Perpendicular lines have negative reciprocal slop to each other.
So first solve for y.
Slope will be the coefficient of x in y=mx+c form.
m is the slope. Then negative reciprocal this slope and in this case the answer is -1/5
Answer:
y = - 2x + 3
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Calculate m using the slope formula
m = ![\frac{y_{2}-y_{1} }{x_{2}-x_{1} }](https://tex.z-dn.net/?f=%5Cfrac%7By_%7B2%7D-y_%7B1%7D%20%20%7D%7Bx_%7B2%7D-x_%7B1%7D%20%20%7D)
with (x₁, y₁ ) = (0, 3) and (x₂, y₂ ) = (1,1) ← 2 points on the line
m =
= - 2
The line crosses the y- axis at (0, 3) ⇒ c = 3
y = - 2x + 3 ← equation of line
Answer:
3.5%
Step-by-step explanation:
The volume of a cylinder = ![\pi r^2h](https://tex.z-dn.net/?f=%5Cpi%20r%5E2h)
<em>r</em> = radius of cylinder,
<em>h</em> = height of cylinder
For the non-optimal can,
<em>r</em> = 2.75/2 = 1.375
<em>h</em> = 5.0
![V = \pi(1.375^2)\times 5.0 = 9.453125\pi](https://tex.z-dn.net/?f=V%20%3D%20%5Cpi%281.375%5E2%29%5Ctimes%205.0%20%3D%209.453125%5Cpi)
<em />
For the optimal can,
<em>d</em>/<em>h</em> = 1,
<em>d</em> = <em>h</em>
2<em>r </em>=<em> h</em>
<em>r</em> = h/2
![V = \pi\left(\dfrac{h}{2}\right)^2\times h = \pi\left(\dfrac{h^3}{4}\right)](https://tex.z-dn.net/?f=V%20%3D%20%5Cpi%5Cleft%28%5Cdfrac%7Bh%7D%7B2%7D%5Cright%29%5E2%5Ctimes%20h%20%3D%20%5Cpi%5Cleft%28%5Cdfrac%7Bh%5E3%7D%7B4%7D%5Cright%29)
They have the same volume.
<em />
<em />
![h^3 = 37.8125](https://tex.z-dn.net/?f=h%5E3%20%3D%2037.8125)
(This is the height of the optimal can)
(This is the radius of the optimal can)
The area of a cylinder is
<em />
<em />
For the non-optimal can,
![A = 2\pi\times\dfrac{2.75}{2}\left(\dfrac{2.75}{2}+5.0\right) = 17.53125\pi](https://tex.z-dn.net/?f=A%20%3D%202%5Cpi%5Ctimes%5Cdfrac%7B2.75%7D%7B2%7D%5Cleft%28%5Cdfrac%7B2.75%7D%7B2%7D%2B5.0%5Cright%29%20%3D%2017.53125%5Cpi)
For the optimal can,
![A = 2\pi\times1.68\left(1.68+3.36\right) = 16.9344\pi](https://tex.z-dn.net/?f=A%20%3D%202%5Cpi%5Ctimes1.68%5Cleft%281.68%2B3.36%5Cright%29%20%3D%2016.9344%5Cpi)
Amount of aluminum saved, as a percentage of the amount used to make the optimal cans = ![\dfrac{17.53125\pi - 16.9344\pi}{16.9344\pi}\times 100\% = 3.5\%](https://tex.z-dn.net/?f=%5Cdfrac%7B17.53125%5Cpi%20-%2016.9344%5Cpi%7D%7B16.9344%5Cpi%7D%5Ctimes%20100%5C%25%20%3D%203.5%5C%25)
6n+n n=5
substitute
6(5)+(5)
30+5
35