Answer:
answer is in explanation
Step-by-step explanation:
Answer:
1) -1/4
2) -1/2
3) 1/3
Step-by-step explanation:
perpendicular lines' slopes are opposite reciprocals so you flip it and change the positive or negative sign
parallel lines have the same slope
Answer: 12
Step-by-step explanation:
2x + 8 = 3x - 4
In order to solve for x, we must isolate x. We can do this by moving all of the numbers with "x" in it to the left side of the equal side, and move everything else to the right of it!
Let's start off by subtracting 8 from both sides. Remember : what you do to one side, you must do it to the other.
2x + 8 - 8 = 3x - 4 - 8
Simplify!
2x = 3x - 12
Now, let's subtract 3x from both sides.
2x - 3x = 3x - 12 - 3x
Simplify!
-x = -12
Divide both sides by -1.
-x ÷ -1 = -12 ÷ -1
Simplify.
x = 12
Answer:
x² + 2x + [3\x - 1]
Step-by-step explanation:
Since the divisor is in the form of <em>x - c</em>, use what is called <em>Synthetic Division</em>. Remember, in this formula, -c gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:
1| 1 1 -2 3
↓ 1 2 0
------------------
1 2 0 3 → x² + 2x + [3\x - 1]
You start by placing the <em>c</em> in the top left corner, then list all the coefficients of your dividend [x² + 5x - 36]. You bring down the original term closest to <em>c</em> then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder, which in this case is a 3, so what you is set the divisor underneath the remainder of 3. Finally, your quotient is one degree less than your dividend, so that 1 in your quotient can be an x², 2 becomes <em>2x</em><em>,</em><em> </em>and the remainder of 3 is set over the divisor, giving you the other factor of <em>x² + 2x + [3\x - 1]</em>.
If you are ever in need of assistance, do not hesitate to let me know by subscribing to my channel [USERNAME: MATHEMATICS WIZARD], and as always, I am joyous to assist anyone at any time.