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Solve log(base6) (x^2+7)=log(base6)(1-5x)
x^2+7=1-5x
x^2+5x+6=0
(x+3)(x+2)=0
x+3=0
x=-3
or
x+2=0
x=-2

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Standard form of equation for a straight line: y=mx+b, m=slope, b=y-intercept
m=3/2 thru (2,6)
equation: y=3x/2+b
solve for b using coordinates of given point
6=3*2/2+b
b=6-3=3
equation: y=3x/2+3
use this same method to solve the following 3 problems:
..
m = -4 thru (-2,3)
m=1/3 thru (-1,3)
m=1/4 thru (2,-4)

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The speed of the boat in still water is 15km/hr.It can go 30 km upstream and return to down stream in 41/2hrs.find the speed of the stream.
**
let x=speed of stream
(15+x)=speed of boat downstream
(15-x)=speed of boat upstream
travel time=distance/speed
30/(15+x)+30/(15-x)=9/2
60/(15+x)+60/(15-x)=9
LCD: (15+x)(15-x)
60(15-x)+60(15+x)=9(15+x)(15-x)
60(15-x+15+x)=9(225-x^2)
60(30)=2025-9x2
9x^2=2025-1800
9x2=225
take sqrt of both sides
3x=15
x=5
speed of stream=5 km/hr

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A fisherman can row a boat 8km downstream and return in 100min. If the speed of the stream is 2km per hour, find the speed of the boat in still water?
**
let x=speed of boat in still water
x+2=downstream speed
x-2=upstream speed
speed=distance/travel time
8/(x+2)+8/(x-2)=100/60
LCD:(x+2)(x-2)
8(x-2)+8(x+2)=(100/60)*(x+2)(x-2)
8(x-2+x+2)=100(x+2)(x-2)/60
240(2x)=100(x^2-4)
480x=100x^2-400
100x^2-480x-400=0
10x^2-48x-40=0
solve by following quadratic formula:

a=10, b=-48,c=-40
ans: x≈5.52
speed of boat in still water≈5.52 km/hr

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THE SUM OF THE AGES OF SON AND FATHER IS 56 YERAS.AFTER 4 YEARS THE AGE OF THE FATHER WILL BE THREE TIMES THAT OF THE SON.FIND THE AGE OF THE FATHER?
**
let x= age of father
56-x=age of son
after 4 yrs
x+4=age of father
(56-x)+4=(60-x) age of son
x+4=3(60-x)
x+4=180-3x
4x=176
x=44
age of father: 44

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Five years ago I was thrice as old as my son and ten years later I shall be twice as old as my son. How old are we now?
let x=son's age now
5 yrs ago:
x-5=son's age
3(x-5)=parent's age
10 yrs later:
x+10=son's age
2(x+10)=parent's age
parent's age 10 yrs later=parent's age 5 yrs ago plus 15
2(x+10)=15+3(x-5)
2x+20=15+3x-15
x=20
3(x-5)+5=50
son's age now: 20
parent's age now: 50

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1.) ((b+4)/(2))^2 . ((3)/(b+4))^3
(b+4)^2/4*27/(b+4)^3
(1/4)*27/b+4
27/4(b+4)
2.) ((b+4)/(b-2)) . ((b^2-4)/(b^2-16))
((b+4)/(b-2)) . (b+2)(b-2))/(b+4)(b-4)
(b+2)/(b-4)

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Find values of a and k for which each parabola contains the given points
y-k = a(x+3)^2 ; (-5,1) (1,7)
**
y-k = a(x+3)^2
y=a(x+3)^2+k
set up two equations with coordinates of given points:
1=a(-5+3)^2+k
7=a(1+3)^2+k
..
1=a(-2)^2+k
7=a(4)^2+k
..
1=4a+k
7=16a+k
..
4=16a+4k
7=16a+k
..
-3=3k
k=-1
..
1=4a-1
4a=2
a=1/2
check:(using point (1,7)
7-(-1)=1/2(1+3)^2
8=8

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What is the radius of the circle given by the equation below?
x^2 + 8x = -y^2 - 16y - 16
**
Standard form of a circle: , (h,k)=(x,y) coordinates of center, r=radius.
To find the radius of given circle, convert given equation to standard form.
x^2 + 8x = -y^2 - 16y - 16
complete the square:
x^2+8x=-y^2-16y-16
x^2+8x+y^2+16y=-16
(x^2+8x+16)+(y^2+16y+64)=-16+16+64

