Answer:
Please if you don't mind, complete your question thanks
A quadratic function whose vertex is the same as the y-intercept has the equation
y=x^2+k (where k is the y-intercept, with vertex (0,k))
Since the vertex coincides with the y-intercept, the axis of symmetry is x=0.
Answer:
Never
Never
Never
Step-by-step explanation:
The equations given are
2x1−6x2−4x3 = 6 ....... (1)
−x1+ax2+4x3 = −1 ........(2)
2x1−5x2−2x3 = 9 ..........(3)
the values of a for which the system of linear equations has no solutions
Let first add equation 1 and 2. Also equation 2 and 3. This will result to
X1 + (a X2 - 6X2) - 0 = 5
And
X1 + (aX2-5X2) + 2X3 = 8
Since X2 and X3 can't be cancelled out, we conclude that the value of a is never.
a unique solution,
Let first add equation 1 and 2. Also equation 2 and 3. This will result to
X1 + (a X2 - 6X2) - 0 = 5
And
X1 + (aX2-5X2) + 2X3 = 8
The value of a = never
infinitely many solutions.
Divide equation 1 by 2 we will get
X1 - 3X2 - 2X3 =3
Add the above equation with equation 3. This will result to
3X1 - 8X2 - 4X3 = 12
Everything ought to be the same. Since they're not.
Value of a = never.
You would simply add the last two equations together. If v=2x+1 and t=4x-1, you can add them up. This would make 2x+1+4x-1. If you combine like terms, 1 and -1 cancel out and you are left with 6x. This means x can be anything.
Answer:
1m cm
10m mm
Step-by-step explanation:
