8GB of built-in storage—enough for 3 VHS movies or 20 LPs; microSD card slot for additional storage; 3” display to preview your content; Built-in speaker and headphone jack; USB port to transfer recorded files to computer; Output video to TV or projector using HDMI or AV Output; Portable size: DWH 2.4 x 4.9 x 0.8” (61 x 125 x 20mm). Java SE 9 Archive Downloads. Go to the Oracle Java Archive page. The JDK is a development environment for building applications using the Java programming language. The JDK includes tools useful for developing and testing programs written in the Java programming language and running on the Java TM platform.
Step 1 :
Checking for a perfect cube :
1.1 x3-x2-9x+9 is not a perfect cube
Trying to factor by pulling out :
1.2 Factoring: x3-x2-9x+9 Thoughtfully split the expression at hand into groups, each group having two terms : Group 1: x3-x2 Group 2: -9x+9 Pull out from each group separately : Group 1: (x-1) • (x2) Group 2: (x-1) • (-9) ------------------- Add up the two groups : (x-1) • (x2-9) Which is the desired factorization Memorytamer 1 4 4 – automatic memory freeing app.
Trying to factor as a Difference of Squares :
1.3 Factoring: x2-9 Theory : A difference of two perfect squares, A2 - B2can be factored into (A+B) • (A-B) Proof : (A+B) • (A-B) = A2 - AB + BA - B2 = A2- AB + AB - B2 = A2 - B2 Note : AB = BA is the commutative property of multiplication. Note : - AB + AB equals zero and is therefore eliminated from the expression. Check : 9 is the square of 3 Check : x2is the square of x1 Factorization is : (x + 3) • (x - 3)
Equation at the end of step 1 :
Step 2 :
Theory - Roots of a product :
2.1 A product of several terms equals zero. When a product of two or more terms equals zero, then at least one of the terms must be zero. We shall now solve each term = 0 separately In other words, we are going to solve as many equations as there are terms in the product Any solution of term = 0 solves product = 0 as well.
Archiver 3 0 9 X 284
Solving a Single Variable Equation :
2.2 Solve : x+3 = 0 Subtract 3 from both sides of the equation : x = -3
Solving a Single Variable Equation :
2.3 Solve : x-3 = 0 Add 3 to both sides of the equation : x = 3
Solving a Single Variable Equation :
2.4 Solve : x-1 = 0 Add 1 to both sides of the equation : x = 1
Archiver 3 0 9 X 2 6
Three solutions were found :
x = 1
x = 3
x = -3
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