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1/2logx+log4=2
log[x^(1/2)(4)]=log(100)
4√x=100
square both sides
16x=10^4=10000
x=10000/16
x=625

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9000 = 5000 (1+r/360)^1440
divide both sides by 1000
9=5(1+r/360)^1440
9/5=(1+r/360)^1440
1.8=(1+r/360)^1440
raise both sides by 1/1440
1.8^(1/1440)=1+r/360
1.8^(1/1440)=(360+r)/360
360*1.8^(1/1440)=(360+r)
r=360*1.8^(1/1440)-360

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e^-2=x^6
-2lne=6lnx
6ln(x)=-2
ln(x)=-2/6=-1/3
x=e^(-1/3)

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Log5x-Log(2x-1)=Log7
log[5x/(2x-1)]=log(7)
5x/(2x-1)=7
5x=14x-7
9x=7
x=7/9

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e^(2x)-2e^x+1 =0 Solve:
let u=e^x
u^2=e^(2x)
..
u^2-2u+1=0
(u-1)^2=0
u=1 (mult. 2)
u=e^x=1
x=0

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Find all solutions in the interval [0,2pi] or [0,360degrees]. may pick degrees or radians. give exact value.
sin(3x+π)+ 1/2 = 1
sin(3x+π)=-1/2
3x+π=inverse sin(1/2)=π/6
3x=π/6-π=-5π/6
x=-5π/18 radians in (quadrant IV)
check:
sin(3x+π)+ 1/2
=sin((-15π/18)+(18π/18))+1/2
=sin(3π/18)+1/2
=sin(π/6)+1/2
=1/2+1/2
=1
..
note: domain of inverse sin: [-π/2, π/2]

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when finding the values of inverse trigonometric functions, how do you get from.. Cos-1 (-0.5) =x cos x = -0.5 and 0_< x _< pi to Cos -1 (-0.5) = 2 pi over 3?
***
Cos -1 (-0.5) = 2 pi over 3
Inverse cos(-0.5)=2π/3
This reads: (2π/3 is an angle between 0 and π whose cos is= (-0.5)
This also means the angle must be in quadrant I or quadrant II.
cos>0 in quadrant I and cos<0 in quadrant II
So the angle must be in quadrant II since the given cos inverse =(-0.5)<0.
If given cos inverse was (0.5)>0, the angle would be in quadrant I=π/3 instead of 2π/3 in quadrant II
By definition, the cos inverse domain is restricted to [0,π]

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How do you solve
7=5+sec(4x+7pi/4)
5+csc(2x+pi/3)=(15+2(sqrt(3)/3)
***
7=5+sec(4x+7pi/4)
7-5=sec(4x+7pi/4)
2=1/cos(4x+7pi/4)
cos(4x+7pi/4)=1/2


LCD:12


...
5+csc(2x+pi/3)=(15+2(sqrt(3)/3)
csc(2x+pi/3)=15+2(sqrt(3)/3)-5
csc(2x+pi/3)=10+2(sqrt(3)/3)≈11.1547
sin(2x+pi/3)=1/11.1547≈0.0896
2x+pi/3=inverse sin(0.0896)≈0.0897
2x=0.0897-pi/3≈-0.9575
x≈-0.4787 radians

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Graph the Exponential Function
f(x)= 4^(x+2)
x+2 is the exponent of the 4.
***
Start with 4^x (red curve) where you have the x-axis as a horizontal asymptote and a y-intercept at (0,1)
4^(x+2) then shifts the curve 2 units to the left.(green curve)
See graphs below of these two curves:

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how long does it take for $900 to double if it is invested at 7% compound continuously?
Formula for continuous compounding: A=Pe^rt, P=initial amt, r=interest rate, A=amt after t years
For given problem:
A/P=2
r=.07
e^rt=2
take log of both sides
rt*lne=ln2
lne=1
rt=ln2
t=ln2/r=ln2/.07≈9.9≈10 years
At 7% compound continuously,$900 to double in approximately 10 yrs

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How many quarts of water must be added to 40 quarts of solution which is 5 percent acid solution to dilute it to a 2 percent solution?
***
let x=amt of water to be added
40+x=amt of final mixture
..
5%*40+0%x=2%(40+x)
2+0=0.8+.02x
.02x=1.2
x=60
amt of water to be added=60 quarts

