Answer:
![\large\boxed{1.\ V=3b^4-15b^3+3b^2-15b}\\\boxed{2.\ V=4\pi n^3-16\pi n^2+16\pi n}\\\boxed{3.\ V=2k^3-k^2-6k}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B1.%5C%20V%3D3b%5E4-15b%5E3%2B3b%5E2-15b%7D%5C%5C%5Cboxed%7B2.%5C%20V%3D4%5Cpi%20n%5E3-16%5Cpi%20n%5E2%2B16%5Cpi%20n%7D%5C%5C%5Cboxed%7B3.%5C%20V%3D2k%5E3-k%5E2-6k%7D)
Step-by-step explanation:
<h3>The picture #1:</h3>
It's a rectangular prism. The formula of a volume of a rectangular prism:
![V=lwh](https://tex.z-dn.net/?f=V%3Dlwh)
l - length
w - width
h - height
We have
![l=b+1,\ w=3b,\ h=b-5](https://tex.z-dn.net/?f=l%3Db%2B1%2C%5C%20w%3D3b%2C%5C%20h%3Db-5)
Substitute:
![V=(b+1)(3b)(b-5)](https://tex.z-dn.net/?f=V%3D%28b%2B1%29%283b%29%28b-5%29)
<em>Use the distributive property </em><em>a(b + c) = ab + ac</em>
<em>and the FOIL:</em><em> (a + b)(c + d) = ac + ad + bc + bd</em>
![V=(3b^3+3b)(b-5)=3b^4-15b^3+3b^2-15b](https://tex.z-dn.net/?f=V%3D%283b%5E3%2B3b%29%28b-5%29%3D3b%5E4-15b%5E3%2B3b%5E2-15b)
<h3>The picture #2:</h3>
It's a cone. The fomula of a volume of a cone:
![V=\dfrac{1}{3}\pi r^2h](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B1%7D%7B3%7D%5Cpi%20r%5E2h)
r - radius
h - height
We have:
![r=n-2,\ h=12n](https://tex.z-dn.net/?f=r%3Dn-2%2C%5C%20h%3D12n)
Substitute:
![V=\dfrac{1}{3}\pi(n-2)^2(12n)](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B1%7D%7B3%7D%5Cpi%28n-2%29%5E2%2812n%29)
<em>Use </em><em>(a + b)² = a² + 2ab + b²</em>
![V=\dfrac{1}{3}\pi(n^2-4n+4)(12n)=(4\pi n)(n^2-4n+4)](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B1%7D%7B3%7D%5Cpi%28n%5E2-4n%2B4%29%2812n%29%3D%284%5Cpi%20n%29%28n%5E2-4n%2B4%29)
<em>Use the distributive property:</em>
![V=4\pi n^3-16\pi n^2+16\pi n](https://tex.z-dn.net/?f=V%3D4%5Cpi%20n%5E3-16%5Cpi%20n%5E2%2B16%5Cpi%20n)
<h3>The picture #3:</h3>
It's a pyramid with a rectangle in the base. The formula of a volume of a rectangular pyramid:
![V=\dfrac{1}{3}abh](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B1%7D%7B3%7Dabh)
a, b - edge of a base
h - height
We have
![a=k-2, b=2k+3, h=3k](https://tex.z-dn.net/?f=a%3Dk-2%2C%20b%3D2k%2B3%2C%20h%3D3k)
Substitute:
![V=\dfrac{1}{3}(k-2)(2k+3)(3k)](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B1%7D%7B3%7D%28k-2%29%282k%2B3%29%283k%29)
<em>Use the distributive property and the FOIL.</em>
![V=\dfrac{1}{3}(k-2)(6k^2+9k)=\dfrac{1}{3}(6k^3+9k^2-12k^2-18k)\\\\=\dfrac{1}{3}(6k^3-3k^2-18k)=2k^3-k^2-6k](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B1%7D%7B3%7D%28k-2%29%286k%5E2%2B9k%29%3D%5Cdfrac%7B1%7D%7B3%7D%286k%5E3%2B9k%5E2-12k%5E2-18k%29%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B3%7D%286k%5E3-3k%5E2-18k%29%3D2k%5E3-k%5E2-6k)
Answer:
Most planets discovered using the transit method have large diameters because planets having large diameters results in a more pronounce (noticeable) dimming effect during their transit between a star and an observer and are therefore readily discovered using the transit method than planets that have a smaller diameter
Explanation:
The transit method involves the detection of a transit (dimming) during the motion of an object between a star and an observer such as the motion of Mercury or Venus between an observer on Earth and the Sun
A large planet gives a higher dimming effect than a small planet and they are therefore more easily detectable using the transit method
The transit method is the method used to discover most of the known exoplanets which are planets that revolve around a star which is not the Sun
As the planet transits in front of a star in an observer's line of sight, it results in the dimming of the stars light by an amount equivalent to the ratio of the planet's to the star's circular areas
Example of planets discovered using the transit method includes the planet HD209458b.
The second one I think, I’m not to sure
One kind of radiation not released by radioactive decay is radiation
Answer:
Eye color is an example of polygenic inheritance. This trait is thought to be influenced by up to 16 different genes. Eye color inheritance is complicated. It is determined by the amount of the brown color pigment melanin that a person has in the front part of the iris.
Explanation: