Q(2, a) is 5 units away from P(–2, –1). Find all possible values of a.

a = 2

Step-by-step explanation:

Here, we want to find all the possible values of a

If Q is 5 units away from P, then the distance between Q and P is 5 units

Mathematically, to find the distance between two points, we use the following formula

D = √( (x2-x1)^2 + (y2-y1)^2

So we have ;

D = √(2-(-2)^2 + (a-(-1)^2

D = √(16) + (a + 1)^2

Since D is 5, we have

5 = √(16) + (a +1)^2

Squaring both sides to remove the square root, we have

25 = 16 + (a + 1)^2

(a + 1)^2 = 25 - 16

(a + 1)^2 = 9

(a + 1)^2 = 3^2

Since power is equal, equate base

a + 1 = 3

a = 3-1

a = 2

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068