Answer: The value of x=1.
Measure of CD =1 unit.
Step-by-step explanation:
Given: Segment EC bisects segment AD at C.
That means , EC divided AD into two equal parts AC and CD,
i.e.
Also it is given that
and ![AC = 4x - 3.](https://tex.z-dn.net/?f=AC%20%3D%204x%20-%203.)
then, we have
![4x-3=\dfrac{6x-4}{2}\\\\\Rightarrow\ 4x-3= 3x-2\\\\\Rightarrow\ 4x-3x=-2+3\\\\\Rightarrow\ x= 1](https://tex.z-dn.net/?f=4x-3%3D%5Cdfrac%7B6x-4%7D%7B2%7D%5C%5C%5C%5C%5CRightarrow%5C%204x-3%3D%203x-2%5C%5C%5C%5C%5CRightarrow%5C%204x-3x%3D-2%2B3%5C%5C%5C%5C%5CRightarrow%5C%20x%3D%201)
So, the value of x=1.
Now, CD = AC = 4(1)-3 =4-3 =1 unit.
Hence, Measure of CD =1 unit.
(2x-1)(x+7) = 0
2x-1=0
2x =1
x = 1/2
x+7=0
x = -7
Answer:
Volume of cylinder =989 ft cube
Step-by-step explanation:
<u>Volume of cylinder</u>
Volume V = πr²h
r - radius and h - height
It is given that,
The radius of a cylindrical construction pipe is 1.5ft and
Length of pipe = 14 ft
<u>To find volume of cylinder</u>
Volume V = πr²h
V = 3.14 x 1.5 x1.5 x 14 =989.1 ≈ 989 ft cube
I'm going to guess that you meant to include parentheses somewhere, so that the ODE is supposed to be
![y'=\dfrac{y^2x}{y^3+x^3}+\dfrac yx](https://tex.z-dn.net/?f=y%27%3D%5Cdfrac%7By%5E2x%7D%7By%5E3%2Bx%5E3%7D%2B%5Cdfrac%20yx)
Then substitute
so that
. Then
![xv'+v=\dfrac{x^3v^2}{x^3v^3+x^3}+v](https://tex.z-dn.net/?f=xv%27%2Bv%3D%5Cdfrac%7Bx%5E3v%5E2%7D%7Bx%5E3v%5E3%2Bx%5E3%7D%2Bv)
![xv'=\dfrac{v^2}{v^3+1}](https://tex.z-dn.net/?f=xv%27%3D%5Cdfrac%7Bv%5E2%7D%7Bv%5E3%2B1%7D)
which is separable as
![\dfrac{v^3+1}{v^2}\,\mathrm dv=\dfrac{\mathrm dx}x](https://tex.z-dn.net/?f=%5Cdfrac%7Bv%5E3%2B1%7D%7Bv%5E2%7D%5C%2C%5Cmathrm%20dv%3D%5Cdfrac%7B%5Cmathrm%20dx%7Dx)
Integrate both sides: on the left,
![\displaystyle\int\frac{v^3+1}{v^2}\,\mathrm dv=\int\left(v+\frac1{v^2}\right)\,\mathrm dv=\dfrac12v^2-\dfrac1v](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Cfrac%7Bv%5E3%2B1%7D%7Bv%5E2%7D%5C%2C%5Cmathrm%20dv%3D%5Cint%5Cleft%28v%2B%5Cfrac1%7Bv%5E2%7D%5Cright%29%5C%2C%5Cmathrm%20dv%3D%5Cdfrac12v%5E2-%5Cdfrac1v)
The other side is trivial. We end up with
![\dfrac12v^2-\dfrac1v=\ln|x|+C](https://tex.z-dn.net/?f=%5Cdfrac12v%5E2-%5Cdfrac1v%3D%5Cln%7Cx%7C%2BC)
Solve in terms of
:
![\boxed{\dfrac{y^2}{2x^2}-\dfrac xy=\ln|x|+C}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cdfrac%7By%5E2%7D%7B2x%5E2%7D-%5Cdfrac%20xy%3D%5Cln%7Cx%7C%2BC%7D)