)(2x^2-50)=0
x+1=0
x=-1
or
2x^2-50=0
x^2-25=0
x^2=25
x=±√25=±5
zeros: -5, -1, and 5

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Solve 2tan theta=2cot theta for all values of theta if theta is measured in degrees
2tanx=2cotx
tanx=cotx
tan45=cot45=1
x=45º+360º*n and 225º+360º*n, n=integer

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Simplify 7sin theta sec theta
=7sinxsecx
=7sinx*1/cosx
=7sinx/cosx
=7tanx

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points (1,4) and (4,2) lie on line k. what is the slope of the line that is perpendicular to k?
slope of straight line=∆y/∆x
slope of line k=(2-4)/(4-1)=-2/3
slope of line perpendicular to line k
=negative reciprocal of slope of line k
=3/2

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What is the slope of the line through the points (2,-6)(9,13)
slope of a straight line=∆y/∆x
=change in y/change in x
=(13-(-6))/(9-2)
=19/7

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the ellipse is center at the origin. the legnth of its horizontal axis is 8,and the length of its vertical axis is 6.what is the equation for the ellipse
**
standard form of equation for an ellipse with 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:
center: (0,0)
length of horizontal major axis=8=2a
a=4
a^2=16
..
length of vertical or minor axis=6=2b
b=3
b^2=9
..
equation:
x^2/16+y^2/9=1

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Solve the following exponential equation. Exact answers only.
π^(1-8x)=e^(3x)
take log of both sides
(1-8x)ln(π)=3xlne
lne=1(log of base=1)
ln(π)-8ln(π)x=3x
3x+8ln(π)x=ln(π)
x(3+8ln(π))=ln(π)
x=ln(π)/(3+8ln(π))

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How do I condense the following expression:
4(log12-log3)-5(log16-log4)
=4(log12/3)-5(log16/4)
=4(log4)-5(log4)
=-log4

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Simplify the expression below and write it as a single logarithim
3log(x+4)-2log(x-7)+5log(x-2)-log(x^2)
3log(x+4)+5log(x-2)-[2log(x-7)+log(x^2)]
log[((x+4)^3*(x-2)^5)/((x-7)^2*x^2)]

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For the conic section described by the equation 9x^2+4y^2-18x+16y-11=0
What is the center?
What is the length of the semimajor axes?
What is the focal length?
**
9x^2+4y^2-18x+16y-11=0
complete the square
9x^2-18x+4y^2+16y-11=0
9(x^2-2x+1)+4(y^2+4y+4)=11+9+16
9(x-1)^2+4(y+2)^2=36
(x-1)^2/4+(y+2)^2/9=1
This is an equation for an ellipse with vertical major axis.
Its standard form: ((x-h)^2/b^2+(y-k)^2/a^2=1, (a>b), (h,k)=(x,y) coordinates of the center
For given equation:
center: (1,-2)
b^2=4
b=2
length of semi major or minor axis=2b=4
..
a^2=9
c^2=a^2-b^2=9-4=5
c=√5
focal length=22c=2√5

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Complete the square, then graph and identify the vertex, focus, directrix, and endpoints of the latus rectum for the equation -14x+2y^2-8y=20
-14x+2y^2-8y=20
complete the square
2(y^2-4y+4)=20+14x+8
2(y-2)^2=14x+28
divide by 2
(y-2)^2=7x+14
(y-2)^2=7(x+2)
This is an equation of a parabola that opens rightwards.
Its standard form: (y-k)^2=4px, (h,k)=(x,y) coordinates of the vertex
For given equation:(y-2)^2=7(x+2)
vertex:(-2,2)
axis of symmetry: y=2
4p=7
p=7/4
Focus: (-2+p,2)=(-2+7/4,2)=(-1/4,2) (p distance to the right of the vertex on the axis of symmetry)
Directrix: x=(-2-p)=(-2-7/4)=-15/4 (p distance to the left of the vertex on the axis of symmetry)
latus rectum:
length of latus rectum=4p=7
2p=7/2
end points: (-1/4,2±2p)
=(-1/4,2±7/2)
=(-1/4,-1.5) and (-1/4,5.5)
see graph below:
y=(7(x+2))^.5+2

