Let the initial point of the vector be (x,y). Then the magnitude of the vector v can be written as:
![|v|= \sqrt{ (-1-x)^{2} + (5-y)^{2} }](https://tex.z-dn.net/?f=%7Cv%7C%3D%20%5Csqrt%7B%20%28-1-x%29%5E%7B2%7D%20%2B%20%285-y%29%5E%7B2%7D%20%7D%20)
The magnitude of vecor v is given to be 10. So we can write:
![10= \sqrt{ (-1-x)^{2} + (5-y)^{2} } \\ \\ 100=(-1-x)^{2} + (5-y)^{2}](https://tex.z-dn.net/?f=10%3D%20%5Csqrt%7B%20%28-1-x%29%5E%7B2%7D%20%2B%20%285-y%29%5E%7B2%7D%20%7D%20%5C%5C%20%5C%5C%20100%3D%28-1-x%29%5E%7B2%7D%20%2B%20%285-y%29%5E%7B2%7D)
Now from the given options, we have to check which one satisfies the above equation. That point will be the initial point of the vector.
The point in option d, satisfies the equation.
Thus, the answer to this question is option D
Answer: x = 16
Step-by-step explanation:
Solve for x by simplifying both sides of the equation, then isolating the variable. :)
Okay I get some of them, your first one - 1 I disagree with. So we can solve with two points by using a equation y2-y1 over x2-x1. So it's 4-8 over - 3-6 with the numerator being - 4 and the denominator being - 9. Any questions on how to solve for slope with two coordinates.
Answer:
144/5
Step-by-step explanation:
![\large \rm \: 9 \frac{2}{3} \times 1 \frac{1}{29} \times \frac{6}{15 } \times 7 \frac{1}{5} \\ = \frac{29}{3} \times \frac{30}{29} \times \frac{6}{15} \times \frac{36}{5} \\ = 10 \times \frac{216}{75} \\ = \frac{2160}{75} \\ = \frac{144}{5}](https://tex.z-dn.net/?f=%5Clarge%20%5Crm%20%5C%3A%209%20%5Cfrac%7B2%7D%7B3%7D%20%5Ctimes%201%20%5Cfrac%7B1%7D%7B29%7D%20%5Ctimes%20%5Cfrac%7B6%7D%7B15%20%7D%20%5Ctimes%207%20%5Cfrac%7B1%7D%7B5%7D%20%20%5C%5C%20%20%3D%20%20%5Cfrac%7B29%7D%7B3%7D%20%20%5Ctimes%20%20%5Cfrac%7B30%7D%7B29%7D%20%20%5Ctimes%20%20%5Cfrac%7B6%7D%7B15%7D%20%20%5Ctimes%20%20%5Cfrac%7B36%7D%7B5%7D%20%20%5C%5C%20%20%3D%2010%20%5Ctimes%20%20%5Cfrac%7B216%7D%7B75%7D%20%20%5C%5C%20%20%3D%20%20%5Cfrac%7B2160%7D%7B75%7D%20%20%5C%5C%20%20%3D%20%20%5Cfrac%7B144%7D%7B5%7D%20%3C%2Fp%3E%3Cp%3E)