Answer:
[tex]\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}[/tex]
Step-by-step explanation:
The picture #1:
It's a rectangular prism. The formula of a volume of a rectangular prism:
[tex]V=lwh[/tex]
l - length
w - width
h - height
We have
[tex]l=b+1,\ w=3b,\ h=b-5[/tex]
Substitute:
[tex]V=(b+1)(3b)(b-5)[/tex]
Use the distributive property a(b + c) = ab + ac
and the FOIL: (a + b)(c + d) = ac + ad + bc + bd
[tex]V=(3b^3+3b)(b-5)=3b^4-15b^3+3b^2-15b[/tex]
The picture #2:
It's a cone. The fomula of a volume of a cone:
[tex]V=\dfrac{1}{3}\pi r^2h[/tex]
r - radius
h - height
We have:
[tex]r=n-2,\ h=12n[/tex]
Substitute:
[tex]V=\dfrac{1}{3}\pi(n-2)^2(12n)[/tex]
Use (a + b)² = a² + 2ab + b²
[tex]V=\dfrac{1}{3}\pi(n^2-4n+4)(12n)=(4\pi n)(n^2-4n+4)[/tex]
Use the distributive property:
[tex]V=4\pi n^3-16\pi n^2+16\pi n[/tex]
The picture #3:
It's a pyramid with a rectangle in the base. The formula of a volume of a rectangular pyramid:
[tex]V=\dfrac{1}{3}abh[/tex]
a, b - edge of a base
h - height
We have
[tex]a=k-2, b=2k+3, h=3k[/tex]
Substitute:
[tex]V=\dfrac{1}{3}(k-2)(2k+3)(3k)[/tex]
Use the distributive property and the FOIL.
[tex]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[/tex]
