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write the equation of a hyperbola in standard form whose center is (-2, -4), a focus at (-2, 6), and eccentricity of 5/4.
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Given hyperbola has a vertical transverse axis. (From given center and focus data, y-coordinates change but x-coordinates do not.)
Its standard form of equation: , (h,k)=(x,y) coordinates of center,
For given hyperbola:
center: (-2,-4)
c=10 (-4 to 6) (distance from center to focus on vertical transverse axis)
c^2=100
eccentricity=5/4=c/a
a=4c/5=40/5=9
a^2=81
c^2=a^2+b^2
b^2=c^2-a^2=100-81=19
Equation of given hyperbola:

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Find the values of x in the interval 0° ≤ x ≤ 360° for which
sin(x) = cos(x)
sin(45º)=cos(45º) (in quadrant I where both sin(x) and cos(x)=√2/2)
sin(225º)=cos(225º) (in quadrant III where both sin(x) and cos(x)=-√2/2)

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How much time does a train 50m long moving at 65 km/hr take to pass another train 75 m long moving at 50km/hr in the same direction?
travel time=distance/speed
distance=50+75=125 m
speed=65-50=15 km/hr
travel time=.125/15*60=0.5 min
ans:0.5 min

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How fast would you have to travel on the surface of earth at the equator to keep up with the sun (that is, so that the sun would appear to remain in the same position in the sky)? Assume the radius of the earth at the equator is 3960 miles.
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The earth makes one revolution every 24 hours.
Each revolution=2π*radius=7920π miles
linear speed of earth=7920π miles/24 hours≈1037 mi/hr
In order to make it appear the sun stays in the same position, one must travel at this speed against the rotation of the earth which is west to east. Traveling at this speed east to west at the equator in effect makes one stationary with respect to the sun. I think.

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A wave is modeled with the function y=1/2 sin3x. Describe the graph of this function, including its period, amplitude, and points of intersection with the x axis.
I have the amp. is 1/2
the period is (2 pi)/3
but don't know how to get the points of intersection. I know one is (0,0)
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Basic equation for sin function: y=ASin(Bx-C), A=amplitude, Period=2π/B, phase shift=C/B.
For given equation: y=1/2 sin3x
Amplitude=1/2
B=3
Period=2π/B=2π/3
C=0
phase shift: none
Because there is no phase shift, the graph is like the basic sin curve with amplitude=1/2 and period=2π/3
As with the basic sin curve, it intersects the x-axis at zero, half the period, and at the end of the period. So, for given function, points of intersection for one period are: (0,0), (π/3,0) and (2π/3,0)

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Solve all equations for (0, 2pi)
sin(x + (pi/4)) + sin(x - (pi/4)) = 1
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Using sin addition identities:
sin(x+π/4)=sinxcos(π/4)+cosxsin(π/4)
sin(x-π/4)=sinxcos(π/4)-cosxsin(π/4)
sinxcos(π/4)+cosxsin(π/4)+sinxcos(π/4)-cosxsin(π/4)=1
2sinxcos(π/4)=1
2sinx√2/2=1
sinx=1/√2=√2/2
x=π/4, 3π/4 (in quadrants I and II where sin>0)

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Find the exact value of x between 0 degrees and 180 degrees for which each is true.
cos x = sqrt(-3)/2
I believe this should be written:cosx=-√3/2
For given domain, [0,180º], reference angle is in quadrant II where cos<0.
cosx=-√3/2
x=150º

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How can do you prove that
secx - (tanx)(sinx) = cosx?
Start with left side:
secx-(tanx)(sinx)
=(1/cosx)-(sinx/cosx)sinx
=(1/cosx)-sin^2x/cosx
=(1-sin^2x)/cos)
=cos^2x/cosx
=cosx
verified: left side=right side

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Find the coordinates of the vertex,focus,ends of the latus rectum and the equation of the directrix.Draw the parabola of:
x^2=12(y+7)
...
This is an equation of a parabola that opens upwards:
Its basic form of equation:(x-h)^2=4p(y-k)
For given equation: x^2=12(y+7)
vertex: (0,-7)
axis of symmetry: x=0
4p=12
p=3
focus: (0,-4) (p-units above vertex on the axis of symmetry)
Ends of latus rectum(focal width):
plugs in y-coordinate of focus(-4) then solve for x
x^2=12(y+7)
x^2=12(-4+7)=12*3=36
x=±√36=±6
ends of latus rectum: (-6,-4) and (6,-4)
see graph below:

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if the point (-2,-7) is on the terminal arm of N, an angle in standard position, what is the value of cos N?
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The value of angles in standard positions is measured by their reference angles.
The reference angle is the angle the terminal arm makes with the horizontal x-axis
The reference angle is in quadrant III where tan>0
The x-value(-2) represents the horizontal(cos) leg of the reference right triangle
The y-value(-7) represents the vertical(sin) leg of the reference right triangle
The hypotenuse=√(2^2+7^2)=√(4+49)=√53
cosN=-2/√53≈-0.2747

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y = 25 -x^2
I will give it a try:
The given equation is that of a parabola that open downward (curve has a maximum).
Its standard form: y=A(x-h)^+k, (h,k)=(x,y) coordinates of the vertex.
For given equation: y=25-x^2
rewrite: y=-x^2+25
Since you don't see h, the x-coordinate of the vertex is=0
The y-coordinate of the vertex is=k=25 which is also the maximum value for that particular parabolic function. A which is assumed=1, affects the slope of steepness of the curve.
...
As for the maximum area is 48.11 square units at x=2.89 units, this must represent a different curve from the one above. If it is a parabola it would be like: y=-(x-2.89)^2+48.11, that is, coordinates of the vertex would be (2.89,48.11)
...
In math and physics applications of the parabola is quite common especially in higher math courses like calculus. To get a better understanding of what the different curves look like, get a graphics calculator and punch in the equation to display the curve.
Hope this helps!

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Write the following in terms of sin θ and cos θ; then simplify if possible. (Leave your answer in terms of sin θ and/or cos θ.)
sin theta cot theta+7 cos theta
..
sin θ*cot θ+7cos θ
sin θ*(cos θ/sin θ)+7cos θ
cos θ+7cos θ=8cos θ

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A pick-up truck is fitted with new tires which have a diameter of 44 inches. How fast will the pick up truck be moving when the wheels are rotating at 285 revolutions per minute?
***
circumference=π*diameter=44π inches
speed of truck=285 rev/min*44π inches/rev=44π*285=1254π inches/min

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sec3x = sqrt2
solve for [0,2pi]
sec3x = √2=1/cos3x
cos3x=1/√2=√2/2
3x=π/4, 5π/4
x=π/12, 5π/12

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tan3x = sqrt 3
solve for [0,2pi]
tan 3x=√3
3x=π/3, 4π/3
x=π/9, 4π/9

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2sin^2x-5sinx=-3
2sin^2x-5sinx+3=0
(2sinx-3)(sinx-1)=0
2sinx-3=0
sinx=3/2 (reject, -1 ≤ sinx ≤ 1)
or
sinx-1=0
sinx=1
x=π/2

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how can I solve this use eccentricity of each ellipse to find its equation is standrad forom eccentricity 2/5 , major axis on the x-axis and of lenth 10 center (0,0)
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Standard form of equation for an ellipse with horizontal major axis:
, (a>b), (h,k) = (x,y) coordinates of the center
For given problem:
center: (0,0)
length of horizontal major axis=10=2a
a=5
a^2=25
eccentricity=c/a=2/5
c=2a/5=10/5=2
c^2=4
c^2=a^2-b^2
b^2=a^2-c^2=25-4=21
Equation of given ellipse:

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write x in terms of a,b and c
log x=2log a-(3logb(1/2)log c)
log x=log[a^2/b^3*c^(1/2)]

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Express as a single logarithm with a coefficient of 1:
3*log(x)-2*log(y)+4*log(z)
log[(x^3*z^4)/y^2]

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Solve: log5x-log5(x-2)=log5(4)
log5[x/(x-2)]=log5(4)
x/(x-2)=4
x=4x-8
3x=8
x=8/3

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write the expression 2logx-(3logY+logz) as a single log
log[(x/(y^3)z]

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2/log(base x)4 + 3/log(base 4)x = 7
convert to Base 10
2/(log4/logx)+3/(logx/log4)=7
2logx/log4+3log4/logx=7
LCD:(logx)(log4)
2(logx)^2+3(log4)^2=7log4logx
log4≈.60206
2(logx)^2+1.08743=4.21442logx
2(logx)^2-4.21442logx+1.08743=0
solve for logx using quadratic formula:
logx=.30102
x=10^.30102
x≈2
or
logx=1.80618
x=10^1.80618
x≈64
..
Check:
for x=2
2/log(base x)4+3/log(base 4)x
=2/log2(4) + 3/log4(2)
=2/2+3/(1/2)
=1+6=7
..
for x=64
2/log(base x)4+3/log(base 4)x
=2/log64(4) + 3/log4(64)
=2/(1/3)+3/3
=6+1=7
ans:
x=2 or 64

