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Find a value of theta in [0, 90](degrees) that satisfies the statement. Leave the answer in decimal degrees rounded to seven decimal places, if necessary.
cosine theta= 0.1879656.
I know the answer is 79.1659173, but I don't the steps to solve it. Can some help me please? Thank you
**
This is how to get the answer with a calculator:
set your calculator to 7 decimal places.
Use the inverse cos key and enter 0.1879656.
You will then see displayed the answer:79.1659173º

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Evaluate the expression.
cot^2(90)-sec(180) both in degrees.
The answer is 1 but I don't know how to solve this. Please help
**
cot(90)=cos(90)/sin(90)=0/1=0
cot^2(90)=0
..
-sec(180)=-1/cos(180)=-1/-1=1
..
cot^2(90)-sec(180)=0+1=1

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clara is one third her sister's age 4 years later clara will be 3less than half her her sister 's age what was clara s age 3 years ago
**
let x=clara's age 3 yrs ago
x+3=clara's present age
3(x+3)=sister's present age
4 years later:
(x+7=clara's age
(3(x+3)+4)=sister's age
3less than half of sister's age=(3(x+3)+4)/2-3
x+7=(3x+9+4)/2)-3
x+10=(3x+13)/2)
2x+20=3x+13
x=7
clara's age 3 yrs ago=7

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Juan is going to deposit $1,000 in an account that will earn him compund interest. The account pays 8% interest, compounded monthly. How much money will he have after 10 years.
**
compound interest formula: A=P(1+r)^n, P=initial investment, r=interest rate per period, A=amt after n periods.
For given problem:
P=1000
r=.08/12
n=10*12=120
..
A=1000(1+.08/12)^120=2219.64
How much money will Juan have after 10 years. $2219.64

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Solve the equation.
(x^3)(9^x) -9^x =0
9^x(x^3-1)=0
9^x≠0
x^3-1=0
x=1

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find vertex, focus, and equation of directrix of y^2 = 36x
This is an equation of a parabola that opens rightwards.
Its standard form of equation: (y-k)=4p(x-h), (h,k)=(x,y) coordinates of the vertex.
For given equation: y^2=36x
vertex:(0,0)
axis of symmetry: y=0 or x-axis
4p=36
p=9
focus: (9,0) (p-distance to the right of the vertex on the axis of symmetry)
directrix: x=9 (p-distance to the left of the vertex on the axis of symmetry)

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Find the standard form of the equation of the hyperbola with the given charateristics and center at the origin Foci:(+-10,0) Asymptotes:y=+-3x
**
This is a hyperbola with horizontal transverse axis.
Its standard form of equation: (x-h)^2/a^2-(y-k)^2/b^2=1
For given hyperbola:
center: (0,0)
slope of asymptotes=±3=b/a (for hyperbolas with horizontal transverse axis)
b=±3a
c=10 (distance from center to foci)
c^2=a^2+b^2=a^2+9a^2=10a^2
100=10a^2
a^2=10
a=√10
b^2=c^2-a^2=100-10=90
equation: x^2/10-y^2/90=1

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graph the ellipse using the proper algebra
x^2+(y^2/12)=1
This is an equation of an ellipse with vertical major axis.
Its standard form: , a>b, (h,k)=(x,y) coordinates of center.
For given equation: x^2+(y^2/12)=1
center: (0,0)
a^2=12
a=√12≈3.46
vertices: (0,0±a)=(0,0±√12)=(0,0±3.46)=(0,-3.46) and (0,3.46) (y-intercepts)
b^2=1
b=1
x-intercepts:(±1,0)=(-1,0) and (1,0)
see graph below as a visual check on the foregoing algebra:
y=±(12-12x^2)^.5

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Find the amplitude, period, and phase shift of the function, and graph one completed period:
y=1+COS (3X+pie/2)
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Equation for the cos function: y=A)cos(Bx-C), A=amplitude, period=2π/B, phase shift=C/B
For given equation: 1+cos(3x+π/2)
Amplitude=1
B=3
period: 2π/B=2π/3
1/4 period=2π/12=π/6
phase shift=C/B=(π/2)/3=π/6 (to the left)
..
Graphing:
I don't have the means to graph the function for you, but I can show you how to develop the coordinates with which you can draw the graph:
Start with the basic function: y=cos x with a period of 2π/3
coordinates: (0,1), (π/6,0), (π/3,-1), (π/2,0), (2π/3,1)
shift to left (π/6): (-π/6,1), (0,0), (π/6,-1), (π/3,0), (π/2,1)
bump up 1 unit: (-π/6,2), (0,1), (π/6,0), (π/3,1), (π/2,2) (final configuration

