Answer:
<em>Numbers: 6 and -2</em>
Step-by-step explanation:
<u>Equations</u>
This question can be solved by inspection. It's just a matter of factoring 12 into two factors that sum 4. Both numbers must be of different signs and they are 6 and -2. Their sum is indeed 6-2=4 and their product is 6*(-2)=-12.
However, we'll solve it by the use of equations. Let's call x and y to the numbers. They must comply:
![x+y=4\qquad\qquad [1]](https://tex.z-dn.net/?f=x%2By%3D4%5Cqquad%5Cqquad%20%5B1%5D)
![x.y=-12\qquad\qquad [2]](https://tex.z-dn.net/?f=x.y%3D-12%5Cqquad%5Cqquad%20%5B2%5D)
Solving [1] for y:

Substituting in [2]

Operating:

Rearranging:

Solving with the quadratic formula:

With a=1, b=-4, c=-12:



The solutions are:


This confirms the preliminary results.
Numbers: 6 and -2
Answer:
14. Quadrant IV
13. Quadrant II
12. Quadrant III
11. Quadrant I
15. -17
16. -1
17. 7
Step-by-step explanation:
/ = divided -1/3(1) = -3 + 5 = 2 (g = 1)
What are the domain and range of the relation (–4, 2), (0, 1), (0, 5), (8, 10)? A.Domain: {–4, 0, 8}; Range: {2, 1, 5, 10} B.Dom
vladimir1956 [14]
It is A because the domain is x and range is the y.
Let p (x) = 4x^4-13x^2-2x
The zero of x-2 is 2
putting x = 2 in p (x), we get,
p (2) = 4×2^4-13×2^2-2×2
= 64 - 52 - 4
= 64 - 56
= 8
Therefore, remainder = 8