Formulas:
c^2=a^2+b^2
Trig Identity:
(sin x)^2+(cos x)^2=1
Circles:
Circum.=2(pi)r
Area=(pi)r^2
Int(dx/x)=ln lxl+C
d(e^(ax))=a(e^(ax))
(a^x)(a^y)=a^(x+y)
sqrt(x)=x^(1/2)
(1/x)=x^(-1)
Eq's. of Lines:
y=mx+b
y-y"=m(x-x")
[m=slope,b=y-int,(x",y")=pt.on line]
ax^2+bx+c=0 (stand.form)
x=((-b)+/-sqrt(b^2-4ac))/2a
log (a^x)=xlog(a)
1+1=2
slope=rise/run
min/max happen where y'=0
Diff. of 2 Squares:
(a^2-b^2)=(a+b)(a-b)
1. Use the addition method to solve each of the following linear systems.
a) 3x-5y= -18
2x+7y= 19
For problems like these, you want to add the two equations together in such a way as to cancel out either the x or y terms. For this problem, I will get rid of the x terms by first multiplying the top row by (2) and the bottom row by (-3):
6x-10y= -36
+-6x-21y= -57
When you add, the 6x and the -6x cancel out, and you are left with:
-31y= -93 therefore, y=3. Now, to find the x that will satisfy both equations as well, you can plug y=3 into either equation to find x.
2. Use synthetic division to perform each division:
a) (2x^3 - x^2 - x - 6) / (x+1)
First, take the negative of the second number in the dividend. In other words, the (x+1) term (denominator), so in this case the negative is (-1), and place it in the corner. Then, starting with the highest powered term, list the coefficients in descending order, making sure to substitute zeros in for missing terms.
-1 | 2 -1 -1 -6
Then, bring down the 2.
-1 | 2 -1 -1 -6
+
2
Now, you take the bottom 2, multiply it by the -1, take the product you get and add that product to the next coeffient number, like this:
-1 | 2 -1 -1 -6
+ -2
2 -3
Keep doing this and you get:
-1 | 2 -1 -1 -6
+ -2 3 -2
2 -3 2 -8
So, then you take the bottom row and incorporate it into your answer. Starting a power lower than you started out with (in this case we started with a power of 3 (x^3)), so we start our answer with x^2:
2x^2 -3x +2 with -8/(x+1) being the remainder.
Also, remember the logarithmic rules:
logx (a/b)=logx a - logx b
log (a^x)=xlog a
logx (a)(b)=logx a + logx b
>
For this problem, you need to know that log A - log B = log (A/B) and that's all.
2.) Find y as a function of x; the constant C is a positive number.
> ln (y-1)+ ln(y+1)= (-x) + C
>
You need to solve for y here. You do this by:
1.) rewrite ln(y-1) + ln(y+1) as ln((y-1)(y+1)) (it's a rule)
2.) notice that y-1 times y+1 is the diff. of two squares and is equal to y^2 - 1^2 = y^2 - 1.
3.) So now you have: ln((y^2)-1)=(-x)+C
4.) to get rid of the ln, you have to use its inverse fxn: e (because ln has a base e) So, write each side of the equation as an exponent of e:
e^(ln((y^2))=e^[(-x)+C]
e raised to ln of anything is just that anything, so the e's and ln's cancel each other out. So, now you have: y^2-1=e^[(-x)+C] and from here, you can solve for y.