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a triangle has vertex coordinates (1,1), (1,5) and (5,3). find the length of each side and decide what type of triangle it is.
Using distance formula: d^2=(x1-x2)^2+(y1-y2)^2
..
Between (1,1) and (1,5)
d^2=(1-1)^2+(5-1)^2=0+16=16
d=4
..
Between (1,5) and (5,3)
d^2=(5-1)^2+(3-5)^2=16+4=20
d=√20
..
Between (5,3) and (1,1)
d^2=(5-1)^2+(3-1)^2=16+4=20
d=√20
..
This is an isosceles triangle with two of the sides=√20 and the third side=4

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3cos^2a+2=3-2cosa
3cos^2a+2cosa-1=0
(3cosa-1)(cosa+1)=0
..
3cosa-1=0
3cosa=1
cosa=1/3
a≈70.5º
or
cosa+1=0
cosa=-1
a=180º

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The endpoint coordinates of the hypotnuse of a right triangle are (1,0) and (6,-5). What are possible coordinates of the point where the legs of this triangle intersect and what is the lenght of the hypotenuse?
**
The best way to see this problem is to roughly plot the points. You will see that the two legs of the right triangle intersect at (0,6). This makes the vertical leg=5 and the horizontal leg=6.
The length of the hypotenuse by Pythagorean Theorem=√(6^2+5^2)=√(36+25)=√61

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Solve the linear inequality and write the solution in set-builder notation.
8x-38˂10
**
8x-38<10
8x<48
x<6
{x|x<6}

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How long will it take $750 to quadruple in an account that pays 4.5% interest compounded continuously?
**
Formula for continuous compounding:A= Pe^rt, P=initial investment, r=interest per period, t=number of periods, A=amount after t periods.
For given problem:
P=750
r=4.5% per year
t=number of years
A=4P
..
A/P=e^rt
4=e^rt=e^.045t
take log of both sides
.045t*lne=ln4
log of base=lne=1
.045t=ln4
t=ln4/.045
t=30.8
ans:
It would take about 40 years to quadruple in an account that pays 4.5% interest compounded continuously?
note: this would be true for any initial investment compounded continuously at 4.5% interest rate.

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What is the focus, vertex, axis of symmetry and directrix of the parabola equation
y2-8y+16x-64=0
complete the square
(y^2-8y+16)+16x-64-16=0
(y-4)^2+16x-80=0
(y-4)^2=-16x+80
(y-4)^2=-16(x-5)
This is an equation for a parabola that open leftwards of the standard form: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex.
For given parabola:
4p=16
p=4
vertex: (5,4)
axis of symmetry: y=4
focus: (1,4) (p units to left of vertex on axis of symmetry)
directrix: x=9 (p units to right of vertex on axis of symmetry)

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Find an equation for the parabola with focus (4,-6) and vertex at (3, -6)
**
Standard form of equation for a parabola that opens rightwards: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of vertex.
For given parabola:
vertex: (3,-6)
axis of symmetry: y=-6
p=1 (distance from focus to vertex on axis of symmetry)
4p=4
Equation of parabola:
(y+6)^2=4(x-3)

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How much Brand A fruit punch (20% fruit juice) must be mixed with 8 L of Brand B fruit punch (50% fruit juice) to create a mixture containing 40% fruit juice?
**
let x=liters of Brand A fruit punch to be mixed with 8 liters of Brand B fruit punch
x+8=liters of the mixture
..
20%x+50%*8=40%(x+8)
.2x+4=.4x+3.2
.2x=.8
x=4 liters
ans:
liters of Brand A fruit punch to be mixed with 8 liters of Brand B fruit punch=4

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(7/3)x^(4/3)-3x^(-2/3)= 0
x^(-2/3)((7/3)x^(6/3)-3)=0
x^(-2/3)=0
1/x^(2/3)≠0
no solution
or
(7/3)x^2-3=0
(7/3)x^2=3
7x^2=9
x^2=9/7
x=±√(9/7)=±3/√7=±3√7/7

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A boy walked over hills at 3 kilometers per hour and on flat roads at 4 kilometers per hour. He walked 18 kilometers in 5 hours. How much of his trip was over hills?
**
let x=distance walked over hills
18-x=distance walked over flat roads
travel time=distance/speed
..
x/3+(18-x)/4=5
LCD:12
4x+54-3x=60
x=6 km
ans:
distance walked over hills=6 km

