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for the equation of a hyperbola, find the standard equation, vertices, foci, asymptotes, and graph
1) (x^2/36)- (y^2/49)=1
2) 25x^2-16y^2-400=0
***
Standard form of equation for a hyperbola with horizontal transverse axis:
, (h,k)=(x,y) coordinates of center
..
1) (x^2/36)- (y^2/49)=1
center:(0,0)
a^2=36
a=√36=6
vertices: (0±a,0)=(0±6,0)=(6,0) and (-6,0)
..
b^2=49
b=√49=7
..
c^2=a^2+b^2=36+49=85
c=√85≈9.2
foci: (0±c,0)=(0±9.2,0)=(9.2,0) and (-9.2,0)
..
slopes of asymptotes=±b/a=±7/6
Equation of asymptote with slope,m=7/6
y=7x/6
..
Equation of asymptote with slope,m=-7/6
y=-7x/6
..
see graph below:
y=±(49x^2/36-49)^.5

I will let you do the 2nd problem.

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Write the standard equation for the parabola with the given set of characteristics. Then graph the parabola
focus (3,8)
vertex (3,2)
***
This is a parabola that opens upward.
Its basic equation: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex.
vertex: (3,2)
axis of symmetry: x=3
p=6 (distance from vertex to focus on the axis of symmetry
4p=24
Equation: (x-3)^2=24(y-2)
see graph below:

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A semicircular arch over a street has a radius 10 feet. How high is the arch at a point whose ground distance is 4 feet from the center? Give the exact answer.
***
Equation of circle with center at (0,0): x^2+y^2=r^2, r=radius
For given problem:
x^2+y^2=5^2
x^2+y^2=25
4 ft from center means x=4
4^2+y^2=25
y^2=25-16=9
y=±√9=±3
reject -3
How high is the arch at a point whose ground distance is 4 feet from the center? 3 ft

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If the equation for a hyperbola is not in the standard form already, how do you change it so you can graph it? I have this equation: 6(x-3)^2-4(y+1)^2=96. Is it necessary to put it in the standard hyperbola form? If so, how would you do this?
***
Standard forms of hyperbola:
For hyperbola with horizontal transverse axis: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center
For hyperbola with vertical transverse axis: , (h,k)=(x,y) coordinates of center
..
For given equation of hyperbola:

change to standard form:
divide both sides by 96

center: (3,-1)
a^2=16
a=4
length of horizontal transverse axis=2a=8
..
b^2=24
b=√24
length of conjugate axis=2b=2√24=2√6
..
slopes of the asymptotes=±b/a=2±√6/4=±√6/2
Equation of asymptote with negative slope:
y=mx+b=-√6x/2+b
solve for b using coordinates of center which are on the asymptote line.
-1=-√6*3/2+b
b=-1+3.6742=2.6742
Equation:y=-√6x/2+2.6742
..
Equation of asymptote with positive slope:
y=mx+b=√6x/2+b
solve for b using coordinates of center which are on the asymptote line.
-1=√6*3/2+b
b=-1-3.6742=-4.6742
Equation:y=√6x/2-4.6742
..
To graph the hyperbola manually, it is best to draw a rectangle around the center:
draw a horizontal line thru the center (3,-1) with end points 4 units from center. (horizontal transverse axis)
draw a vertical line thru the center (3,-1) with end points√24/2 units from center.(conjugate axis)
draw lines thru the end points to form a rectangle
Asymptotes go thru the corners of this rectangle and the center.
You now should be able to graph the hyperbola knowing the coordinates of the center, asymptotes, and the fact that hyperbola has a horizontal transverse axis

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Write the standard equation for the parabola with the given set of characteristics. Then graph the parabola
axis of symmetry: y=0
focus: (4,0)
vertex: (0,0)
***
This is a parabola that opens rightward:
Its basic equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of vertex
For given parabola:
p=4 (distance from vertex to focus on the axis of symmetry)
4p=16
Equation: y^2=16x
see graph below as a visual check:

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If the line passing through the points (a,2) and (7,7) is parallel to the line passing through the points (4,8) and (a+3,2) what is the value of a?
***
parallel lines have equal slopes
slope=∆y/∆x
(7-2)/(7-a)=(2-8)/((a+3)-4)
5/(7-a)=-6/(a-1)
-42+6a=5a-5
a=37

