This is a blog for all of Mr. James' math classes at HTHS. This is for weekly updates for AP Calculus, Calculus, and Pre-AP Pre-Calculus. This blog is also used to discuss anything pertaining to these classes.
Friday, August 21, 2009
3rd Period Test 1
Everyone from 3rd period, this is where you need to comment with your solutions to the problems.
11 comments:
This is the solution from Kasey:
Solve:
x2-9+18
(x-6)(x-3)
x=6 x=3
that is my solution : )
Simplify:
1/(x+2) + 1/(x^2-4) - 2/(x^2-x-2)
(x+1)(x-2)/(x+2)(x-2)(x+1) + (x+1)/(x+2)(x-2)(x+1) - (2)(x+2)/(x+2)(x-2)(x+1)
(x+1)(x-2) + (x+1)-2(x+2)/(x+1)(x+2)(x-2)
x^2-x-2+x+1-2x-4/(x+1)(x-2)(x+2)
x^2-2x-5/(x+1)(x-2)(x+2)
Factor the expression completely:
x^6 - 1
= (x^3+1)(x^3-1)
28)
(a^-2)(b^- 4)(c^6)/
(2^-2)(a^ -6)(b^ 8)
4(a^6)(c^6)/
(a^2)(b^12)
4(a^4)(c^6)/
b^12
#2. State the property of real numbers being used.
(a+b)(a-b)=(a-b)(a+b)
...Commutative Property - When we add two numbers, order doesn't matter. When we multiply two numbers, order doesn't matter.
this is austin justice just so you know :)
find all real solutions of the equation:
1/x + 2/x-1 = 3
find a common denominator and apply what you did to the denominator to the numerator.
x^2-x/x(x-1) + 2x^2-2x/x(x-1) = 3x^2+3x/x(x-1)
then simplify the numerator
x(x-1)/x(x-1) + 2x(x-1)/x(x-1) = 3x(x-1)/x(x-1)
now simplify the fractions
1 + 2 = 3
3 = 3
0
Emily Moore, Period 3
\sqrt[4]{4}\sqrt[4]{324}
\sqrt[4]{4}\sqrt[4]{(4)(9)(9)}
\sqrt[4]{(4)(4)(9)(9)}
\sqrt[4]{(16)(81)}
(2)(3)
6 is the solution
Emily Moore
Sorry mine looks weird! Couldn't figure out how to add the pic....the directions for my problem are "evaluate the expression"
Brittany and Zach, both of your problems will factor out smaller.
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