Answer:
The angle between the given vectors u and v is ![\theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]](https://tex.z-dn.net/?f=%5Ctheta%3Dcos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B3%7D%7B%5Csqrt%7B10%7D%7D%5Cright%5D)
Step-by-step explanation:
Given vectors are
and 
Now compute the dot product of u and v:




Now find the magnitude of u and v:









To find the angle between the given vectors

![\theta=cos^{-1}\left[\frac{\overrightarrow{u}.\overrightarrow{v}}{|\overrightarrow{u}|\overrightarrow{v}|}\right]](https://tex.z-dn.net/?f=%5Ctheta%3Dcos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B%5Coverrightarrow%7Bu%7D.%5Coverrightarrow%7Bv%7D%7D%7B%7C%5Coverrightarrow%7Bu%7D%7C%5Coverrightarrow%7Bv%7D%7C%7D%5Cright%5D)
![=cos^{-1}\left[\frac{15}{5\times \sqrt{10}}\right]](https://tex.z-dn.net/?f=%3Dcos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B15%7D%7B5%5Ctimes%20%5Csqrt%7B10%7D%7D%5Cright%5D)
![=cos^{-1}\left[\frac{15}{5\times \sqrt{10}}\right]](https://tex.z-dn.net/?f=%3Dcos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B15%7D%7B5%5Ctimes%20%5Csqrt%7B10%7D%7D%5Cright%5D)
![\theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]](https://tex.z-dn.net/?f=%5Ctheta%3Dcos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B3%7D%7B%5Csqrt%7B10%7D%7D%5Cright%5D)
Therefore the angle between the vectors u and v is
![\theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]](https://tex.z-dn.net/?f=%5Ctheta%3Dcos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B3%7D%7B%5Csqrt%7B10%7D%7D%5Cright%5D)
Let "a", "b" and "с" be sides of the triangle ("с" is the longest side). The triangle will be right if a² + b² = c²
A.
a² + b² = 12² + 13² = 144 + 169 = 313
c² = 25² = 625
313 ≠ 625
B.
a² + b² = 36² + 77² = 1296 + 5929 = 7225
c² = 85² = 7225
7225 = 7225 ← right triangle
C.
45, 45, 45 ← this is an equilateral triangle (all angles = 60°)
<span>
D.
</span>a² + b² = 45² + 56² = 2025 + 3136 = 5161
c² = 67² = 4489
5161 ≠ 4489
Answer is B
Answer:
C
Step-by-step explanation:
39-7=32
40-7=33
41-7=34
42-7=35