keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above
![\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cqquad%20%5Cqquad%20y%20%3D%20%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B3%7Dx-5)
well then therefore
![\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7Bperpendicular%20lines%20have%20%5Cunderline%7Bnegative%20reciprocal%7D%20slopes%7D%7D%20%7B%5Cstackrel%7Bslope%7D%7B3%5Cimplies%20%5Ccfrac%7B3%7D%7B1%7D%7D%20~%5Chfill%20%5Cstackrel%7Breciprocal%7D%7B%5Ccfrac%7B1%7D%7B3%7D%7D%20~%5Chfill%20%5Cstackrel%7Bnegative~reciprocal%7D%7B-%5Ccfrac%7B1%7D%7B3%7D%7D%7D)
so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )
![(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1})\implies y+3=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y=-\cfrac{1}{3}x+\cfrac{1}{3}-3\implies y=-\cfrac{1}{3}x-\cfrac{8}{3}](https://tex.z-dn.net/?f=%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B-3%7D%29%5Cqquad%20%5Cqquad%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20-%5Ccfrac%7B1%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-%5Cstackrel%7By_1%7D%7B%28-3%29%7D%3D%5Cstackrel%7Bm%7D%7B-%5Ccfrac%7B1%7D%7B3%7D%7D%28x-%5Cstackrel%7Bx_1%7D%7B1%7D%29%5Cimplies%20y%2B3%3D-%5Ccfrac%7B1%7D%7B3%7Dx%2B%5Ccfrac%7B1%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20y%3D-%5Ccfrac%7B1%7D%7B3%7Dx%2B%5Ccfrac%7B1%7D%7B3%7D-3%5Cimplies%20y%3D-%5Ccfrac%7B1%7D%7B3%7Dx-%5Ccfrac%7B8%7D%7B3%7D)
Answer:
2/3 (B)
Step-by-step explanation:
The y values increase by 2 every time and the x values increase by 3. The rule for slope is rise(y)/run(x) so you get b, 2/3
Answer:
A. 5.16 s.
B. 5.66 s.
Step-by-step explanation:
A.
For a simple harmonic motion,
T = 2pi (sqrt * (l/g))
Given:
L1 = 3 cm
T1 = 4 s
L2 = 5 cm
T2 = ?
4 = 2pi*sqrt(3/g)
g = 7.4
At, L2,
T2 = 2pi*sqrt(5/7.4)
= 5.16 s.
B.
M1 = M1
M2 = 2*M1
For a simple harmonic motion,
T = 2pi (sqrt * (m/k))
4 = 2pi (sqrt * (M1/k))
M1/k = 0.405
Inputting the above values,
T2 = 2pi (sqrt * (2*M1/k))
= 2pi (sqrt * (2 * 0.405))
= 5.66 s.
Answer:
30
Step-by-step explanation:
following the bodmas rule
we will get the answer 30
You have to dive the x and the Y so R=7