Answer:
Part a) [tex]h(g(140))=\frac{4,900}{\pi}\ cm^2[/tex]
Part b) [tex]f(j(5.18))=2\pi\sqrt{\frac{5.18}{\pi}}\ cm[/tex]
Step-by-step explanation:
we know that
The circumference of a circle is
[tex]C=2\pi r[/tex]
Solve for r
[tex]r=\frac{C}{2\pi}[/tex]
so
[tex]f(r)=2\pi r[/tex] ---> represents the circumference (in cm) of a circle whose radius is r cm
[tex]g(C)=\frac{C}{2\pi}[/tex] ---> represents the radius (in cm) of a circle whose circumference is C cm
The area of a circle is
[tex]A=\pi r^{2}[/tex]
solve for r
[tex]r=\sqrt{\frac{A}{\pi}}[/tex]
so
[tex]h(r)=\pi r^{2}[/tex] ---->represents the area (in cm2) of a circle whose radius is r cm
[tex]j(A)=\sqrt{\frac{A}{\pi}}[/tex] ---> represents the radius (in cm) of a circle whose area is A cm^2.
Part a. Use function notation to represent the area of a circle whose circumference is 140 cm
[tex]h(g(C))=\pi (\frac{C}{2\pi})^{2}=\frac{C^2}{4\pi}[/tex]
substitute the value
C=140 cm
[tex]h(g(140))=\frac{140^2}{4\pi}[/tex]
[tex]h(g(140))=\frac{4,900}{\pi}\ cm^2[/tex]
Part b. Use function notation to represent the circumference of a circle whose area is 5.18 cm^2.
[tex]f(j(A))=2\pi\sqrt{\frac{A}{\pi}}[/tex]
substitute the value
A=5.18 cm^2
[tex]f(j(5.18))=2\pi\sqrt{\frac{5.18}{\pi}}\ cm[/tex]
Answer:
A. H(g(140))
B. F(j(5.18)
Step-by-step explanation:
That’s how you write a FUNCTION NOTATION ...
