Math Student Help

I took algebra 1 last year, and this year I borrowed an algebra 2 textbook from my teacher to better prepare for next year. However, after flipping through the whole book, I found it to be the same as algebra 1. Is algebra 2 just reviewing algebra 1? I though there would be many new things to learn, but it seems like everything in algebra 2 I already learned last year.

Algebra 1, as I understand it, is like "Basic (High school / advanced) Algebra. Algebra 2 is the start of College Algebra. You will DEFINITELY notice a difference as the course goes on.

I understand that you need to create equations in computer science, but what I’m wondering is are these just basic equations that you learn in beginning algebra? For example, in algebra there was problem solving that got progresseively harder like some using quadradic formulas and such. Do you need to create the more difficult equations from word problems or is it more basic? Thanks.

There is a branch called Boolean Algebra

rewrite each expression so that all exponents are positive.

1. 2x^-3 2. 2v^-2w^-3 3. x^2y^-3/z^-5

The important thing to remember about negative exponents is that they mean to move the number or variable to the OTHER SIDE of the dividing line. Once it crosses over, the sign changes.

1. 2x^-3
= (2 x^-3 ) / 1
= 2 / x^3

2. 2v^-2w^-3
= 2(v^-2)(w^-3) / 1
= 2 / (v^2w^3)

3. x^2y^-3/z^-5
= x^2(y^-3) / (z^-5)
= x^2 / (y^3)(z^-5)
= (x^2)(z^5) / (y^3)

9 - s^2 / 3rs
divided by
s^2 + 2s - 15 / 5rs

[(9 - s^2) / 3rs] / [(s^2 + 2s - 15) / 5rs]
= [ (9 - s^2) / 3rs ] * [ 5rs / (s^2 + 2s - 15) ]
= [ [(3-s)(3+s)] / 3rs ] * [ 5rs / [(s+5)(s-3)] ]
*canceling factors that are present in both numerator and denominator, we get:
[ [(3-s)(3+s)] / 3 ] * [ 5 / [(s+5)(s-3)] ]
but 3-s = -1(s-3), so
[ [(-1)(s-3)(3+s)] / 3 ] * [ 5 / [(s+5)(s-3)] ]
*canceling s-3,
[ [(-1)(3+s)] / 3 ] * [ 5 / [(s+5)] ]
= [(-5)(3+s)] / [(3)(s+5)]

hey guys im stuck on these 4 algebra questions i need some help so if you can please help me. 10points and best answer

it says to find the first, fourth, and eighth terms of each sequence.

A(n)=2*3^n-1

A(n)=3*4^n-1

A(n)=3*2^n-1

A(n)=-1*5^n-1
yea i know they are lol that why im asking for help.

*=mulitplication

^= to that power i guess

Whichever term you want to find, you substitute that number for n. (If it’s the first term n = 1, fourth term n = 4 and so on)

I’ll do the first sequence for you and then I’m sure you’ll be able to do the others.

A(n) = 2*3^n-1:

A(1) = 2*3^1-1
A(1) = 2*3^0
A(1) = 2*1 = 2

A(4) = 2*3^4-1
A(4) = 2*3^3
A(4) = 2*9 = 18

A(8) = 2*3^8-1
A(8) = 2*3^7
A(8) = 2*2187 = 4374

Access full lesson containing this video at: http://www.yourteacher.com/algebra1/addingandsubtractingrationalexpressions.php Students learn that when adding rational expressions, the first step is to factor each of the denominators, if possible, then give each term a common denominator by multiplying the numerator and denominator of each term by the appropriate value. Next, add across the numerators and keep the denominators the same. Finally, factor the resulting numerator, if possible, to determine if the rational expression can be reduced.

Duration : 0:3:11

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I understand that you need to create equations in computer science, but what I’m wondering is are these just basic equations that you learn in beginning algebra? For example, in algebra there was problem solving that got progresseively harder like some using quadradic formulas and such. Do you need to create the more difficult equations from word problems or is it more basic? Thanks.

Math in general is very important in computer science and algebra is pretty much important too especially in the security field. How advanced? I think it is a various matter but as for my college they are teaching Geometric/Linear Algebra for CS students..

1. 2x^3 - 5x^2 - 6x + 4 = 0
2. x^4 - 5x^3 + 5x^2 + 5x - 6
3. l x^2 - 5 l = 4
4. A purse contains $5.90 in nickels, dimes and quarters. The numbers of nickels is twice as many as the dimes, and there are two more quarters than dimes. How many coins of each type are in the purse? (Nickel = 5 cents, dime = 10 cents, quarter = 25 cents)

3.
|x^2 - 5| = 4
x^2 - 5 = +/-4
x^2 - 5 = 4
x^2 = 9
x = +/- 3

x^2 - 5 = -4
x^2 = 1
x = +/- 1

Access full lesson containing this video at: http://www.yourteacher.com/prealgebra/leastcommondenominator.php Students learn to add unlike fractions by first finding a common denominator. For example, to add 1/2 + 1/3, first find the common denominator of 2 and 3, or the Least Common Multiple of 2 and 3, which is 6. Then multiply the numerator and denominator of 1/2 by 3, to get the equivalent fraction 3/6, and multiply the numerator and denominator of 1/3 by 2, to get the equivalent fraction 2/6. Next, add the like fractions, 3/6 + 2/6, to get 5/6. So 1/2 + 1/3 = 5/6. Students also learn to subtract unlike fractions by following the same process that is used when adding unlike fractions.

Duration : 0:1:18

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Algebra Word Problem: Work. This lesson is appropriate for any algebra 1 or Algebra 2 student

Duration : 0:4:8

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