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A blend of coffee was made by using coffee worth $4.25 a pound and coffee worth $3.49 a pound. The blend contained seven pound less than three times as much of the $3.49 coffee than of the $4.25 coffee. How much of each kind was used if the total value of the blend was $196.37?
***
let x=lbs of $3.49 coffee used
3x-7=lbs of $4.25 coffee used
..
3.49x+4.25(3x-7)=196.37
3.49x+12.75x-29.75=196.37
16.24x=226.12
x≈13.92
3x-7≈34.77
..
lbs of $3.49 coffee used≈13.92
lbs of $4.25 coffee used≈34.77

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Almonds worth $4.95 a pound were mixed with walnuts worth $3.95 a pound. How many pounds of each were used if there were 11 more pounds of almonds than walnuts and if the value of the mixture was $116.75?
***
let x=lbs of walnuts used
x+11=lbs of almonds used
..
4.95(x+11)+3.95x=116.75
4.95x+54.45+3.95x=116.75
8.9x=62.30
x=7
x+11=18
lbs of walnuts used=7
lbs of almonds used=18

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Prove this identity:
2cos^2(x/2)=(sin^2x)/(1-cosx)
..
left side:
use cos half-angle formula:
2[√((1+cosx)/2)]^2=2((1+cosx)/2)=1+cosx
..
right side:(sin^2x)/(1-cosx)=(1-cos^2x)/(1-cosx)=
verified:
left side=right side=1+cosx

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given that cosA=1/3 where A is an acute angle, the exact value of cos A/2 equals
***
use half-angle formula for cos:
cos(A/2)=√[(1+cosA)/2]=√(1+(1/3))/2)=√((4/3)/2)=√(4/6)=√(2/3)
Check with calculator:
cosA=1/3
A≈70.5288º
A/2≈35.2644º
cos(A/2)≈cos35.2644º≈0.8165
√(2/3)≈0.8165

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8 years ago the father's age was 8 times the age of his son.after 10 years,the father's age will be twice the age of the son.what are their present age
***
8 yrs ago:
let x= son's age
8x=father's age
At present:
x+8=son's age
8x+8=father's age
After 10 years:
x+18=son's age
8x+18=father's age
father's age twice son's age
8x+18=2(x+18)
8x+18=2x+36
6x=18
x=3
Present ages:
x+8=11=son's age
8x+8=32=father's age

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A train running between two towns arrives at its destination 10 minutes late when it
goes 40 miles per hour and 16 minutes late when it goes 30 miles per hour. The
distance between the two towns is.
***
x=on-time travel time
x+(1/6)=travel time at 40 mph
x+(4/15)=travel time at 30 mph
distance=travel time*speed
10 min=1/6 hr
16 min=16/60=4/15 hr
..
40(x+(1/6))=30(x+(4/15))
40x+40/6=30x+8
10x=8-40/6=48/6-40/6=8/6=4/3
x=4/30=2/15
40(x+(1/6))=40*.3=12
or
30(x+(4/15))=30*6/15=12
distance between the two towns is=12 mi

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Determine the equation of any vertical asymptotes and the value of x for any holes in the graph:
***

Equation:
There is a hole at x=2
There are no asymptotes. Equation is that of a line, y=x-9

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A cyclist rode the first 22-mile portion of his work at a constant speed. For the 14- mile cool down portion of his workout, he reduced his speed by 4 miles per hour. Each portion of the workout took the same time. Find the cyclist's speed during the first portion and find his speed during the cool down portion.
***
let x=speed during first portion(22 mi)
x-4=speed during cool down portion(14 mi)
travel time=distance/speed
..
22/x=14/(x-4)
14x=22x-88
8x=88
x=11
x-4=7
speed during first portion(22 mi)=11 mph
speed during cool down portion(14 mi)=7 mph

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John and Mary leave their house at the same time and drive in opposite directions. John drives at 90 mi/h and travels 35 mi farther than Mary, who drives at 40 mi/h. Mary's trip takes 15 min longer than John's. For what length of time does each of them drive?
***
let x=John's travel time
15 min=1/4 hr
(x+1/4)=Mary's travel time
distance=travel time*speed
..
90x-40(x+1/4)=35
90x-40x-10=35
50x=45
x=45/50=9/10 hr or 54 min
x+1/4=18/20+5/20=23/20 hr or 69 min
John's travel time=54 min
Mary's travel time=69 min