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(4x+12)/(x-2) - (2x+8)/(-x+2)
(4x+12)/(x-2) + (2x+8)/(x-2)
4x+12+2x+8/(x-2)
(6x+20)/(x-2)
2(3x+10)/(x-2)

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5/ (3) sqrt(5) + sqrt (3)
5/(3√5+√3)
multiply top and bottom by the conjugate of the denominator=(3√5-√3)
5(3√5-√3)/(3√5+√3)(3√5-√3)
(15√5-5√3)/(45-3)
(15√5-5√3)/(42)

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4/ (6-2 (sqrt (6)
4/(6-2√6)
multiply top and bottom by conjugate of denominator=(6+2√6)
4(6+2√6)/(6-2√6)(6+2√6)
(24+8√6)/(36-24)(difference of squares)
(24+8√6)/12
24/12+8√6/12
2+2√6/3

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why isn't (1,3) (2,2) (1,5) (2,6) a function
For an expression to be a function, for each x there must only be only one y.
You will note that for points (1,3) and (1,5), x has two different values y.
Similarly, (2,2 and (2,6) show the same. Therefore, these points are not from a function.
If you graph these points, you will get a curve similar to a parabola opening rightwards and it would fail the vertical line test which allows only one intersection for a function.

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Decide whether or not the functions are inverses of each other.
1] f(x)= 4x+16 and g(x)= 1/4x-4
2] f(x)= sqrt(x+8), domain [-8, infinity); g(x)=x^2+8, domain (-infinity, infinity)
and last one:
Determine
i) the domain of the function,
ii) the range of the function,
iii) the domain of the inverse, and
iv) the range of the inverse.
f(x) = 2x + 1
**
1] f(x)= 4x+16 and g(x)= 1/4x-4
(fog)(x)=f[g(x)]=4[(x/4)-4)]+16=x-16+16=x
(gof)(x)=g[f(x)]=(1/4)(4x+16)-4=x+4-4=x
Therefore functions are one-to-one and inverses to each other.
..
2] f(x)= sqrt(x+8), g(x)=x^2+8
f(x) is a one-to-one function but g(x) is not. g(x) is a parabola which is bumped 8 units up and it would fail the horizontal line test as it would have two intersections, that is for a given y, you could have two different x's. These two functions are not inverses of each other.
..
f(x)=2x+1
i) domain:all real numbers or (-∞,∞)
ii) range: (-∞,∞) (This is a straight line with a slope=2 and y-intercept=1.
iii) the domain of the inverse.
x=2y+1
2y=x-1
y^-1=(x-1)/2=x/2-1/2 (This is a straight line with slope=1/2 and y-intercept=-1/2)
domain: (-∞,∞)
iv) range of inverse: (-∞,∞)

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Please help and show work. Thanks in advance.
a. Does the function f(x)=x^2 have an inverse? If so, what is it? If not, why not?
b. the function f(x)= x+3/x-5 has an inverse. Please find it.
**
The method used to find the inverse: interchange x and y, then solve for y.
symbol used for inverse: y^-1
..
a. y=x^2
This function does not have an inverse. If you graph this function, you will see that is is a parabola that opens upwards so for a given y you will have two different x's. This means it is not a one-to-one function and cannot have an inverse. You can also use the horizontal line test to show it will have two intersections which means the function is not one-to-one
..
b. y=(x+3)/(x-5)
This function is one-to-one and its inverse can be found by interchanging x and y, then solving for y.
x=(y+3)/(y-5)
xy-5x=y+3
xy-y=5x+3
y(x-1)=5x+3
y^-1=(5x+3)/(x-1)

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I don't understand how to do problems that include three or four problems together such as x^3+5x^2-16x-80 can you help or explain.
**
Usually, but not always, problems like this with 4 terms are set up so you can factor the expression by a method called grouping:
x^3+5x^2-16x-80
Group terms to get a common factor
=x^2(x+5)-16(x+5)
factor out common term (x+5)
=(x+5)(x^2-16)
expand difference of squares
=(x+5)(x+4)(x-4)

