Answer:
Yes if Angle 7 and angle 6 are linear pairs in angles created by a straight line that cuts through parallel lines.
Step-by-step explanation:
Remember that two parallel lines that are cut by a transveral will create 8 angles, which will be similar to each other. Making four pairs of congruent angles means that they will be exactly the same angle located in different parts of the system. So if angles 6 and 7 are congruent, and the lines are cut by a transversal, then the lines are parallel.
This may look a little confusing but all you have to do is plug the equation given for h(x) which is 2x-5 in to the h(x) area in the equation where it says h(x) + g(x).
So far that’s (2x-5) + g(x)
Then,
Do the same with equation given for g(x) which is now,
(2x-5) + (3x+1)
Then solve,
2x - 5 + 3x + 1
2x + 3x- 5 + 1
5x - 4
Therefore the answer is 5x - 4.
8 units that Is the answer
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