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A car travels 400 miles in 7 hours. Before noon the driver averages 60 miles per hour, but after noon he averages only 50 miles per hour. At what time did he leave?
***
let x=distance traveled before noon
400-x=distance traveled after noon
Travel time=distance/speed
..
travel time before noon plus travel time after noon=7 hrs

50x+24000-60x=7*50*60
-10x+24000=21000
10x=3000
x=300
travel time before noon=x/60=300/60=5 hrs
At what time did he leave? 5 hrs before noon=7 am

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Find an asymptote of this conic section..
9x^2-36x-4y^2+24y-36=0
complete the square:
9(x^2-4x+4)-4(y^2-6y+9)=36+36-36
9(x-2)^2-4(y-3)^2=36

This is an equation of a hyperbola with horizontal transverse axis:
Its standard form of equation: , (h,k)=(x,y) coordinates of center
For given equation:
center: (2,3)
a^2=4
a=2
b^2=9
b=3
Slopes of asymptotes for hyperbolas with horizontal transverse axis=±b/a=±3/2
Asymptotes are equations of straight lines that go thru the center of the hyperbola: y=mx+b, m=slope, b=y-intercept
..
Equation of asymptote with negative slope, m=-3/2
y=-3x/2+b
solve for b using coordinates of the center (2,3)
3=-3*2/2+b
b=6
equation:y=-3x/2+6
..
Equation of asymptote with positive slope, m=3/2
y=3x/2+b
solve for b using coordinates of the center (2,3)
3=3*2/2+b
b=0
equation:y=3x/2
..
Equations of asymptotes: y=-3x/2+6 and y=3x/2

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solve for an equation from just a vertex at a origin and with the directrix of y= -3/4?
***
Given directrix (-3/4) shows this is a parabola that opens downward:
Basic form of equation for such a parabola with vertex at (0,0):
x^2=-4py
axis of symmetry:x=0
p=3/4 (distance from vertex to directrix on the axis of symmetry)
4p=3
equation: x^2=-3y

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This is an equation of a hyperbola with horizontal transverse axis.
Its standard form:,(h,k)=(x,y) coordinates of center.
For given equation:
center:(0,0)
a^2=25
a=5
b^2=49
b=7
c^2=a^2+b^2=25+49=74
c=√74≈8.60
You got everything right except a and b which you had in reverse.
For hyperbolas, a and b don't change positions like in ellipses
For hyperbolas with vertical transverse axis, the y-term is placed ahead of the x-term, but a and b remain in the same position as that for a hyperbola with horizontal transverse axis

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If f(x)= cos(2x)-3sin(4x), find the exact value of f'(PIE/6)
***
For[0,2π]
f(x)= cos(2x)-3sin(4x)
f(π/6)=cos(2π/6)-3sin(4π/6)
..
cos(2π/6)=cos(π/3)=1/2
sin(4π/6)=sin(2π/3)=√3/2 (in Q2 where sin>0)
..
f(π/6)=cos(2π/6)-3sin(4π/6)
=cos(π/3)-3sin(2π/3)
=1/2-3√3/2
=(1-3√3)/2

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Solve the equation on the given interval, expressing the solution for x in terms of inverse trigonometric functions. (Enter your answers as a comma-separated list.)
7(cos x)^2-cos x−6=0 on [π/2, π]
..
7(cos x)^2-cos x−6=0
(7cosx+6)(cosx-1)=0
..
7cosx+6=0
7cosx=-6
cosx=-6/7
x=arccos(-6/7)
..
cosx-1=0
cosx=1
x=arccos(1)
..
solutions: x=arccos(-6/7), arccos(1)

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verify the identity
cos^2 B-sin^2 B=1-2sin^2 B
start with left side:
cos^2 B-sin^2 B
1-sin^2B-sin^2B=1-2sin^2B
verified: left side=right side

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sin^2x-1=0
sin^2x=1
take sqrt of both sides
sinx=1
x=π/2

