Check the picture below.
since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.
1)
![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=4 \end{cases}\implies V=\cfrac{\pi (3)^2(4)}{3}\implies V=12\pi \implies V\approx 37.7](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D3%5C%5C%20h%3D4%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B%5Cpi%20%283%29%5E2%284%29%7D%7B3%7D%5Cimplies%20V%3D12%5Cpi%20%5Cimplies%20V%5Capprox%2037.7)
2)
now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.
![\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20sphere%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D3%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B4%5Cpi%20%283%29%5E3%7D%7B3%7D%5Cimplies%20V%3D36%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bhalf%20of%20that%20for%20a%20semisphere%7D%7D%7BV%3D18%5Cpi%20%7D%5Cimplies%20V%5Capprox%2056.55)
3)
well, you'll be serving the cone and top combined, 12π + 18π = 30π or about 94.25 in³.
4)
pretty much the same thing, we get the volume of the cone and its top, add them up.
Answer: the correct option is
(D) The imaginary part is zero.
Step-by-step explanation: Given that neither a nor b are equal to zero.
We are to select the correct statement that accurately describes the following product :
We will be using the following formula :
From product (i), we get
![P\\\\=(a+bi)(a-bi)\\\\=a^2-(bi)^2\\\\=a^2-b^2i^2\\\\=a^2-b^2\times (-1)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }i^2=-1]\\\\=a^2+b^2.](https://tex.z-dn.net/?f=P%5C%5C%5C%5C%3D%28a%2Bbi%29%28a-bi%29%5C%5C%5C%5C%3Da%5E2-%28bi%29%5E2%5C%5C%5C%5C%3Da%5E2-b%5E2i%5E2%5C%5C%5C%5C%3Da%5E2-b%5E2%5Ctimes%20%28-1%29~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7Di%5E2%3D-1%5D%5C%5C%5C%5C%3Da%5E2%2Bb%5E2.)
So, there is no imaginary part in the given product.
Thus, the correct option is
(D) The imaginary part is zero.
9.56 because you don't round up when two is the next number. (The hundredths place is the second number after the decimal, where the 6 is in the original number.)
Answer:
B
Step-by-step explanation:
Answer:
x = $5.49 cost of the bananas
Step-by-step explanation:
x = cost of the bananas
$3.89 = cost of the peanut butter
$ 9.38 is the cost of both
Bananas + peanut butter = $9.38
x + $3.89 = $9.38
<u> - $ 3.89 = -$3.89 </u> Subtract $3.89 from both sides
x = $5.49