r^2=64
r=8

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Hyperbola:
9x^2-16y^2-18x-32y-151=0
9x^2-18x-16y^2-32y-151=0
complete the square: (you already did with a small error which I have corrected)
This is a hyperbola with horizontal transverse axis: (x-term listed first)
Its standard form of equation: , (h,k)=(x,y) coordinates of center.
Center:(1,-1)
a^2=16
a=√16=4
Vertices: (1±a,-1)=(1±4,-1)=(-3,-1) and (5-1)
..
b^2=9
b=√9=3
..
c^2=a^2+b^2=16+9=25
c=√25=5
Foci: (1±c,-1)=(1±5,-1)=(-4,-1) and (6-1)
..
Asymptotes:(straight line equations that go thru center, of the form y=mx+b, m=slope, b=y-intercept)
For hyperbolas with horizontal transverse axis: slope=±b/a=±3/4
Equation for asymptote with negative slope:
y=-3x/4+b
solve for b using coordinates of center.
-1=-3*1/4+b
b=-1/4
equation:y=-3x/4-1/4
..
Equation for asymptote with positive slope:
y=3x/4+b
solve for b using coordinates of center.
-1=3*1/4+b
b=-7/4
equation:y=3x/4-5/4
..
Graph:
9(x^2-2x+1)-16(y^2+2y+1)=151-16+9
9(x-1)^2-16(y+1)^2=144
(x-1)^2/16-(y+1)^2/9=1
..
See graph below:
y=±((9(x-1)^2/16)-9)^.5-1

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Find the foci of the ellipse with the equation: ((x^2)/36))+((y^2)/(64))=1
This is an ellipse with vertical major axis: (denominator of y-term>denominator of x-term)
Its standard form of equation: , a>b, (h,k)=(x,y) coordinates of center.
..
For given equation:
center: (0,0)
a^2=64
b^2=36
c^2=a^2-b^2=64-36=28
c=√28≈5.3
foci: (0,0±c)=(0,0±√28)=(0,0±5.3)=(0,-5.3) and (0,5.3)

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How do I change this into H-K form to find the properties of an elipse:
x^2+4y^2-6x-16y+21=0
complete the square
x^2-6x+4y^2-16y+21=0
(x^2-6x+9)+4(y^2-4y+4)=-21+9+16
(x-3)^2+4(y-2)^2=4
(x-3)^2/4+(y-2)^2/1=1
This is an ellipse with horizontal major axis and center at (3,2)
length of horizontal major axis=2a=4
length of minor axis=2b=2

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what is the equation for this conic section with An ellipse with vertices (-4, 0) and (4, 0) and a minor axis of length 6.
**
This is an ellipse with horizontal major axis. (x-coordinate changes but y-coordinates do not)Its standard form of equation: , a>b, (h,k)=(x,y) coordinates of center
..
For given ellipse:
center: (0,0)
length of horizontal major axis=8=2a
a=4
a^2=16
length of minor axis=6=2b
b=3
b^2=9
Equation of given ellipse:

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y=1/2tan2(x-5π/6)+1
Equation for tan function:
y=tan(Bx-C), period=π/B, phase shift=C/B
Rewrite given equation in this form:
y=1/2tan(2x-10π/6)+1
B=2
period=π/B=π/2
1/4 period=π/8=3π/24
phase shift=C/B=(10π/6)/2=5π/6 (to the right)
vertical asymptote at x=13π/12+2πk,k=any integer (function undefined at this point)
horizontal asymptotes: none

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For the function:
f(x): (5x-3)(x+4)/(-3x-4)(4x-3)
What are the x and y intercepts?
**
y-intercepts:
set x=0
y=(5x-3)(x+4)/(-3x-4)(4x-3)
y=(-3)(4)/-4)(-3)
=-12/12
=-1
..
x-intercepts
set y=0
(5x-3)(x+4)=0
x=3/5,-4

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what are the vertical, horizontal, and oblique asymptotes of
F(x)=10/(x-3)(x^2-49)
F(x)=10/(x-3)(x+7)(x-7)
Horizontal Asymptote: y=0 or x-axis (degree of numerator less than degree of denominator)
Vertical Asymptotes:x=-7, 3 and 7 (set denominator=0, then solve for x)
Oblique asymptotes: none (To exist, numerator must be 1 degree higher than degree of denominator)

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Solve 1-log(3x+1)=log(x-2)
log base =10
1-log(3x+1)=log(x-2)
1=log(x-2)+log(3x+1)
log(10)=log[(x-2)(3x+1)]
(x-2)(3x+1)=10
3x^2-5x-2=10
3x^2-5x-12=0
(x-3)(3x+4)=0
x=3
or
x=-4/3 (reject, (x-2)>0

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The expression 2x^3+x^2-7x-6 factorizes into a product of three factors.
Given that two of the factors are (x+1) and (x-2), write down its third factor.
**
use synthetic division:
.....2....|...2......1.......-7.....-6.......
.......................4........10.....6....
..............2.......5.........3......0...
(x-2)(2x^2+5x+3)=0
(x-2)(x+1)(2x+3)=0
third factor: (2x+3)

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Factorise x^3+3x^2-4x-12 completely.
x^3+3x^2-4x-12
x^2(x+3)-4(x+3)
(x^2-4)(x+3)
(x+2)(x-2)(x+3)

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Write the equation of a line that passes through the points (4,4) and (-2,1).
Your answer should be in slope intercept form (y = mx + b)
**
equation of a line: y=mx+b, m=slope, b=y-intercept
slope=∆y/∆x=(4-1)/(4-(-2))=3/6=1/2
equation y=(1/2)x+b
solve for b using coordinates of one of the given points(4,4) (either point can be used)
4=(1/2)4+b
b=2
equation of given line: y=x/2+2