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Find the exact value of the trigonometric functions.
a. 13pi/2=6π+π/2=π/2
This is like rotating 3 revolutions (6π) counter-clockwise +π/2
b.(-11pi/2)=5π+π/2=4π+π+π/2=π/2
This is like rotating 2 revolutions (4π) clockwise+π+π/2

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How do you solve
2sinx cosx = -sinx?
divide both sides by sinx
2cosx=-1
cosx=-1/2
x=2π/3 or 4π/3 (in quadrants II and III where cos<0)

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Sec^2x=4
secx=2=1/cosx
cosx=1/2
x=π/3 or 60º

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sin(8pi)=?
on a unit circle, 8π is like making 4 revolutions counter-clockwise starting from zero and returning to zero. So, sin(8π)=0

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Find the exact value of each of the following under the given conditions:





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let o=opposite side
let a=adjacent side
let h=hypotenuse
..
π/2 < A < π (quadrant II)
cotA=-24/7=a/o
a=-24, o=7
h=√(o^2+24^2)=√(49+576)=√625=25
sinA=o/h=7/25
cosA=a/h=-24/25
tanA=o/a=-7/24
...
0 < B < π/2 (quadrant I)
cosB=5/6=a/h
a=5, h=6
o=√(h^2-a^2)=√(36-25)=√11
sinB=o/h=√11/6
cosB=a/h=5/6
tanB=o/a=√11/5
...
Identity: sin(A+B)=sinAcosB+cosAsinB
=7/25*5/6+(-24/25*√11/6)
=35/150-24√11/150=(35-24√11)/150
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Identity: sin(A-B)=sinAcosB-cosAsinB
=7/25*5/6-(-24/25*√11/6)
=35/150+24√11/150
=(35+24√11)/150
...
Identity:cos(A-B)=cosAcosB+sinAsinB
=-24/25*5/6+7/25*√11/6
=-120/150+7√11/150
=-120+7√11/150
...
Identity:tan(A+B)=(tanA+tanB)/(1-tanAtanB)
=(-7/24+√11/5)/(1-(-7/24*√11/5)
=(-7/24+√11/5)/(1+7√11/120)

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find the exact value of the remaining trigonometric functions:
1. cotθ = 12/5, cosθ < 0
let O=opposite side
let A=adjacent side
H=hypotenuse
..
cot>0 and cos<0 only in quadrant III
cotθ = 12/5=A/O
A=-12, O=-5
H=√(O^2+A^2)=√(144+25)=√169=13
sinθ=-5/13
cosθ=-12/13
tanθ=5/12
cscθ=-13/5
secθ=-13/12
cotθ=12/5
..
2. cosθ = -3/5, θ in quadrant III
sin and cos<0
cosθ=-3/5=A/H
A=-3, H=5
O=√(H^2-A^2)=√(25-9)=√16=-4
sinθ=-4/5
cosθ=-3/5
tanθ=4/3
cscθ=-5/4
secθ=-5/3
cotθ=3/4

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I must find the horizontal and vertical asymptotes of : f(x) = x/ x^2 - 9
..
When the degree of the numerator is 1 less than the degree of the denominator, as in this case, the horizontal asymptote is the x-axis or y=0.
Vertical asymptotes are found by setting the denominator=0, then solving for x-values which makes the function undefined.
x^2-9=0
(x+3)(x-3)=0
x≠3
and
x≠-3
horizontal asymptotes: x-axis
Vertical asymptotes: x=3, x=-3)

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|x-9|=x^2-9x
solve for 2 possibilities:
for (x-9)>0
x-9=x^2-9x
x^2-10x+9=0
(x-9)(x-1)=0
x=9
or
x=1
..
for (x-9)<0
-x+9=x^2-9x
x^2-8x-9=0
(x-9)(x+1)=0
x=9
or
x=-1
Check:
for x=9
|x-9|=x^2-9x
0=81-81
0=0 (x=9 ok)
..
for x=1
|x-9|=x^2-9x
|1-9|=1-9
8≠-8 (reject x=1)
..
for x=-1
|x-9|=x^2-9x
|-1-9|=x^2-9x
10=10 (x=-1 ok)
solution: x=9,-1

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√(x + 7) - 2√(x) = -2
√(x+7)=2√x-2
square both sides
x+7=4x-8√x+4
-3x+3=-8√x
-3(x-1)=-8√x
square both sides again
9(x-1)^2=64x
9(x^2-2x+1)=64x
9x^2-18x+9-64x=0
9x^2-82x+9=0
solve for x using quadratic formula:
x≈0.1111..(reject, extraneous root)
x=9