Find the angle between vector u=i+sqrt 7 j and vector v=-i-4j to the nearest degree.

a. 173 degrees
b. 145 degrees
c. 115 degrees
d. 97 degrees

I just took the test and got 100%

1. C, The vector will not change direction but will increase in magnitude.

2. C, 56 ft lbs

3. C, (-2,-3)

4. A, u = -4 sqrt 41/41 i - 5 sqrt 41/41 j

5. A, 56

6. C, 11.4 ft/sec

7. Do not copy my exact answer please (: They are equal. You know because if you take the points of each segment and subtract them, you will get the same points for both segments.

8a. A, (-10,-16)

8b. D, 2 sqrt 89

9a. C, (-14,-21)

9b. B, 7 sqrt 13

10a. C, (-18,-26)

10b. B, 10 sqrt 10

11. B, 5,144 degrees

12. A, 173 degrees

13. C, (1/6, -1/5)

14. D, 163 mph

15. C, 10j

16. D, 2 sqrt 3

17. A, 98.1 degrees

18. Do not copy my exact answer please (: The angle between the two vectors is 26.75 degrees.

Feel free to mark as brainliest (:

use this formula:

p(a\b) = p(a and b)/p(b).

take it from here.

no solution

step-by-step explanation:

there is not enough information in the problem to solve.