Midpoint of JN is at coordinates (3,2)
Step-by-step explanation:
Coordinates of j= (10,-2)
Coordinates of N = (-4,6)
We need to find midpoint of JN.
The formula used to find midpoint is:
![Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})](https://tex.z-dn.net/?f=Midpoint%3D%28%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D%2C%5Cfrac%7By_1%2By_2%7D%7B2%7D%29)
We are given:
![x_1=10,x_2=-4,y_1=-2,y_2=6](https://tex.z-dn.net/?f=x_1%3D10%2Cx_2%3D-4%2Cy_1%3D-2%2Cy_2%3D6)
Putting values and finding midpoint of JN
![Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})](https://tex.z-dn.net/?f=Midpoint%3D%28%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D%2C%5Cfrac%7By_1%2By_2%7D%7B2%7D%29)
![Midpoint=(\frac{10-4}{2},\frac{-2+6}{2})](https://tex.z-dn.net/?f=Midpoint%3D%28%5Cfrac%7B10-4%7D%7B2%7D%2C%5Cfrac%7B-2%2B6%7D%7B2%7D%29)
![Midpoint=(\frac{6}{2},\frac{4}{2})](https://tex.z-dn.net/?f=Midpoint%3D%28%5Cfrac%7B6%7D%7B2%7D%2C%5Cfrac%7B4%7D%7B2%7D%29)
![Midpoint=(3,2)](https://tex.z-dn.net/?f=Midpoint%3D%283%2C2%29)
So, midpoint of JN is at coordinates (3,2)
Keywords: Midpoint of line
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The solution to the problem is as follows:
<span> 2^HCF(100,120) -1 = 2^20 -1 (Ans)
</span>
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Answer:
The smallest box dimensions are 4 x 4 cm.
Find the diagonal and this would be the diameter of the smallest circle.
Using the Pythagorean theorem:
4^2 + 4^2 = c^2
16 + 16 = c^2
c^2 = 32
c = √32
c = 5.658 cm ( Round answer as needed.)
You must have the distances mixed up when you typed it because it does not work out equation-wise the way you have it. The distance comes out negative, and distance NEVER should come out negative. If you switch the distances traveled you get that the passenger train is going 60 and the freight train is going 40, which makes sense because if the passenger train is going 60 and the freight train is going 40 which is the 20 mph difference we were told exists between the 2 trains.
Answer:
0
Step-by-step explanation:
First we need to find the value of constant k in equation. To find k we use the formula:
![\int\limits^{100}_0 {ke^{-0.01x}} \, dx =1\\Integrating:\\k\int\limits^{100}_0 {e^{-0.01x}} \, dx =1\\\frac{k}{-0.01}[e^{-0.01x}]_0^{100}=1\\-100k [e^{-0.01x}]_0^{100}=1\\-100k[e^{-0.01*100}-e^{-100*0}]=1\\-100k[e^{-1}-e^0]=1\\-100k(-0.632)=1\\63.2k = 1\\k=0.0158](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B100%7D_0%20%7Bke%5E%7B-0.01x%7D%7D%20%5C%2C%20dx%20%3D1%5C%5CIntegrating%3A%5C%5Ck%5Cint%5Climits%5E%7B100%7D_0%20%7Be%5E%7B-0.01x%7D%7D%20%5C%2C%20dx%20%3D1%5C%5C%5Cfrac%7Bk%7D%7B-0.01%7D%5Be%5E%7B-0.01x%7D%5D_0%5E%7B100%7D%3D1%5C%5C-100k%20%5Be%5E%7B-0.01x%7D%5D_0%5E%7B100%7D%3D1%5C%5C-100k%5Be%5E%7B-0.01%2A100%7D-e%5E%7B-100%2A0%7D%5D%3D1%5C%5C-100k%5Be%5E%7B-1%7D-e%5E0%5D%3D1%5C%5C-100k%28-0.632%29%3D1%5C%5C63.2k%20%3D%201%5C%5Ck%3D0.0158)
the probability that a person age 20 will live for at least 20 more years = P(40 ≤ x <∞).
The person would live for at least 40 years
Therefore:
P(40 ≤ x <∞) =
![\int\limits^{\infty}_{40} 0.0158e^{-0.01x}} \, dx \\Integrating:\\0.0158\int\limits^{\infty}_{40} {e^{-0.01x}} \, dx =\frac{0.0158}{-0.01}[e^{-0.01x}]_{40}^{\infty}\\=-1.58 [e^{-0.01x}]_{40}^{\infty}=-1.58[e^{-\infty}-e^{-100*40}]\\=-1.58[e^{-\infty}-e^{-4000}]=-1.58(0-0)=0](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B%5Cinfty%7D_%7B40%7D%200.0158e%5E%7B-0.01x%7D%7D%20%5C%2C%20dx%20%5C%5CIntegrating%3A%5C%5C0.0158%5Cint%5Climits%5E%7B%5Cinfty%7D_%7B40%7D%20%7Be%5E%7B-0.01x%7D%7D%20%5C%2C%20dx%20%3D%5Cfrac%7B0.0158%7D%7B-0.01%7D%5Be%5E%7B-0.01x%7D%5D_%7B40%7D%5E%7B%5Cinfty%7D%5C%5C%3D-1.58%20%5Be%5E%7B-0.01x%7D%5D_%7B40%7D%5E%7B%5Cinfty%7D%3D-1.58%5Be%5E%7B-%5Cinfty%7D-e%5E%7B-100%2A40%7D%5D%5C%5C%3D-1.58%5Be%5E%7B-%5Cinfty%7D-e%5E%7B-4000%7D%5D%3D-1.58%280-0%29%3D0)