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What is the standard form of the ellipse with the equation
16x^2+25y^2=150y+175?
complete the square:
16x^2+25(y^2-6y+9)=175+225
16x^2+25(y-3)^2=400
x^2/25+(y-3)^2/16=1
This is an equation of an ellipse with horizontal major axis.
Its standard form: (x-h)^2/a^2+(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center

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logx^e^5 = 5
e^5logx=5
logx=5/e^5
x=10^(5/e^5)
x≈1.08

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Solve the equation:
ln x + ln 2x=2.4
ln(x*2x)=2.4
ln(2x^2)=2.4
e^2.4=2x^2
x^2=e^(2.4)/2
x=√(e^(2.4)/2)
x≈2.348

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5^4x+3=29
5^4x=29-3=26
4xlog5=log26
x=log26/4log5
x≈.506

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Log base 3(x-4)=2
convert to exponential form
3^2=x-4=9
x=13

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lne=1
1.) e^x=148.413
xlne=ln(148.413
x=ln(148.413)≈4.999...≈5
..
2.) 25=e^3x
3xlne=ln25
3x=ln25
x=ln25/3≈1.073
3.) 9e^5x=5
e^5x=5/9
5xlne=ln(5/9)
x=ln(5/9)/5≈-0.117
..
4.) 1/4e^x=30
e^x=120
xlne=ln120
x=ln120≈4.787

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Can you please help me solve and simplify?
3log of 27(x-1)=2
log of 27(x-1)=2/3
convert to exponential form
10^2/3=27(x-1)=27x-27
27x=10^2/3+27
x=(10^2/3+27)/27
using calculator
x≈1.172

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x^2/36-y^2/16=1
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of the center
For given equation:x^2/36-y^2/16=1
center: (0,0)
..
a^2=36
a=√36=6
vertices: (0±a,0)=(0±6,0)=(-6,0) and (6,0)
..
b^2=16
b=√16=4
conjugate points: (0,0±b)=(0,0±4)=(0-4) and (0,4)
..
c^2=a^2+b^2=36+16=52
c=√52≈7.2
foci: (0±c,0)=(0±7.2,0)=(-7.2,0) and (7.2,0)
..
slope of asymptotes for hyperbolas with horizontal transverse axis=±b/a=±4/6=±2/3
asymptotes are straight lines that intersect at the center.
Equation for asymptote with negative slope:
y=-2x/3+b
since y-intercept is at 0, b=0
equation of asymptote: y=-2x/3
..
Equation for asymptote with positive slope:
y=2x/3+b
since y-intercept is at 0, b=0
equation of asymptote: y=2x/3

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x^2/25+y^2/16=1
What is the foci?
This is an equation of an ellipse with horizontal major axis
Its standard form: (x-h)^2/a^2+(y-k)^2/b^2=1, (a>b), (h,k)=(x,y) coordinates of the center
For given equation:
center: (0,0)
a^2=25
b^2=16
c^2=a^2-b^2=25-16=9
c=√9=3
foci: (0±c,0)=(0±3,0)=(-3,0) and (3,0)
- -
Given 36x^2+9y^2-324=0
36x^2+9y^2=324
divide by 324
x^2/9+y^2/36=1
This is an ellipse with vertical major axis
center: (0,0)
..
a^2=36
a=√36=6
vertices:(0,0±a)=(0,0±6)=(0,-6) and (0,6)
..
b^2=9
b=3
co-vertices:(0±b,0)=(0±3,0)=(-3,0) and (3,0)
..
c^2=a^2-b^2=36-9=25
c=√25=5
foci:(0,0±c)=(0,0±5)=(0,-5) and (0,5)

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3¼/2 = B/two-thirds
3 and 1/4=13/4
rewrite equation:
(13/4)/2=B/(2/3)
cross-multiply
2B=(2/3)(13/4=26/12
divide by 2
B=26/24=13/12
If an exact answer is wanted, you must leave it in this fraction form.
If a decimal answer is wanted, you will get 13/12=1.08333...
so, you would probably round it off depending on how many decimal places are wanted.
one decimal place=1.1
two decimal places=1.08
three decimal places=1.083
etc.
hope this helps