Answer:
there you go
Step-by-step explanation:
The length of a train car is 50.6 feet. This is 5.8 feet less than 6 times the width. What is the width?
l = 50.6
We are also given:
6w - 5.8 = 50.6
Answer:
Wassup Obito
Step-by-step explanation:
I would have done it if you had asked for only 1 question but you have asked someone to do more than 10 q
You must be mad to think someone will do the entire thing
Mark me brainliest
A) x + y = 110
x - y = 40
B) Melvin swims for 35 minutes every day.
C) It is not possible
Step-by-step explanation:
Step 1:
Let x denote the duration for which Melvin plays tennis and y denote the duration for which he swims
Step 2 :
Part A :
Given that the total duration for which he plays and tennis is 110 minutes
so we have x + y = 110
Also given that he plays tennis for 40 minutes more than he swims
So, x = y+ 40 =>x-y = 40
So the linear pair of equations are
x + y = 110
x-y = 40
Step 3 :
Part B
Solving the above 2 equations we have
y+40+y = 110 = > 2y+ 40 = 110 = > y = 2y = 70 = >35
x = y+40 = 35+ 40 = 75
So Melvin plays tennis for 75 minutes and swims for 35 minutes every day.
Step 4 :
Part C
Given he plays and swims for 110 mins exactly, i.e x + y = 110
If Melvin plays tennis for 70 minutes , then x = 70, then the time for which he can swim is 110 - 70 = 40 mins.
He gets only 40 mins for swimming which is not 40 mins more than he plays tennis .
So this case is not possible.
Answer:
3 1/4
Step-by-step explanation:
3 5/8 is too long
3 is too short so is 3 1/8
youre welcome
Answer:
Therefore the value of y(1)= 0.9152.
Step-by-step explanation:
According to the Euler's method
y(x+h)≈ y(x) + hy'(x) ....(1)
Given that y(0) =3 and step size (h) = 0.2.
Putting the value of y'(x) in equation (1)
Substituting x =0 and h= 0.2
![y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]](https://tex.z-dn.net/?f=y%280%2B0.2%29%5Capprox%20y%280%29%2B0.2%5B0%5Ctimes%20y%280%29-%5Cfrac12%20%28y%280%29%29%5E2%5D)
![\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]](https://tex.z-dn.net/?f=%5CRightarrow%20y%280.2%29%5Capprox%203%2B0.2%5B-%5Cfrac12%20%5Ctimes3%5D)
[∵ y(0) =3 ]
Substituting x =0.2 and h= 0.2
![y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]](https://tex.z-dn.net/?f=y%280.2%2B0.2%29%5Capprox%20y%280.2%29%2B0.2%5B%280.2%29%5E2%5Ctimes%20y%280.2%29-%5Cfrac12%20%28y%280.2%29%29%5E2%5D)
![\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]](https://tex.z-dn.net/?f=%5CRightarrow%20y%280.4%29%5Capprox%20%202.7%2B0.2%5B%280.2%29%5E2%5Ctimes%202.7-%20%5Cfrac12%282.7%29%5E2%5D)
Substituting x =0.4 and h= 0.2
![y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]](https://tex.z-dn.net/?f=y%280.4%2B0.2%29%5Capprox%20y%280.4%29%2B0.2%5B%280.4%29%5E2%5Ctimes%20y%280.4%29-%5Cfrac12%20%28y%280.4%29%29%5E2%5D)
![\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]](https://tex.z-dn.net/?f=%5CRightarrow%20y%280.6%29%5Capprox%20%201.9926%2B0.2%5B%280.4%29%5E2%5Ctimes%201.9926-%20%5Cfrac12%281.9926%29%5E2%5D)
Substituting x =0.6 and h= 0.2
![y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]](https://tex.z-dn.net/?f=y%280.6%2B0.2%29%5Capprox%20y%280.6%29%2B0.2%5B%280.6%29%5E2%5Ctimes%20y%280.6%29-%5Cfrac12%20%28y%280.6%29%29%5E2%5D)
![\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]](https://tex.z-dn.net/?f=%5CRightarrow%20y%280.8%29%5Capprox%20%201.6593%2B0.2%5B%280.6%29%5E2%5Ctimes%201.6593-%20%5Cfrac12%281.6593%29%5E2%5D)
Substituting x =0.8 and h= 0.2
![y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]](https://tex.z-dn.net/?f=y%280.8%2B0.2%29%5Capprox%20y%280.8%29%2B0.2%5B%280.8%29%5E2%5Ctimes%20y%280.8%29-%5Cfrac12%20%28y%280.8%29%29%5E2%5D)
![\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]](https://tex.z-dn.net/?f=%5CRightarrow%20y%281.0%29%5Capprox%20%200.8800%2B0.2%5B%280.8%29%5E2%5Ctimes%200.8800-%20%5Cfrac12%280.8800%29%5E2%5D)
Therefore the value of y(1)= 0.9152.
