First you must acknowledge that you are dealing with a line therefore you must write linear equation or linear function in this case.
Linear function has a form of,
![y=mx+n](https://tex.z-dn.net/?f=y%3Dmx%2Bn)
Then calculate the slope <em>m</em> using the coordinates of two points. Let say <em>A(x1, y1)</em> and <em>B(x2, y2)</em>,
![m=\dfrac{\Delta{y}}{\Delta{x}}=\dfrac{y_2-y_1}{x_2-x_1}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B%5CDelta%7By%7D%7D%7B%5CDelta%7Bx%7D%7D%3D%5Cdfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
Now pick a point either <em>A</em> or <em>B</em> and insert coordinates of either one of them in the linear equation also insert the slope you just calculated, I will pick point <em>A</em>.
![y_1=mx_1+n](https://tex.z-dn.net/?f=y_1%3Dmx_1%2Bn)
From here you solve the equation for n,
![y_1=mx_1+n\Longrightarrow n=y_1-mx_1](https://tex.z-dn.net/?f=%3C%2Fp%3E%3Cp%3Ey_1%3Dmx_1%2Bn%5CLongrightarrow%20n%3Dy_1-mx_1%3C%2Fp%3E%3Cp%3E)
So you have slope <em>m</em> and variable <em>n</em> therefore you can write down the equation of the line,
![f(x)=m_{slope}x+n_{variable}](https://tex.z-dn.net/?f=f%28x%29%3Dm_%7Bslope%7Dx%2Bn_%7Bvariable%7D)
Hope this helps.
r3t40
Answer:
Acute
Step-by-step explanation:
This answer is acute because obtuse is arm width a right angle is a straight L a straight angle is just a line and acute angle is a little pinch
Answer:
The volume of a cylinder is 62.8 cm and the volume of a cone is 20.9 cm³ .
Step-by-step explanation:
Formula
![Volume\ of\ a\ cylinder = \pi r^{2} h](https://tex.z-dn.net/?f=Volume%5C%20of%5C%20a%5C%20cylinder%20%3D%20%5Cpi%20r%5E%7B2%7D%20h)
![Volume\ of\ a\ cone = \pi r^{2} \frac{h}{3}](https://tex.z-dn.net/?f=Volume%5C%20of%5C%20a%5C%20cone%20%3D%20%5Cpi%20r%5E%7B2%7D%20%5Cfrac%7Bh%7D%7B3%7D)
Where r is the radius and h is the height .
As given
A cylinder and cone have the same height and radius. The height of each is 5 cm, and the radius is 2 cm.
![\pi = 3.14](https://tex.z-dn.net/?f=%5Cpi%20%3D%203.14)
Thus
![Volume\ of\ a\ cylinder =3.14\times 5\times 2\times 2](https://tex.z-dn.net/?f=Volume%5C%20of%5C%20a%5C%20cylinder%20%3D3.14%5Ctimes%205%5Ctimes%202%5Ctimes%202)
= 62.8 cm³
Thus the volume of a cylinder is 62.8 cm³ .
![Volume\ of\ a\ cone = 3.14\times 2\times 2\times frac{5}{3}](https://tex.z-dn.net/?f=Volume%5C%20of%5C%20a%5C%20cone%20%3D%203.14%5Ctimes%202%5Ctimes%202%5Ctimes%20frac%7B5%7D%7B3%7D)
![Volume\ of\ a\ cone =\frac{5\times 3.14\times 2\times 2}{3}](https://tex.z-dn.net/?f=Volume%5C%20of%5C%20a%5C%20cone%20%3D%5Cfrac%7B5%5Ctimes%203.14%5Ctimes%202%5Ctimes%202%7D%7B3%7D)
![Volume\ of\ a\ cone =\frac{62.8}{3}](https://tex.z-dn.net/?f=Volume%5C%20of%5C%20a%5C%20cone%20%3D%5Cfrac%7B62.8%7D%7B3%7D)
= 20.9 (Approx) cm³
Thus the volume of a cone is 20.9 cm³ .
Therefore the volume of a cylinder is 62.8 cm and the volume of a cone is 20.9 cm³ .
If m =2 insert m so 4+4-3+5
answer is 5
Answer:
5/9
Step-by-step explanation:
What we have in this question are four white balls and 5 red balls.
This is the sample space
[(4w,0R) (4w,1R)(4w,2R)(4w,3R)(4w,4R)(0w,5R)(1w,5R)(2w,5R)
So we have 9 possible sets of events
The event number with the last ball being white = 5
Probability of the last ball drawn being white = 5/9
= 0.56