Help Please..!! Will Fan and Medal:)) The diameter of z is 5 in.What is its Area in terms of Pie??
A)2.5 in^2 B)5in.^2 C)6.25 in.^2 D)25 in ^2
\(\large 5\times .5=2.5\)
And 2.5 sqrd is 6.25 so A.
Omggg thank you:)) could you help me with like 2 more??:))
\(\bf \textit{area of a circle}=\pi r^2\qquad r=radius=\frac{diameter}{2}\to \frac{5}{2}\) solve for "r" to find the radius in terms of \(\Large \pi\) , as opposed to pie, though yours sounds more delicious
So its still A right?? @jdoe0001
well... what did you get for "r"?
hmmmm actually...you're just supposed to .... set it \(\pi \) terms... so \(\bf \textit{area of a circle}=\pi r^2\qquad r=radius=\frac{diameter}{2}\to {\color{brown}{ \frac{5}{2}}} \\ \quad \\ \textit{area of a circle}=\pi \left({\color{brown}{ \frac{5}{2}}}\right)^2\to \square ?\)
Hold on Whattt?? I dont understand any of that xD
well.. that's the equation for the Area of a circle... in this case, circle "z"
Ohhhh I see now...it showed a different equation because my OS is acting up
\(\bf \textit{area of a circle}=\pi r^2\qquad r=radius=\frac{diameter}{2}\to {\color{brown}{ \frac{5}{2}}} \\ \quad \\ \textit{area of a circle}=\pi \left({\color{brown}{ \frac{5}{2}}}\right)^2\to \pi (2.5)^2\to 6.5\pi\)
woops darn \(\bf \textit{area of a circle}=\pi r^2\qquad r=radius=\frac{diameter}{2}\to {\color{brown}{ \frac{5}{2}}} \\ \quad \\ \textit{area of a circle}=\pi \left({\color{brown}{ \frac{5}{2}}}\right)^2\to \pi (2.5)^2\to 6.25\pi\)
So its not 2.5 its 6.5??
:/
just got a typo =(
So it is 2.5??:))
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