Answer:
1 hour and 15 minutes thats how long the documentary was
Answer:
<h2>90°</h2>
Step-by-step explanation:
First you must calculate the module or the magnitude of both vectors
The module of u is:
![|u|=\sqrt{(8)^2 + (-3)^2} \\\\|u|=\sqrt{64 + 9}\\\\|u|=8.544](https://tex.z-dn.net/?f=%7Cu%7C%3D%5Csqrt%7B%288%29%5E2%20%2B%20%28-3%29%5E2%7D%20%5C%5C%5C%5C%7Cu%7C%3D%5Csqrt%7B64%20%2B%209%7D%5C%5C%5C%5C%7Cu%7C%3D8.544)
The module of v is:
![|v|=\sqrt{(-3)^2 + (-8)^2} \\\\|u|=\sqrt{9 + 64}\\\\|u|=8.544](https://tex.z-dn.net/?f=%7Cv%7C%3D%5Csqrt%7B%28-3%29%5E2%20%2B%20%28-8%29%5E2%7D%20%5C%5C%5C%5C%7Cu%7C%3D%5Csqrt%7B9%20%2B%2064%7D%5C%5C%5C%5C%7Cu%7C%3D8.544)
Now we calculate the scalar product between both vectors
![u*v = 8*(-3) + (-3)*(-8)\\\\u*v = -24+ 24=0](https://tex.z-dn.net/?f=u%2Av%20%3D%208%2A%28-3%29%20%2B%20%28-3%29%2A%28-8%29%5C%5C%5C%5Cu%2Av%20%3D%20-24%2B%2024%3D0)
Finally we know that the scalar product of two vectors is equal to:
![u*v = |u||v|*cos(\theta)](https://tex.z-dn.net/?f=u%2Av%20%3D%20%7Cu%7C%7Cv%7C%2Acos%28%5Ctheta%29)
Where
is the angle between the vectors u and v. Now we solve the equation for ![\theta](https://tex.z-dn.net/?f=%5Ctheta)
![0 = 8.544*8.544*cos(\theta)\\\\0 = cos(\theta)\\\\\theta= arcos(0)\\\\\theta=90\°](https://tex.z-dn.net/?f=0%20%3D%208.544%2A8.544%2Acos%28%5Ctheta%29%5C%5C%5C%5C0%20%3D%20cos%28%5Ctheta%29%5C%5C%5C%5C%5Ctheta%3D%20arcos%280%29%5C%5C%5C%5C%5Ctheta%3D90%5C%C2%B0)
the answer is 90°
Whenever the scalar product of two vectors is equals to zero it means that the angle between them is 90 °
Answer:
Linear functions are those whose graph is a straight line. A linear function has the following form. y = f(x) = a + bx. A linear function has one independent variable and one dependent variable. The independent variable is x and the dependent variable is y.
Carrie is 2 and Kim is just born.
you can have,
Carrie = 3 4 5 6
Kim = 1 2 3 4
Or something like that, thats a table btw