Answer: A
Step-by-step explanation:
The first thing we need to do is to make sure our equation is in standard form. The given equation is not in standard form.
3x²-2x=0
Now that the equation is in standard form, we can find our A, B, C to see which quadratic equation is correct.
A=3
B=-2
C=0
We can automatically eliminate D because the first value is -3 when it is supposed to be -(-2). Also, the denominator is supposed to be 2a. We know that A=3. The denominator should be 2(3), not 2(-2).
Next, we can eliminate B and C. For the 4ac, we know it has to be 4(3)(0) because A=3 and C=0. B and C have given 4(3)(2) and 4(3)(1), respectively. This does not match A=3 and C=0.
Therefore, A is the correct answer.
Y-x =17
y = -x + 17
substitute this into the second equation
y = 4x + 2
-x + 17 = 4x + 2
0= 5x -15
15/5 = 5x/5
3 = x
Answer:
Your teacher is right, there is not enough info
Step-by-step explanation:
<h3>Question 1</h3>
We can see that RS is divided by half
The PQ is not indicated as perpendicular to RS or RQ is not indicates same as QS
So P is not on the perpendicular bisector of RS
<h3>Question 2</h3>
We can see that PD⊥DE and PF⊥FE
There is no indication that PD = PF or ∠DEP ≅ FEP
So PE is not the angle bisector of ∠DEF
Simplify it to 6v=12
divied 12 by 6
v=2