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Find sin x/2, cos x/2, and tan x/2
from the given information.
sec x = 10/9 ,270° < x < 360°
sin x/2=
cos x/2=
tan x/2=
**
sec x=10/9=hypotenuse/adjacent side
hypotenuse=10
adjacent side=9
opposite side=√(10^2-9^2)=√(100-81)=√19
sin x=-√19/10 (in quadrant IV where sin<0)
cos x=9/10
..
Identities:
sin x/2=±√[(1-cos x)/2]
choose positive root because x/2 is in quadrant II where sin>0
sin x/2=√[(1-cos x)/2]=√[(1-9/10/2]=√.1/2=√.05
..
cos x/2=±√[(1+cos x)/2]
choose negative root because x/2 is in quadrant II where cos<0
cos x/2=-√[(1+cos x)/2]=-√[(1+9/10/2]=-√1.9/2=-√.95
..
tan x/2=sin x/(1+cos x)=-(√19/10)/(1+9/10)=-(√19/10)/1.9
..
How to check answers with calculator:
sec x=10/9
cos x=9/10
cos^-1=(9/10)≈25.84º (reference angle in specified quadrant IV)
standard position of angle=360-25.84=334.16
x/2=334.16/2=167.08
reference angle=180-167.08=12.92º
..
sin x/2=sin 12.92≈0.2236..(in quadrant II where sin>0)
√.05=0.2236..
..
you can check cos x/2 and tan x/2 in the same way

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Can someone assist me to express log[2](A)+log[2](B)-3log[2](C)as a single logarithm.
log2[AB/C^3]

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Suppose that ln5 =S and ln11 =T. Use the properties of logarithm in terms of S and T
log5+log11=log[5*11]=S+T
log(55)=S+T

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Find the domain of the function.
f(x)= log8 (x+3)
domain: (-3,∞)

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g(x) = In (x-6) find the domain of the function.
domain: (6,∞)

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5log x - (1/3)log (x^2+1) + 4log (x-1)
log[(5(x-1)^4)/(x^2+1)^1/3]

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1.) Solve each equation.
a.) log7(2x)^2+log7 (3)=2
log7[(2x)^2*3]=2
convert to exponential form
7^2=4x^2*3=12x^2
12x^2=49
x^2=49/12
x=±√(49/12)≈±2.02
..
b.) log4 (8)+log4 (x^2-8)=8
log4[8*(x^2-8)]=8
4^8=8*(x^2-8)=8x^2-64
8x^2-64=65536
8x^2=65536+64=65600
x^2=8200
x≈±90.55
..
c.) log12 (-3a-8)+log12 (a+4)
log12[(-3a-8)(a+4)]=log12[(-3a^2-20a-32)]=log12[-(3x+8)(x+4)]
..
d.) ln-3k)=ln(-2k-4)=?
ln argument mssing

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The following problems give an expression for one of the three functions sin , cos, or tan ,with (theta) in the first quadrant. Find expressions for the other two functions. Your answers will be algebraic expressions of x.
(a) sin (theta) = 3/x
(b) x = 4 cos (theta)
(c) tan (theta) = x
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Use the following notation: O=opposite side, A=adjacent side, H=hypotenuse
..
(a) sin (theta) = 3/x=O/H
O=3
H=x
A=√(H^2-O^2)=√(x^2-3^2)=√(x^2-9)
cos (theta)=A/H=√(x^2-9)/x
tan (theta)=O/A=3/√(x^2-9)
..
b) x = 4 cos (theta)
cos (theta)=x/4=A/H
A=x
H=4
O=√(H^2-A^2)=√(4^2-x^2)=√(16-x^2)
sin(theta)=O/H=√(16-x^2)/4
tan (theta)=O/A=√(16-x^2)/x
(c) tan (theta) = x=O/A
O=x
A=1
H=√(O^2+A^2)=√(x^2+1^2)=√(x^2+1)
cos (theta)=A/H=1/√(x^2+1)
sin(theta)=O/H=x/√(x^2+1)

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solve: log base a of x + log base a of (x-2)= log base a of (x+4)
loga[x(x-2)]=loga(x+4)
x(x-2)=(x+4)
x^2-2x=x+4
x^2-3x-4=0
(x-4)(x+1)=0
x=4
or
x=-1 (reject, (x-2)>0)

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Solve log 2x + log 12 = 3. Round to the nearest hundredth if necessary.
log 2x + log 12 = 3
log(2x*12)=log(10^3)
2x*12=10^3=1000
2x=1000/12
x=1000/24
x≈41.67

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Write the expression as a single natural logarithm.
2 ln x – 5 ln c
ln[x^2/c^5]

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Solve the exponential equation.
1/16= 64^(4x–3)
1/4^2=4^3^(4x-3)
4^-2)=4^(12x-9)
12x-9=-2
12x=7
x=7/12

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Solve the logarithmic equation. Round to the nearest ten-thousandth if necessary.
3 log(2x) = 4
log(2x)=4/3
convert to exponential form
10^(4/3)=2x
2x≈21.5443
x≈21.5443/2
x≈10.7722

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Solve ln(5x + 7) = 8. Round to the nearest thousandth.
ln(5x + 7) = 8
convert to exponential form: base(e) raised to log of the number(8)=number(5x+7)
e^8=5x+7
5x=e^8-7
x=(e^8-7)/5
x≈594.792

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Use natural logarithms to solve the equation. Round to the nearest thousandth.
5e^(2x + 11)=30
divide by 5
e^(2x + 11)=6
(2x+11)lne=ln6
lne=1
2x+11=ln6
2x=ln6-11
x=(ln6-11)/2
x≈-4.604

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solve each equation for x by first rewriting both sides as powers of the same base.
32^x=64
(2^5)^x=2^6
2^5x=2^6
5x=6
x=6/5

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Whats (5x)^(-1/2)=10
(5x)^(-1/2)=10
1/(5x)^(1/2)=10
√(5x)=1/10
square both sides
5x=1/100
x=1/500

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Log(base 9)(x-6)+log (base 9)(x-6)=1
log9[(x-6)(x-6)]=log9(9)
x^2-12x+36=9
x^2-12x+25=0
solve for x by completing the square
x^2-12x+25=0
(x^2-12x+36)=-25+36
(x-6)^2=11
x-6=±√11
x=6±√11
x=6±3.317
x≈2.683(reject, (x-6)>0)
or
x≈9.317

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what is the relationship between cotangent theta and cotangent NEGATIVE theta?
cot(x)=-cot(-x) (this is called an odd function, compared to an even function)
Examples:
pick angle of 45º in quadrant I
this gives you a reference angle of 45º where cot 45º=1
-45º will also give you a reference angle of 45º but in quadrant IV where cot -45=-1
..
pick angle 135º in quadrant II
this gives you a reference angle of 45º where cot135º=-1
-135º will also give you a reference angle of 45º but in quadrant III where cot-135º=1
..
pick angle 225º in quadrant III
this gives you a reference angle of 45º where cot225º=1
-225º will also give you a reference angle of 45º but in quadrant II where cot-225º=-1
..
note: I used a reference angle of 45º for the examples but this will hold true for any reference angle.
This will also hold true for the tan function since cot is the reciprocal of tan.
A negative angle means rotating in a clockwise direction and a positive angle in a counter-clockwise direction.The value for the trig functions and their reciprocals is computed from the reference angle and the sign depends on which of the 4 quadrants the angle is in. See if you can apply this method to the sin and cos functions.
..
Hope this helps!

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simplify: -10 sin (2x+1) cos (2x+1)
use Identity:sin2x=2sinxcosx
-10 sin (2x+1) cos (2x+1)
=-5*2 sin (2x+1) cos (2x+1)
=-5sin[2(2x+1)]
=-5sin(4x+2)

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find all solutions of tanx + 1= 0, where 0 < (or equal to) X < (or equal to) pi
0 ≤ x ≤ π
tanx+1=0
tanx=-1
x=3π/4 (In quadrant II where tan<0)