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From 8x + 30 > 4x + 38,
we subtract 4x from both sides to get
4x + 30 > 38
then we subtract 30 from both sides and get
4x > 8
now divide both sides by four to get
x > 2

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For a problem like this, we multiply every term in the first polynomial by every term in the second, thus we should get nine terms when we're done...like this:
(a^2 + 2ab - b^2)(a^2 - 7ab + b^2)=
a^4 - 7a^3b + a^2b^2
+2a^3b - 14a^2b^2 + 2ab^3
-a^2b^2 + 7ab^3 - b^4
Now we combine like terms to get
a^4 - 5a^3b - 14a^2b^2 + 9ab^3 - b^4

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Okay, we have
a1 = 4
a2 = 3(4) - 5 = 7
a3 = 3(7) - 5 = 16
a4 = 3(16) - 5 = 43

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Take the ln of both sides and get
13^x = 127
ln(13^x) = ln(127)
x * ln(13) = ln(127)
x = ln(127) / ln(13)

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In general, the nth term of an arithmetic sequence is given by
a(n) = a(1) + (n-1)d
In our given sequence, a(1) = 3, n = 17, and d = 6 (the common difference)...thus we get
a(n) = 3 + (n-1)(6)
a(17) = 3 + (17-1)(6)
a(17) = 3 + 96 = 99

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We are given the formula for the volume
V = 1/3 * s^2 * h
Now this formula gives us the volume in terms of BOTH the side and the height.
If we know the height is equal to half the side, we can write
h = (1/2)s
That allows us to substitute (1/2)s in for h in the original formula so that we have the volume in terms of ONLY the side...thus we get
V = 1/3 * s^2 * h
V = 1/3 * s^2 * (1/2)s
V = (1/6)s^3

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Okay from
4 + 6 ln x = 16
we get
6 ln x = 12
ln x = 2
x = e^2

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Okay from
x^2 + x - 20 = 0
we get
x = (-1 ± sqrt(1 - (-80))) / 2
x = (-1 ± sqrt(81)) / 2
x = (-1 ± 9) / 2
x = -5 or 4

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The sum of any infinite geometric series is given by
S = a1 / (1-r)
where a1 is the first term and r is the common ratio...
Thus in
1 + 1/5 + 1/25 + 1/125 + ...,
1 is our first term and 1/5 is our ratio...thus the sum is
s = 1 / (1-(1/5)) = 1 / (4/5) = 5/4

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Let's look for the greatest common factor first and try to factor what's left...we get
12x^4 y^3 + 16x^3 y^3 - 2*x^3 y^4 =
2x^3y^3(6x + 8 - y)
Since we cannot factor what's left, we're done...

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Okay, using the notation C(n,r) for binomial coefficients, we get
(2x + 1)^4 =
C(4,0)*(2x)^4 + C(4,1)*(2x)^3*1 + C(4,2)*(2x)^2*1^2 + C(4,3)(2x)*1^3 + C(4,4)*1^4 =
16x^4 + 32x^3 + 24x^2 + 8x + 1

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If the price is given by
p = 2x^2 - 100
then the revenue generated by the sale of x items is just p times x, or
Revenue = px = 2x^3 - 100x

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The "intercept" form of a line is not often well-taught...it is
x/a + y/b = 1
where a and b are the x- and y- intercepts, respectively.
Thus we get
x/5 + y/7 = 1
To put this in standard form, we multiply by 35 and get
7x + 5y = 35

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Okay, we exponentiate both sides...as in "3 to the"...since it is to the base three...we get
log(bottom side 3) 1/81 = x
1/81 = 3^x
1/(3^4) = 3^x
since 81 = 3^4 and then
3^(-4) = 3^x
x = -4

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Any two lines which have slopes which are NEGATIVE RECIPROCALS of each other are perpendicular...
Thus
y = (3/5)x + 6 and
y = (-5/3)x - 2
are perpendicular...
so are
y = -7x + 9 and
y = (1/7)x - 3

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Okay, using the laws of exponents, we add them when we multiply like bases and subtract them when we divide like bases, so we get
(x^(2/5) * x^(4/5)) / x^(3/5) =
x^(6/5) / x^(3/5) =
x^(3/5)

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Sure...think of it this way...
If you have three quarters, how much money have you? Well, 75 cents.
If you have ten quarters, how much money have you? Well, 250 cents.
If you have q quarters, how much money have you? Well, 25q cents.
In every case you multiply the number of quarters times 25 to get its total value.
The same is true for 5 cents per nickel...so....
The value of q quarters and q - 5 nickels is $3.05 becomes
25q + 5(q-5) = 305
Notice how it's 305 cents...we could also write this as
.25q + ,05(q-5) = 3.05 if we are talking dollars...
The equations are equivalent...

