If u wanna go by a technical term then yes, it can. The positive and negative side of a number. 4 example, -5. It's absolute value is 5. Positive 5's absolute value is 5 as well.
Now if u don't wanna be technical and don't want to use negative numbers, then no, it cannot.
Answer:
oh dang.. imma head out, no big brain time for me..
Step-by-step explanation:
<span>Let R=A AND (NOT B)] OR (NOT B)
If A=T, B=T
then ~B=F,
A and ~B = F
So R=F
If A=T,
B=F
~B=T
A and ~B=T
so (A and ~B) or ~B = T
etc.</span>
Answer:
C. Kalena made a mistake in Step 3. The justification should state: -x²
+ x²
Step-by-step explanation:
Given the function x(x - 1)(x + 1) = x3 - X
To justify kelena proof
We will need to show if the two equations are equal.
Starting from the RHS with function x³-x
First we will factor out the common factor which is 'x' to have;
x(x²-1)
Factorising x²-1 using the difference of two square will give;
x(x+1)(x-1)
Note that for two real number a and b, the expansion of a²-b² using difference vof two square will give;
a²-b² = (a+b)(a-b) hence;
Factorising x²-1 using the difference of two square will give;
x(x+1)(x-1)
Factorising x(x+1) gives x²+x, therefore
x(x+1)(x-1) = (x²+x)(x-1)
(x²+x)(x-1) = x³-x²+x²-x
The function x³-x²+x²-x gotten shows that kelena made a mistake in step 3, the justification should be -x²+x² not -x-x²
