Answer:
B² - 4AC < 0
Step-by-step explanation:
Check for:
B² - 4AC
The equations for which
B² - 4AC < 0
have complex roots
For A, B and C, use this form of the quadratic:
Ax² + Bx + C = 0
<span>The system of linear equations
−x − y = 1
y = x + 3
can be solved in various ways. This problem hints that you should graph both lines and find the coordinates of their point of intersection.
But we can also solve this system using substitution. Note that y = x + 3. We can subst. x + 3 for y in the 1st equation, as follows:
-x - (x+3) = 1. This simplifies to -x -x - 3 = 1, or -2x = 4, or x = -2.
Since y = x + 3, subbing -2 for x produces y: y = -2 + 3 = 1
Thus, the solution is (-2, 1). </span>
Answer:
k = 5
Step-by-step explanation:
The given line is y = 4x - 14.
The given point is (k, k + 1).
For this point, x = k and y = k + 1.
You substitute these equations into the line's equation to find the value of k.
(x = k)
(y = k + 1)
y = 4x - 14
(k + 1) = 4(k) - 14
Solve for k.
k + 1 = 4k - 14
Subtract k from both sides.
1 = 3k - 14
Add 14 to both sides.
15 = 3k
Divide both sides by 3.
15/3 = k
Simplify.
5 = k
And that's the answer!
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