Compare two similar sides:
24:32 this reduces to 3:4
Answer: D) 3:4
we have
![f(x)=\frac{1}{4} x^{2} +bx+10](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B1%7D%7B4%7D%20x%5E%7B2%7D%20%2Bbx%2B10)
This is a vertical parabola open upward
The axis of symmetry is equal to the x-coordinate of the vertex
The vertex is the point (h,k)
the equation of the axis of symmetry is ![x=h](https://tex.z-dn.net/?f=x%3Dh)
In this problem
![x=6](https://tex.z-dn.net/?f=x%3D6)
so the x-coordinate of the vertex is ![h=6](https://tex.z-dn.net/?f=h%3D6)
<u>Convert the quadratic equation in vertex form</u>
Group terms that contain the same variable, and move the constant to the opposite side of the equation
![f(x)-10=\frac{1}{4} x^{2} +bx](https://tex.z-dn.net/?f=f%28x%29-10%3D%5Cfrac%7B1%7D%7B4%7D%20x%5E%7B2%7D%20%2Bbx)
Factor the leading coefficient
![f(x)-10=\frac{1}{4}(x^{2} +4bx)](https://tex.z-dn.net/?f=f%28x%29-10%3D%5Cfrac%7B1%7D%7B4%7D%28x%5E%7B2%7D%20%2B4bx%29)
Complete the square. Remember to balance the equation by adding the same constants to each side
![f(x)-10+b^{2}=\frac{1}{4}(x^{2} +4bx+4b^{2})](https://tex.z-dn.net/?f=f%28x%29-10%2Bb%5E%7B2%7D%3D%5Cfrac%7B1%7D%7B4%7D%28x%5E%7B2%7D%20%2B4bx%2B4b%5E%7B2%7D%29)
Rewrite as perfect squares
![f(x)+(b^{2}-10)=\frac{1}{4}(x+2b)^{2}](https://tex.z-dn.net/?f=f%28x%29%2B%28b%5E%7B2%7D-10%29%3D%5Cfrac%7B1%7D%7B4%7D%28x%2B2b%29%5E%7B2%7D)
![f(x)=\frac{1}{4}(x+2b)^{2}-(b^{2}-10)](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B1%7D%7B4%7D%28x%2B2b%29%5E%7B2%7D-%28b%5E%7B2%7D-10%29)
remember that
![h=6](https://tex.z-dn.net/?f=h%3D6)
so
![(x+2b)=(x-6)](https://tex.z-dn.net/?f=%28x%2B2b%29%3D%28x-6%29)
![2b=-6](https://tex.z-dn.net/?f=2b%3D-6)
![b=-3](https://tex.z-dn.net/?f=b%3D-3)
substitute in the equation
![f(x)=\frac{1}{4}(x-6)^{2}-((-3)^{2}-10)](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B1%7D%7B4%7D%28x-6%29%5E%7B2%7D-%28%28-3%29%5E%7B2%7D-10%29)
![f(x)=\frac{1}{4}(x-6)^{2}+1](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B1%7D%7B4%7D%28x-6%29%5E%7B2%7D%2B1)
therefore
<u>the answer is</u>
![b=-3](https://tex.z-dn.net/?f=b%3D-3)
Answer:
the third one
Step-by-step explanation:
Without any more information, I can only assume you're trying to find the volume of this bounded region. In that case, you're computing
![\displaystyle\iiint_E\mathrm dV=\iiint_E1\,\mathrm dx\,\mathrm dy\,\mathrm dz](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciiint_E%5Cmathrm%20dV%3D%5Ciiint_E1%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dz)
This can be written as
![\displaystyle\int_{x=-2}^{x=2}\int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}\int_{z=x^2+y^2}^{z=4}\mathrm dz\,\mathrm dy\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7Bx%3D-2%7D%5E%7Bx%3D2%7D%5Cint_%7By%3D-%5Csqrt%7B4-x%5E2%7D%7D%5E%7By%3D%5Csqrt%7B4-x%5E2%7D%7D%5Cint_%7Bz%3Dx%5E2%2By%5E2%7D%5E%7Bz%3D4%7D%5Cmathrm%20dz%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dx)
Converting to cylindrical coordinates will make this easier.
![\begin{cases}\mathbf x(r,\theta,z)=r\cos\theta\\\mathbf y(r,\theta,z)=r\sin\theta\\\mathbf z(r,\theta,z)=z\end{cases}\implies\dfrac{\partial(\mathbf x,\mathbf y,\mathbf z)}{\partial(r,\theta,z)}=\left|\det\begin{bmatrix}\mathbf x_r&\mathbf y_r&\mathbf z_r\\\mathbf x_\theta&\mathbf y_\theta&\mathbf z_\theta\\\mathbf x_z&\mathbf y_z&\mathbf z_z\end{bmatrix}\right|=r](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cmathbf%20x%28r%2C%5Ctheta%2Cz%29%3Dr%5Ccos%5Ctheta%5C%5C%5Cmathbf%20y%28r%2C%5Ctheta%2Cz%29%3Dr%5Csin%5Ctheta%5C%5C%5Cmathbf%20z%28r%2C%5Ctheta%2Cz%29%3Dz%5Cend%7Bcases%7D%5Cimplies%5Cdfrac%7B%5Cpartial%28%5Cmathbf%20x%2C%5Cmathbf%20y%2C%5Cmathbf%20z%29%7D%7B%5Cpartial%28r%2C%5Ctheta%2Cz%29%7D%3D%5Cleft%7C%5Cdet%5Cbegin%7Bbmatrix%7D%5Cmathbf%20x_r%26%5Cmathbf%20y_r%26%5Cmathbf%20z_r%5C%5C%5Cmathbf%20x_%5Ctheta%26%5Cmathbf%20y_%5Ctheta%26%5Cmathbf%20z_%5Ctheta%5C%5C%5Cmathbf%20x_z%26%5Cmathbf%20y_z%26%5Cmathbf%20z_z%5Cend%7Bbmatrix%7D%5Cright%7C%3Dr)
Now, in cylindrical coordinates the integral is given by
![\displaystyle\iiint_E\mathrm dV=\iiint_Er\,\mathrm dr\,\mathrm d\theta\,\mathrm dz](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciiint_E%5Cmathrm%20dV%3D%5Ciiint_Er%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%5C%2C%5Cmathrm%20dz)