X^2 - 6x + 9 = -1 + 9. We need to solve for x.
First, add one on each side of the equation:
x^2 - 6x + 9 + 1 = -1 + 9 + 1
x^2 - 6x + 10 = 9
Then subtract 9 from each side:
x^2 - 6x + 10 - 9 = 9-9
x^2 - 6x + 1 = 0
Now we got an equation = 0. This an equation from the second degree, it represented by a parabola which turns up since a>0.
This equation is the developed form as ax^2 + bx + c with a=1; b = -6; and c=1.
Now, to find the zeroes of this equation, we need to find delta Δ.
Δ = b^2 - 4ac.
If Δ>0, the equation admits 2 zeroes: x=(-b-√Δ)/2a and x = (-b+√Δ)/2a.
If Δ<0, the equation doesn't admits any zero.
If Δ=0, the equation admits one zero x = -b/2a
Δ = (-6)^2 - 4(1)(1)
Δ = 36 - 4
Δ = 32
Δ>0
So the zeroes of the equation are:
x = (-b-√Δ)/2a = (6-4√2)/2
x = (-b+√Δ)/2a = (6+4√2)/2
Hope this Helps! :)
Answer:
No solution
Step-by-step explanation:
There are no values of Z that makes this equation true.
Answer:
I think it might be B
i hope this helps!
(if B is wrong then it is definitely C)
For 13. the answer is w = -1.
Here are the steps
1. subtract 5 from both sides ( 3w = 2 - 5)
2. simplify 2 - 5 to -3 (3w = -3)
3. divide both sides by 3 ( finally the answer is w = -1.
)
Hello!
This is real simple!
So let's look at your first ordered pair, (-2,0)
-2 is the x-coordinate, 0 is the y-coordinate. 0 is the origin - or the center - of your graph. Don't put a point yet!
Next, -2 is to the left, so look for -2 on your graph (or count two to the left) Now put a point there. You have plotted (-2,0) on your graph!
Let's try another.
(0,4)
Look at 0 on your graph. The 4 is the y-coordinate, indicating that you'll go up on the y-axis.
Go up 4 on your y-axis
Put a point, you plotted (0,4)
One more?
(2,0)
x. y
Go to positive 2 on your x-axis (since it is the x-coordinate). Notice the y-coordinate is 0, you don't need to go up! Plot your point
You plotted (2,0)
I hope I helped. Please ask if you still need help.