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verify all solutions to the equations in the interval [0,2pie)
2sin^2x=2+cosx
2(1-cos^2x)=2+cosx
2-2cos^2x=2+cosx
2cos^2x+cosx=0
cosx(2cosx+1)=0
..
cosx=0
x=1
..
2cosx+1=0
cosx=-1/2
x=2π/3, 4π/3 (in Q2 and Q3 where cos<0)
..
solutions: x=1, 2π/3, 4π/3

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find all solutions to the equations [0,2pie)
2sin^2x+3sinx+1=0
(2sinx+1)(sinx+1)=0
..
2sinx+1=0
sinx=-1/2
x=7π/6, 11π/6 (in Q3 and Q4 where sin<0)
..
sinx+1=0
sinx=-1
x=3π/2
solutions: x=7π/6, 11π/6, 3π/2

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Use the trigonometric function values to evaluate.
cos90° + 3sin270°
**
cos90º=0
3sin270º=-1*3=-3
cos90°+3sin270°=0+(-3)=-3

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Determine the signs of the trigonometric functions for 218°
This places the angle 218º in Q3 where:
sin<0
cos<0
tan>0
csc<0
sec<0
cot>0

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For: [0,π/2]
sin2theta-costheta=0
sin2x-cosx=0
2sinxcosx-cosx=0
cosx(2sinx-1)=0
..
cosx=0
x=1
..
2sinx-1=0
sinx=1/2
x=π/6

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Sketch an angle θ in standard position and find the values of the six trig functions for the angle.
(-4,3)
***
Given coordinates places θ in Q2 where sin>0, cos<0, tan<0.
We are working with a (3-4-5) reference right triangle.
sin θ=3/5
cos θ=-4/5
tan θ=-3/4
csc θ=5/3
sec θ=-5/4
cot θ=-4/3

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The diagonal of a square is 56 centimeters long. Find the perimeter of the square to the nearest tenth.
***
let x=side of the square
The diagonal and two of the sides form a right triangle
By the pythagorean theorem,

perimeter of the square=4x=158.4 cm

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Angular velocity : a trucker drives 55 miles per hour. His truck's tires have a diameter of 26 inches. What is the angular velocity of the wheels in revolutions per second.
***
1 revolution(rev)=circumference of truck's tires=π*diameter=π*26 inches

cancel out:mi,hr,ft,inches

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if sin s = -1/4 and s is in quadrant 4 and if cos t = -2/5 and t is in quadrant 2, find the following:
cos(s + t)
sin(s - t)
***
let O=opposite side of reference right triangle
let A=adjacent side of reference right triangle
let H=hypotenuse of reference right triangle
sin s=-1/4 (in Q4)=O/H
O=-1, H=4


..
cos t = -2/5(in Q2)=A/H
A=-2, H=5


..
cos(s + t)=cos s cos t-sin s sin t
=√15/4*-2/5-(-1/4)*√21)/5
=-2√15/20+√21/20
=(√21-2√15)/(20)
..
sin(s-t)=sin s cos t-cos s sin t
=-1/4*-2/5-√15/4*√21/5
=2/20-√315/20
=(2-√315)/20
..
Check with calculator:
sin(s)=-1/4(in Q4)
s≈345.52º
cos(t)=-2/5 (in Q2)
t≈113.58
..
s+t≈459.1
reference angle=80.9º(in Q2 where cos<0)
Cos(s+t)=cos(99.1)≈-0.1581..
(√21-2√15)/20≈-0.1581..
..
(s-t)≈231.94
reference angle≈51.94 (in Q3 where sin<0)
sin(s-t)=sin(51.94)≈-0.7874...
(2-√315)/20≈-0.7874...

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A pulley with a radius of 8 inches rotates three times every five seconds. Find the angular velocity of the pulley in radians/sec (round to the nearest hundredth). Find the linear velocity to the nearst ft/hr.
***
let C=circumference of 8 inch pulley


cross out: rev, sec, inches

..

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if cotx + cosecx = 1.5, then prove that cosx = 5/13
let O=opposite side
let A=adjacent side
let H=hypotenuse
..
cosx=5/13=A/H
A=5, H=13




This proves that if cotx+cscx=1.5, cosx = 5/13, because both equations come from the same 5-12-13 right triangle

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take sqrt of both sides



x=30º, 330º (in Q1 and Q4 where cos>0)