N(t) = 211e0.0109t
For 1980, t=0 and N(t) = 211. This is the population in 1980. We want to know when the population will be twice as big, so we want N(t) = 2*211. Let's plug that into the population equation and solve for t:
2*211 = 211e0.0109t
2...

Cos and cos-1 are inverses of one another, so cos(cos-1(4/5)) = 4/5
Same for sin and sin-1.
(BTW, it's "pi" not "pie")

2log5(A) - log5(B) + (1/3)log5(C) - 5log5(D)
Log property: b*log(a) = log(ab)
log5(A2) - log5(B) + log5(C1/3) - log5(D5)
Rearranging:
log5(A2) + log5(C1/3) - (log5(B) + log5(D5))
Log...

f(x)= √x
g(x)= √(x-2) - 3
The -2 in √(x-2) translates the graph 2 units to the right
The -3 at the end shifts the graph down 3 units

0.45a - 0.35a - 0.9
Combine like terms:
0.1a - 0.9
That's the most you can do with it.

log16 √(x+1) = 1/4
log16 (x+1)1/2 = 1/4
(1/2)*log16 (x+1) = 1/4
log16 (x+1) = 1/2
161/2 = (x+1)
4 = x + 1
3 = x

P(t) = P0*(1 - r)t
P(t) = Population at time t = 80% of P0 = 0.8P0
P0 = initial population = 10,000
r = rate of decrease = 3% = 0.03
t = years
P(t) = P0*(1 - r)t
0.8*(10,000) = (10,000)*(1 - 0.03)t
0.8 = (0.97)t
log(0.8) = log(0.97t)
log(0.8) = t*log(0...

n² + 6n + 9 = 7
n² + 6n = -2
Apply c2 = (b/2)2, where b=6. Since we add (6/2)2 to the right side of the equation, we need to add it to the right side as well to keep things equal.
n² + 6n + (6/2)2 = -2 + (6/2)2
n2 + 6n +...

(1-2y)/(y+3) = x
1-2y = x(y+3)
1-2y = xy + 3x
1 = xy + 3x + 2y
1 - 3x = xy + 2y
1 - 3x = y(x + 2)
(1-3x)/(x+2) = y

Let S equal your salary before the COLA raise:
S + (0.021)*S = $32,850
S*(1 + 0.021) = $32,850
S*(1.021) = $32,850
S = $32,851/1.021
Can you finish?

x4 + 29x2 + 100 = 0
This is "quadratic-like" polynomial. Let u = x2, then u2 = x2*x2 = x4:
x4 + 29x2 + 100 = 0
u2 + 29u + 100 = 0
Factors to:
...

Complex roots always come in conjugate pairs, so if 2-7i is root so is 2+7i. The two linear factors then would be:
(x - (2-7i))*(x - (2+7i))

I'm not sure exactly what you're asking, but the percent of respondents is:
Part Part%
---------- = ----------- where the Part = 250 and the Whole = 5750
Whole ...

Not sure if it's 70 years or 710 years - your question lists both. I've provided the answer for 710 years. If it's 70 years, replace the 710 with 70.
A(t) = A0*(1/2)t/710 where:
A(t) = the amount left in t years
A0 = the initial amount = 30 g
t...

Well if it's a 3/4" screw and one part is 3/4", then the "other two" parts are zero. I'm guessing you meant one part is 3/16 of an inch, rather than 3/4 because it's the only way to get 9/16 as the answer for the sum of the other two parts:
3/4" - 3/16"...

P(t) = P0ert where:
P(t) = the value of the investment at time t = $46,800
P0 = the initial investment = $2000
r = the interest rate expressed as a decimal = 10.1% = 0.101
t = time = what you're trying to find
P(t) = P0ert
$46,800 = $2000e0.101t
46,800/2000...

P=P0ert
Where:
P = value after 3 years = $11.979
P0 = initial value = $21,005
r = depreciation rate ( decimal form)
t= years = 3
11,979 = (21,005)e3r
11,979/21,005 = e3r
...

The three angles add up to 180o. So add them up and set them equal to 180:
2√(2x + 5 √(2x) + 2√(2x) = 180
9√(2x) = 180
√(2x) = 20
2x = 202
...

P(h) = P0*(1 - r)h/1000
Where:
P(h) is the pressure at altitude h
P0 = the pressure at sea level (h=0 m) = 9898 kp
r = rate of decrease = 11.7% (?)
h = altitude above sea level = 3000 m (?)
Plug the numbers into the equation and use your calculator to...

There is a lot of information missing from the problem statement. I am going to assume:
The cannon is aimed horizontally (no vertical component)
The John falls from 4000 feet to sea level.
With the first assumption, the speed with which John is moving as he...