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A construction worker drops a hammer from a height of 50 feet. After how many seconds will the hammer land on the ground? Round your answer to the nearest hundredth.
***
Equation for falling objects: t=√(2y/g), t=time in seconds, y=displacement, g=gravity=32ft/sec^2
t=√(2*50)/g))=√(100/32)=1.77 (ans B)
After how many seconds will the hammer land on the ground? 1.77 sec (ans B)

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Freddie leaves home and rides his bicycle out into the country for 2.7 hours. On his return trip along the same route, it takes him 0.9 of an hour longer. If his rate on the return trip was 3 miles per hour slower than on the trip out into the country, find the total roundtrip distance.
***
let x=rate of speed of initial trip
x-3=rate of speed of return trip
distance=travel time*speed (same for both trips)
2.7x=(2.7+.9)(x-3)
2.7x=3.6(x-3)
2.7x=3.6x-10.8
.9x=10.8
x=12
roundtrip distance=2*2.7*12=64.8 mi

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Writing an equation for a parabola in the form y= a(x-h)^2 + k with given information.
1. It gives you : with y- intersept 10, x- intercept 2, and equation of axis of symmetry x-3=0
how would i find a and k?
***
Standard form of equation for a parabola: y=a(x-h)^2+k, (h,k)=(x,y) coordinates of vertex, a=multiplier which affects slope or steepness of curve
Equation of axis of symmetry: x-3=0
x=3
so, h=3
Equation: y=a(x-3)^2+k
points given:
y-intercept: (0,10)
x-intercept: (2,0)
using points to form a system of two equations to solve for a and k
..
10=a(0-3)^2+k
0=a(2-3)^2+k
..
10=9a+k
0= a +k
subtract
10=8a
a=10/8=5/4
k=-a=-5/4
Equation:
y=(5/4)(x-3)^2-5/4

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Robert is 2 years older than Jazmine. Jazmine is 3 years older than Chris, who is one year less than 1/2 of Michael's age. Together, the sum of their ages is 30. How old is each child? Show your work and explain your reasoning
***
Michael's age=x
Chris' age=(x/2)-1
Jasmine's age=(x/2)-1+3=(x/2)+2
Robert's age=(x/2)+2+2=(x/2)+4
sum of their ages=30
x+(x/2)-1+(x/2)+2+(x/2)+4=30
(5x/2)+5=30
5x/2=25
5x=50
x=10
(x/2)-1=4
(x/2)+2=7
(x/2)+4=9
Michael's age=10
Chris' age=4
Jasmine's age=7
Robert's age=9

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Vertices at (-2,-3) and (8,-3), one end of the minor axis at (3,-7)
**
Given data shows this is an ellipse with horizontal major axis. (x-coordinates of vertices change but y-coordinates do not)
Its standard form of equation: , a>b, (h,k)=(x,y) coordinates of center
For given ellipse:
x-coordinate of center=(-2+8)/2=6/2=3 (midpoint formula)
y-coordinate of center=-3
center: (3,-3)
length of horizontal major axis=10 (-2 to 8)=2a
a=5
a^2=25
b=4 (-3 to -7)
b^2=16
Equation of given ellipse:

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Let
sin A =12/13
with A in Q2 and
sin B = − 15/17
with B in Q3. Find sin(A + B), cos(A + B), and tan(A + B).
**
let o=opposite side
let a=adjacent side
let h=hypotenuse
..
sin A =12/13=o/h
o=12, h=13
a=-√h^2-o^2)=-√(13^2-12^2)=-√(169-144)=-√25=-5
cos A=a/h=-5/13
tan A=o/a=-12/5
..
sinB =-15/17=o/h
o=-15, h=17
a=-√h^2-o^2)=-√(17^2-15^2)=-√(289-225)=-√64=-8
cos B=a/h=-8/17
tan B=o/a=-15/-8=15/8
..
sin(A+B)=(sinA*cosB)+(cosA*sinB)
=(12/13)*(-8/17)+(-5/13*-15/17)
=-96/221+75/221
=-21/221
..
cos(A+B)=cosA*cosB)-(sinA*sinB)
=(-5/13)*(-8/17)-(12/13*-15/17)
=40/221+180/221
=220/221
..
tan(A+B)=(tanA+tanB)/(1-tanA*tanB
=(-12/5+15/8)/(1-(-12/5)(15/8))
=(-96/40+75/40)/(1+180/40
=(-21/40)/(220/40)
=-21/220
..
Check with calculator:
sinA=12/13 in Q2
A≈112.62º
sin B=-15/17 in Q3
B≈241.93º
A+B≈354.55º
reference angle=5.45º in Q4
..
sin(A+B)=sin(354.55)=sin(5.45) in Q4≈-0.0945...
-21/221=≈-0.0945...
..
cos(A+B)=cos(354.55)=cos(5.45) in Q4≈0.9954...
220-221≈0.9954...
..
tan(A+B)=tan(354.55)=tan(5.45) in Q4≈-0.0954...
-21/220≈-0.0954...

