Lines that are parallel have the same slopes. The slope is the number in front of x, so the slope of y = 3x-4 is 3. So our new line's slope will also be 3
Then we can use the point-slope formula for a line:
y - y1 = m(x-x1) where m is the slope and (x1,y1)...

f = number of $5 bills
t = number of $10 bills
f + t = 15
5f + 10t = 130
t = 15-f
5f + 10(15-f) = 130
5f + 150 - 10f = 130
-5f = -20
f = 4
t = 11
There are 11 $10 bills and 4 $5 bills.

x = amount invested at 6%
y = amount invested at 9%
x+y = 25,000
interest income from 6% investment: .06x
interest income from 9% investment: .09y
total interest income = 1,860 = .06x + .09y
two equations, two variables
x+y=25000...

A = P(1+r/n)^(nt)
A = new amount
P = principal (original amount)
r = annual interest rate (as a decimal)
n = number of compounding periods / year
t = number of years
P = 1000
r = .06
n = 4 (compounding quarterly)
t = 1/4 (3 months = 1/4...

(x-1)-2 = 25
(x-1)-2 is the same as 1/(x-1)2
1/(x-1)2=25
1=25(x-1)2
1/25 = (x-1)2
1/5 = √(x-1)2
+ or - 1/5 = x-1
x-1 = 1/5 or x-1 = -1/5
x = 1 1/5 or x = 4/5

The problem does not mention compounding interest, so we can assume simple interest.
Interest = Principal * Rate * Time
12%: I12=x*.12*1=.12x
6%: I6=(6300-x)*.06*1=378-.06x
Total Yield = I12+I6=660
.12x+378-.06x=660...

False, that is called the Poisson distribution

Use an exponential decay equation:
A(t) = Ao e^(-kt) where Ao is the starting amount
at time t = 3 hours, we have half of what we started with, so 0.5Ao:
.5Ao = Aoe^(-k*3)
.5=e^(-3k)
ln(.5)=-3k
k=ln(.5)/-3
k=.231
A(t) =...

sin2x = sinx
use the identity: sin2x=2sinxcosx
2sinxcosx = sinx
2sinxcosx - sinx = 0
sinx(2cosx-1) = 0
sinx = 0 OR 2cosx-1=0
sin x = 0 when x is an integer multiple of pi, or x = npi (n = 0, +-1,...

5x2y - 3xy2 = 14
(2,1):
5(2)2(1) - 3(2)(1)2 = 14
5(4)(1) - 3(2)(1) = 14
20 - 6 = 14
14 = 14
d/dx (5x2y - 3xy2 = 14)
5x2(dy/dx) + 10xy - [3x(2y dy/dx) + 3y2] = 0
5x2(dy/dx) + 10xy - 6xy(dy/dx) - 3y2...

We can factor 9-x on the bottom as (3-√x)(3+√x).
Then we can cancel out the 3-√x s. We are left with 1/(3+√x). Now if we plug in 9 for x. We get a value of 1/6, so it is continuous at x=9.

We can rewrite the term of the infinite series like this:
2 * (4/5)^n
Whenever you have a number raised to n like this [r^n], it's a geometric series. A geometric series converges when the absolute value of the number being raised to a power is less than 1. It...

To find max and min, we need the first derivative.
f(x) = xsqrt(x2+4) = x(x2+4)1/2
f ' (x) = x*1/2(x2+4)-1/2 * 2x + (x2+4)1/2 * 1 = x2/sqrt(x2+4) + sqrt(x^2+4)
= x2/sqrt(x2+4) + (x2+4)/sqrt(x2+4) =...

Formula for sum of geometric series:
Sum = a/(1-r)
where a is the first term and r is what we multiply by each time.
a = 32
To find r: -8/32 = -1/4; 2/-8 = -1/4. So r = -1/4
Sum = 32/(1-(-1/4)) = 32/(1+1/4) = 32/(5/4)...

For interest compounding continuously, we need this formula:
A = Pert
A is Amount at some time t
P is the initial amount
r is the interest rate (as a decimal)
t is number of years
For our problem, we would plug in:
A...

The problem discusses two conditions about the field: the area and the length of fencing. First, the entire field has an area of 1,000,000 ft^2. Since we don't know the dimensions yet, let's call those x and y. Since it's a rectangle, the area equation would be:
1,000,000...

You're on the right track
If you extend the height and slant height, we have a "missing" cone. Let's call the missing cone's slant height x.
So now we can think of a proportion: smaller missing cone to bigger overall cone (actual + missing)
small radius/big radius...

Right and left behavior refer to what graph approaches as x goes to infinity (right) and negative infinity (left)
if the function is 5 minus 7/(2x-3x^2), then when x gets big positive or big negative 7/(2x-3x^2) goes to 0
and we are just left with 5. So the graph approaches y = 5 (as...

This polynomial is prime

when you have a limit where the base and exponent both have a variable, we need to use a log, then L'hospitals Rule
so first let's do ln of our limit to get x out of the exponent
ln [lim x->oo (x/1+x)^x] = x * ln[lim x->oo x/1+x] = lim x->oo[x*ln(x/1+x)]
then we...