![\bf ~~~~~~~~~~~~\textit{angle between two vectors } \\\\ cos(\theta)=\cfrac{\stackrel{\textit{dot product}}{u \cdot v}}{\stackrel{\textit{magnitude product}}{||u||~||v||}} \implies \measuredangle \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||~||v||}\right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} u=i+\sqrt{7}j\implies &\\\\ v=-i+9j\implies & \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bangle%20between%20two%20vectors%20%7D%20%5C%5C%5C%5C%20cos%28%5Ctheta%29%3D%5Ccfrac%7B%5Cstackrel%7B%5Ctextit%7Bdot%20product%7D%7D%7Bu%20%5Ccdot%20v%7D%7D%7B%5Cstackrel%7B%5Ctextit%7Bmagnitude%20product%7D%7D%7B%7C%7Cu%7C%7C~%7C%7Cv%7C%7C%7D%7D%20%5Cimplies%20%5Cmeasuredangle%20%5Ctheta%20%3D%20cos%5E%7B-1%7D%5Cleft%28%5Ccfrac%7Bu%20%5Ccdot%20v%7D%7B%7C%7Cu%7C%7C~%7C%7Cv%7C%7C%7D%5Cright%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20u%3Di%2B%5Csqrt%7B7%7Dj%5Cimplies%20%26%3C1%2C%5Csqrt%7B7%7D%3E%5C%5C%5C%5C%20v%3D-i%2B9j%5Cimplies%20%26%3C-1%2C9%3E%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf u\cdot v\implies (1)(-1)~+~(\sqrt{7})(9)\implies -1+9\sqrt{7}\implies 9\sqrt{7}-1~\hfill dot~product \\\\[-0.35em] ~\dotfill\\\\ ||u||\implies \sqrt{1^2+(\sqrt{7})^2}\implies \sqrt{1+7}\implies \sqrt{8}~\hfill magnitudes \\\\\\ ||v||\implies \sqrt{(-1)^2+9^2}\implies \sqrt{1+81}\implies \sqrt{82} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20u%5Ccdot%20v%5Cimplies%20%281%29%28-1%29~%2B~%28%5Csqrt%7B7%7D%29%289%29%5Cimplies%20-1%2B9%5Csqrt%7B7%7D%5Cimplies%209%5Csqrt%7B7%7D-1~%5Chfill%20dot~product%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%7C%7Cu%7C%7C%5Cimplies%20%5Csqrt%7B1%5E2%2B%28%5Csqrt%7B7%7D%29%5E2%7D%5Cimplies%20%5Csqrt%7B1%2B7%7D%5Cimplies%20%5Csqrt%7B8%7D~%5Chfill%20magnitudes%20%5C%5C%5C%5C%5C%5C%20%7C%7Cv%7C%7C%5Cimplies%20%5Csqrt%7B%28-1%29%5E2%2B9%5E2%7D%5Cimplies%20%5Csqrt%7B1%2B81%7D%5Cimplies%20%5Csqrt%7B82%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
make sure your calculator is in Degree mode.
Option D:
A = 1, B = –3 and C = 4.
Solution:
Given equation is x – 3y + 4 =0.
General equation of a line format:
<em>Ax + By + C = 0</em>
<u>To identify A, B and C in the given equation:</u>
x – 3y + 4 =0 is a equation of a line which is in the form of Ax + By + C = 0.
That is,
The coefficient of x is A ⇒ A = 1
The coefficient of y is B ⇒ B = –3
The constant term is C ⇒ C = 4
A = 1, B = –3 and C = 4.
Hence Option D is the correct answer.
Answer:
C. F(x) = (x+2)²
Step-by-step explanation:
A horizontal shift to the right by "h" units is achieved by replacing x with (x-h). The shift we want is to the left by 2 units, so h = -2, and we have ...
f(x) = g(x -(-2)) = g(x+2)
f(x) = (x+2)²
I think 12 might be the second one, I'm not sure though!