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write the equation of the parabola in standard form
x^2 + 2x + 8y + 25 = 0
Standard form of equation for a parabola: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex.
complete the square:
-8y=(x^2+2x+1)+25-1
-8y=(x+1)^2+24
divide by -8
y=-(1/8)(x+1)^2-3
This is an equation of a parabola that opens downwards with vertex at (-1,-3)

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an airplane flying into a headwind travels the 1800-mile flying distance between two cities in 3 hours and 36 minutes. On the return flight, the distance is traveled in 3 hours. Find the air speed of the plane and the speed of the wind, assuming that both remain constant.
**
let x=speed of plane
let c=speed of wind
x-c =net speed into wind
x+c=net speed with wind
speed*travel time=distance
3 hrs and 36 min=3.6 hrs
..
(x-c)*3.6=1800
(x+c)*3=1800
..
3.6x-3.6c=1800
3.0x+3.0c=1800
..
multiply first equation by 3 and second equation by 3.6
10.8x-10.8c=5400
10.8x+10.8c=6480
add equations to eliminate c
21.6x=11880
x=11880/21=550 mph
..
3c=1800-3x=1800-1650=150
c=150/3=50 mph
..
ans:
speed of plane=550 mph
speed of wind=50 mph

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Please help me solve this equation:
log^7 100-log^7(y-5)=log^7 10
I will assume you mean base 7 instead of exponent 7:
log7(100)-log7(y-5)=log7(10)
log7(100)-log7(y-5)-log7(10)=0
log7(100)-(log7(y-5)+log7(10))=0
place under single log
log7[(100)/(y-5)(10)]=0
convert to exponential form: base(7) raised to log of number(0)=number(100)/(y-5)(10)
7^0=(100)/(y-5)(10)=1
(y-5)(10)=100
10y-50=100
10y=150
y=15

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Solve the equation. (Do not use mixed numbers in the answer.)
ln(5x-6)+ln(x) = ln(e)
ln(5x-6)+ln(x) =1
ln[(5x-6)(x)]=1
e^1=5x^2-6x=e
5x^2-6x-e=0
solve by following quadratic formula:

a=5, b=-6, c=-e

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Write the trigonometric expression as an algebraic expression in u.
15) cos (sin-1 u)
**
(sin-1 u) is an angle whose sin=u/1 (opposite side/hypotenuse)
The adjacent side for this angle would then be=√(1^2-u^2)=√(1-u^2) (by pythagorean theorem)
Thus, cos (sin-1 u)=√(1-u^2)/1 =√(1-u^2)

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A drain can empty 3/4 of a sink in 1 minute and the faucet can fill 5/6 of the sink in 1 minute. If the faucet is on and the drain is open, how long will it be before the sink overflows?
**
let x=minutes sink overflows
5x/6-3x/4=1
LCD:12
10x-9x=12
x=12
ans:
The sink will overflow in 12 minutes

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Given the polynomial function f(x),find the rational zeros,then the other zeros(that is,solve the equation f(x)=0),and factor f(x) into linear factors.
F(x)=x^4+9x^3+17x^2-9x-18
**
Using Rational Roots Theorem:
....0...|......1......9......17......-9......-18
....1...|......1.....10.....27......18........0 (1 is a zero)
....2...|......1.....11.....37......65......112 (2 is upper bound)
=======================
....0...|......1.....10.....27......18
..-1...|......1.....9.......18......0 (-1 is a zero)
F(x)=(x-1)(x+1)(x^2+9x+18)
F(x)=(x-1)(x+1)(x+6)(x+3)
zeros are: -6, -3, -1, and 1

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solve 4+ln(27x) = 25-2lnx
ln(27x)+2lnx=21
ln[(27x)(x^2)]=21
ln(27x^3)=21
convert to exponential form
e^21=27x^3
take cube root of both sides
3x=e^7
x=e^7/3

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use the change of base formula to approximate log5 (6) to 3 decimal places
log5(6)=log(6)/log(5)≈1.113

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THe quesion is 3^Log base 3 of 2 and there is an option to there not being an answer.
3^log3(2)
by definition: base raised to log of number=number (exponential form)
so 3^log3(2)=2

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log5(1.25*4)
log5(5)=1

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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