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Solve for the general solution for x:
4 cos^2 x-1=0
cos^2 x=1/4
cosx=1/2
x=π/3

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I need to find the exact value of tan40°-tan10°/1+tan40°tan10°
Identity: tan(s-t)=(tan s-tan t)/(1+tan s*tan t)
tan40°-tan10°/1+tan40°tan10°=tan(40º-10º)=tan30º=√3/3

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what is the slope of a line that is perpendicular to y=2
y=2 is a horizontal line, so a line perpendicular to it is a vertical line whose slope is defined as undefined

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evaluate the fucntion with out using a calculator sin(-13pi/6)
***
-13π/6=-2π-π/6
This is like going one revolution clockwise,+π/6, terminating in quadrant IV with a reference angle of π/6.
sin(-13pi/6)=sin(π/6)=-1/2 (In quadrant IV where sin<0)

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cos x + 1 = √3 sin x where 0 ≤ x < 2π
cos x + 1 = √3 sin x
square both sides
cos^2x+2cosx+1=3sin^2x
cos^2x+2cosx+1=3(1-cos^2x)
cos^2x+2cosx+1=3-3cos^2x)
4cos^2x+2cosx-2=0
(4cosx-2)(cosx+1)=0
..
4cosx-2=0
cosx=1/2
x=π/3, 5π/3 (In quadrants I and IV where cos>0)
or
cosx+1=0
cosx=-1
x=π

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solve for x: log[5](x2(squared)+ x + 4= 2

convert to exponential form: base(5) raised to log of number(2)=number(x^2+x+4)
5^2=x^2+x+4=25
x^2+x-21=0
solve for x by quadratic formula:
x=-5.11
or
x=4.11

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Csc(3x-15)=sec(10+2x)
1/sin(3x-15)=1/cos(2x+10)
[cos(2x+10)/sin(3x-15)]=1
For this equation to be true, the angles for both sin and cos must be=π/4
..
2x+10=π/4
2x=π/4-10
x=(π/4-10)/2
x≈-4.6073
..
3x-15=π/4
x=π/4+15)/3
x≈5.2618
Check:with calculator
cos(2x+10)=cos(2*-4.6073+10)≈0.7071
sin(3x-15)=sin(3*5.2618-15)=0.7071
[cos(2x+10)/sin(3x-15)]=0.7071/0.7071=1

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Simplify:
(1-sin^2x)/(csc^2x-1)
(cos^2x)/((1/sin^2x)-1)
(cos^2x)/((1-sin^2x)/(sin^2x))
(cos^2x)/((cos^2x)/(sin^2x))
sin^2x

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how long it will take a savings acct bal to reach 1,000,000 if i deposited 2947 at 9% interest compounded quarterly
***
A=P(1+i)^n
P=initial investment=2947
i=interest per period(quarter)=.09/4=0.0225
n=number of periods
A=amount after n periods=10^6
..
10^6=2497(1+.0225)^n
10^6/2497=(1.0225)^n
take log of both sides
log(10^6)-log(2497)=nlog(1.0225)
log(10^6)=6
6-log(2497)/(log(1.0225))=n
n≈269 periods(quarters) or ≈ 67 yrs

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find f(1/4) where
f(x) = 12 sin{pi x + 3pi/4}
f(1/4)=12*sin(π/4+3π/4)
=12*sin(π)
=12*0
=0