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Okay, using the laws of exponents...we multiply when raising a power to another power and get
(81k^4m^-8)^1/4 =
81^(1/4) * k * m^(-2) =
3k / m^2

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Yes, you are almost there...notice from
x(x-4)+2y(x-4)
there is a common factor in each term...that is, the (x-4)...
you can pull it out in front just like any other factor and get
(x-4) times what's left from each term...so we get
(x-4)(x+2y)

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Okay from
log9 (x-3) + log9 (x-3)= 1
we combine to get
2*log9 (x-3) = 1
log9 [(x-3)^2] = 1
now exponentiate "nine to the..." and get
[(x-3)^2] = 9
x - 3 = ±3
and x = 6 or x = 0
but x cannot be zero, so
x = 6

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Just posted this solution...

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Okay, we FOIL it and get
(4-2i)^2 =
(4 - 2i)(4 - 2i) =
16 - 8i - 8i + 4i^2 =
16 -16i - 4 =
12 - 16i

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Here you must pull out the i's first...so you get
sqrt -81 multiply by sqrt-49 =
i*sqrt(81) * i*sqrt(49) =
9i * 7i =
-63

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Okay first we square both sides...we get
sqrt( y^2-3y+18 ) = y + 2
y^2 - 3y + 18 = y^2 + 4y + 4
now collect terms and solve...we get
-3y + 18 = 4y + 4
14 = 7y
y = 2
you have to check these by plugging 2 in for y in the original...turns out okay...

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Not quite...the setup is like this
x^2 + (x + 1)^2 = 85
(the sum of the squares)
Now we get
x^2 + x^2 + 2x + 1 = 85
2x^2 + 2x - 84 = 0
x^2 + x - 42 = 0
(x + 7)(x - 6) = 0
and
x = -7 or x = 6
but they must be positive, hence your numbers are
6 and 7

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Okay from
2x^2 + 4x + 1 = 0
the first thing we do is divide by 2 and get
x^2 + 2x + 1/2 = 0
then complete the square
x^2 + 2x + 1 = 1/2
(x + 1)^2 = 1/2
now take the root
x + 1 = ± sqrt(1/2)
x = -1 ± sqrt(1/2)
x = -1 ± [sqrt(2) / 2]

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If you multiply the top equation by minus two, you go from
x + 5y =10
-2x -10y =-20
to
-2x -10y =-20
-2x -10y =-20
You can see that these are the very same line.
Thus the system is dependent and there ar an infinite number of solutions.

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The slope intercept form is
y - y1 = m(x - x1)
and we get
y - 2 = (3/4)(x + 8)
y - 2 = (3/4)x + 6
y = (3/4)x + 8

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Okay from
x^2 + 4x + 4 = 7
subtract 7 and get
x^2 + 4x - 3 = 0
now apply the quadratic formula and get
x = (-4 ± sqrt(16 + 12)) / 2
x = (-4 ± sqrt(28)) / 2
x = (-4 ± 2*sqrt(7)) / 2
x = -2 ± sqrt(7)

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Okay, from
(5t-2)^2 = 12
take the root of both sides
5t - 2 = ±sqrt(12)
5t = 2 ± sqrt(12)
t = (2 ± 2*sqrt(3)) / 5

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Okay from
t^2 + 6t + 10 = 0
we use the quadratic formula and get
t = (-6 ± sqrt(36 - 40)) / 2
t = (-6 ± sqrt(-4)) / 2
t = (-6 ± 2i) / 2
t = -3 ± i
Technically speaking, these are complex solutions, not purely imaginary...