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PRECAL STUDY GUIDE COMMENTS
1. When distributing the exponent, distribute to the coefficients as well as the variables.
2. Multiply everything by LCD 3. Have the Absolute Value isolated on one side, solve for the negative of the
other side and the positive 4. Perpendicular=negative inverse. Plug given point into point-‐slope
(y-‐y1)=m(x-‐x1) 5. F-‐1 Switch x and y, solve for Y 6. DISTANCE FORMULA: √ {(x2 – x1) ² + (y2 – y1) ²}
7. MIDPOINT FORMULA: ( x1 + x22
, y1 + y22
)
8. Set y to 0, solve for x. 9. Vertex from Vertex Form ( y = a(x − h)2 + k ) is (h,k) 10. Graph it to find the limits on the factors of C (constant term) to test. Then, put
it outside, coefficients inside, bring down the first one, multiply by the constant thingy outside, then add that to the next coefficient, repeat until you’re done. Then, magic. Or scary stuff. I’ll just get this wrong.
11. If Degree of Numer. > Degree of denom., no horizontal asymptote If Degree of Numer. = Degree of denom, take fraction of coefficients, = asymptope If Degree of Numer. < Degree of Denom, Horizonal asymptope =0
12. Variation Direct Variation: y=kx // Inverse Variation: y = kx //
Combined Variation: Just throw em’ together, y = kxz
13. Equation of an Ellipse (w/vertex) (x − h)2
a2+(y− k)2
b2=1// c =distance from
vertex to foci. c2 + b2 = a2 So long as a > b. otherwise switch them.
14. EQUATION OF A HYPERBOLA: x2
a2−y2
b2=1OR x
2
a2−y2
b2=1
REMEMBER, with hyperbolas, free will is an illusion. You are trapped to rules. A always comes first. Minus signs end your choice. Minus is love. Minus is life. If the endpoints of transverse axis have the same absolute value for x (one is – one is +), the hyperbola opens sideways. If the number is the same for y, opens vertically. THEY ARE EQUAL TO A.
So you’ve got that. Now, if it opens horizontally, the asymptote is y = ± bax . If
it opens vertically, the asymptote is y = ± abx . That should give you your
equation. Bitch.
15. OH THANK FUCKING GOD THIS IS SHORTER. Ok, so, you've got this shit, and you’re going to want to get the clearly easier one on one side and have the rest of the shit on the other. STANDARD FORM FOR PARABOLAS: (y− k)2 = 4p(x − h) or (x − h)2 = 4p(y− k)This one will be easier with an example. y2 − 2y−10x + 21= 0 LOOK, y is already squared. So solve for that. y2 − 2y =10x − 21 put the useless shit on the other side, then COMPLETE THE SQUARE. y2-‐2y+1=10x-‐21+1 Then you get y2 − 2y+1=10x − 20 FACTOR (y−1)2 =10(x − 2) BAM ANSWER
16. REMEMBER THE PARABOLA FORM I TAUGHT YOU A MINUTE AGO? Well you need it again. Remember, VERTEX: (h,k). FOCUS: (h+p, k) DIRECTRIX: x=h-‐p
17. So in reality, I copied most of my work from someone else, so this is me translating. If you look at equation one, and acknowledge that x has a coefficient of 1, you can change the equation into parabola standard form, which means that this graphs into a parabola. Now, if you look at equation two, it's a circle. Both x and y are over 1, and they add up to equal one. So I guess, since parabola ≠ circle (that's some hardcore fucking math right there) there is no solution. This seems better than trying to work out substitution or some shit.
18. FUCK. Trig. Deep breaths. Ok, this is simple I guess. Remember the unit circle? It had 4(5?) main points that super mattered and were easy. They
were 0 = (1, 0) , π2= (0,1) , π = (−1,0) , 3π
2= (0,−1) , & 2π = (1, 0) Remember,
cos=x, and sin=y. So from this, we can simply plug in the t=, translate the trig function, and solve for (x,y). Not so bad after all. Trig is alright I guess
19. I lied. Trig sucks. Theres these identities or whatever, they’re no fun at all. This one requires a bit more effort. So apparently, tan2 t +1= sec2 t . This is essential to solving #19. But if you’ve got that, it's the simple matter of taking the original x= and y=, solving for the trig function of t, and then plugging them into the trig identity (Idk I think that's what theyre called. And since you have two squares and a 1, chances are this is going to be a circle (if you’re lucky), an ellipse, or a hyperbola. Have fun.
20. Same shit.
21. (i−1)!(i+ 2)!i=1
5
∑ Fuck this. Although I personally like summation. It’s tedious but it
makes sense. Look at the small numbers next to sigma. i=1, and 5. This means you plug in 1, 2, 3, 4, and 5 for i. Then, you find that, and do the !. Which quite simply means, take the number, and multiply it by all numbers before it. For example, 5!=5x4x3x2x1. Get it? Now, once you get all 5 answers, add them together (use a calculator finding a common denominator sucksss) And BAM answer. Also for some reason 0!=1 don't ask me.
22. Patricia has her shit together for this one. I haven’t started planning for my future at all. So we have what she starts with, we have what she adds each
year, and we are trying to find the amount she deposited in her 20th year and the total amount deposited in that time. So we use two equations. an = a1 + (n−1)dWill give us our first answer. D is the amount she adds each year, and n is the year #. Plug everything in and you’ll get your answer.
REMEMBER IT FOR THE NEXT PART. The next equation is sn =n2(a1 + an ) , or
at least I think it is, all I have is the equation with the numbers already plugged in. Anyhow again plug in the numbers, solve, get an answer.
23. 5(−0.2)i−1i=1
∞
∑ Infinity whaaaaaaaaaat. So my sources (the person I copied
from) Tell me that the magical thing in the parenthesis is R, a1 is just the
equation minus the sigma with i=1, and to solve this, I just use 51+ R
. WAIT
a1=5. So theres an equation. a11+ R
. Use that.
24.
OH SHIT THIS IS THE THING THAT USES THE WEIRD FRACTION WITHOUT A LINE. Now, we solve. This btw uses Binomial Theorem wow fancy big words. So, The r th term of the expansion
of (a+b)n is:
(stole this from the internet)
Your calculator can do the first part. Replace the variables with numbers, go to math, and press the nCr thingy (its under the PRB tab) Plug in n and r, and that gives you your coefficient. The rest is easy.
25. 2 salads. 7 Main courses. 3 desserts. Should the indecisiveness not kill everyone, how many choices are there, including individual items. Well, there are 12 individual items, then we can pair salads with main courses, salads with desserts, and main courses with desserts. Lets give those variables. x, y, and z, respectively. So we can make this equation: (coefficients_ added)+ (x• y)+ (x• z)+ (y• z) I’m sure there’s a simpler way, but I don't know it, and this works.
26. GIVEN A SEQUENCE… so much stuff, but whatever. Let me google how to do this then write it down in a simplified manner. Ok so in this example, theres 36 numbers, 5-‐40. So, what are the odds its greater than 10 and less than 29? 29-‐10=19. So 19/36. BUT THERES AN OR, meaning this is a probability with two events, so we have the other one, it being an odd number, which has off 18/36. How many numbers are both odd and between 10 and 26? Count on your fingers, it's a 30 question test and you have 90 minutes. You’ve got time.
There’s 9 in this problem btw. So, we have (1936
+1836)− 936
=2736
=34 Magic.
27.