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the ratio of two no is 3:5, and their difference is 46. find the larger no.
let x=larger no.
y=smaller no.
3/5=y/x
x-y=46
y=x-46
3/5=(x-46)/x
3x=5x-230
2x=230
x=115
larger no.=115

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A plane leaves the airport and flies East at 200 mph. 2 hours later,
another plane leaves and flies West at 150 mph. Write and solve a
system of linear equations that can be used to determine when the
planes are 750 miles apart.
**
let x-2=travel time of plane flying East
x=travel time of plane flying West(when planes are 750 miles apart)
distance=*travel time*speed
equation:
150(x-2)+200x=750
150x-300+200x=750
350x=1050
x=3
planes are 750 miles apart after 3 hrs

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The speed of a stream is 3 km/h. A boat travels 4 km upstream in the
same time it takes to travel 10 km downstream. What is the speed of
the boat.
**
let x=speed of the boat in still water
x-3=speed upstream
x+3=speed downstream
travel time=distance/speed (same for upstream and downstream)
4/(x-3)=10/(x+3)
10x-30=4x+12
6x=42
x=6
speed of the boat in still water=6 km/hr

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The ratio of the ages of Rajesh to that of Ramesh is 5 : 8. After 10 years the new ratio of their ages will be 7 : 10. What is the age of Ramesh at present?
**
let y=Rajesh's present age
let x=Ramesh's present age
y/x=5/8
y=5x/8
..
After 10 years:
y+10=Rajesh's age
x+10=Ramesh's age
(y+10)/(x+10)=7/10
7x+70=10y+100
7x+70=10(5x/8)+100
7x+50x/8=30
56x+50x=240
6x=240
x=40
Ramesh's present age=40

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A bus leave Sacramento traveling South at 40 mph. A half hour hour
later, a car leaves from the same location on the same route at 60 mph.
How far from Sacramento will the car overtake the bus?
**
let x=travel time of car
(x+.5)=travel time of bus
distance=speed*travel time(the same for car and bus)
..
60x=40(x+.5)
60x=40x+20
20x=20
x=1
distance car overtakes bus=60x=60 miles

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find the exact value of cos 23pi/12 using the properties of special triangles
Use cos half-angle identity
cos(x/2)=±√[(1+cos(x)/2]
cos(23π/12)=cos(23π/6)/2)=√[(1+cos(23π/6)/2]
For cos(23π/6) use reference angle π/6 in quadrant IV where cos>0
cos(23π/6)=cos(π/6)=√3/2
cos(23π/12)
=cos(23π/6)/2)
=√[(1+cos(23π/6)/2]
=√[(1+√3/2/2]
calculator check:
cos(23π/12)≈0.9659..
√[(1+√3/2/2]≈0.9659..

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|4x-2 divided by 5| = |6x+3 divided by 2|
|(4x-2)/5|=|(6x+3)/2|
solve for 2 equations
(4x-2)/5=(6x+3)/2
5(6x+3)=2(4x-2)
30x+15=8x-4
22x=-19
x=-19/22
check:
|(4x-2)/5|=|(6x+3)/2|
|(4(-19/22)-2)/5|=|(6(-19/22)+3)/2|
|-1.0909|=|-1.0909|
1.0909=1.0909
check ok
..
(4x-2)/5=-(6x+3)/2
-5(6x+3)=2(4x-2)
-30x-15=8x-4
-38x=11
x=-11/38
check:
|(4x-2)/5|=|(6x+3)/2|
|(4(-11/38)-2)/5|=|(6(-11/38)+3)/2|
|-.6316|=|.6316|
.6316=.6316
check ok
solution: x=-19/22, -11/38

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|10x+65|=7x+25
solve for 2 possibilities
..
10x+65=7x+25
3x=-40
x=-40/3
check:
(10x+65)=7x+25
|10*-40/3+65|=7*-40/3+25
|10*-40/3+65|=7*-40/3+25
|-68.33|=-68.33
68.33≠-68.33
does not check, no solution
..
-(10x+65)=7x+25
-10x-65=7x+25
-17x=90
x=-90/17
check:
|10x+65|=7x+25
|10*-90/17+65|=7*-90/17+25
|12.06|=-12.06
12.06≠-12.06
does not check, no solution

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how do you solve these equations
1) e^-x=5
-xlne=ln5
x=-ln5
..
2) 11^5x=33
5xlog11=log33
5x=log33/log11
x=(log33/log11)/5
x≈0.2916
..
3) 4e^2x=17
ln4+2xlne=ln17
2x=ln17-ln4
x=(ln17-ln4)/2
x≈.7235

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what is the vertex, axis of symmetry, and direction of opening for
y=x^2-18x+79?
complete the square
y=(x^2-18x+81)+79-81
y=(x-9)^2-3
This is an equation of a parabola that opens upwards. (lead coefficient positive)
Its standard form: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex.
For given equation: y=(x-9)^2-3
vertex:(9,-3)
Axis of symmetry: x=9