find the shaded area. When appropriate, use the pie key on your calculator and round your answer to the nearest hundredth. 5in 10 in-shaded area
i'm here! what is the question?
yes!
the shape is an flipped U ad the shaded part is 10in ------- 5in ---------- 10in
okay, you want the area of that shape? what part is shaded?
boy, there's no one on here tonight! SO many unanswered questons!
yes i know
its the inner part and like i said its a U SHAPE
upside-down U....what is 5, the side or the radius?
Yes upside down the space is open and is like this ( ) | | |--- 5in ----- | |----- 10in |
ahmmm....still doesn't help.....the bottom is open? are u trying to give me the shape?
yes
does it look like an archway with distance across the bottom 5 and height 10??
and is the entire shape shaded?
just the 10inch part
where's the 5?
the 5 is above the 10
hmm....you have an arch shaped figure....with a line across it near the top? is the shaded part below the line?
its basically an inner layer is shaded and |-----------| 10 in
still not seeing it...do you have skype??
no i dont have that but my text book is online
Oh!! what's the address for it and I will look!
ok can u shoot me ur email
yes use lchapman@odu.edu
wait ok laurie
what it is is a half of circle
ok....like the end of track?
but inside the half of circle theres an lil circle like a donut shape if u think of a donut the also a hole like a lil circle shapin the inside
inside the circle is shaded which is 10 in the middle circle is 5 in
http://digitalbookshelf.southuniversity.edu/#/books/0558542549/pages/17256184
#28
this is taking a bit to get the site, etc....shouldn't be much longer
ok
what's the name of the book?
theres no name
did it take u to the book there u see angels?
I can't see the book....I created an account , but don't know where to go to from there....it wouldn't take me straight to the page etc.....do you have an account of something to get into the site? Or did you ahve to pay for the book?
think I have your picture though.....ok, is like a washer or donut cut in half with the "hole" being 5 inches across and the entire donut being 10 inches across?
no what is it taking u to a sign in?
the "book" site says I don;'t have a site license for A Survey of Mathematics with Applications
o i should have took u through the login but u pictured it correct
yes, I set up an account and did the download, but I don't have site license for this...okay...the picture is trght??
So from side to side is 10 inches, shown by -----10------- at the bottom of the picture, right?
yes to the pic
ok, it hung up for a minute.....start it this way, draw a complete donut or washer with a hole in the middle.....you
need to find the are of the donut .....it's a circle, so that is Area=pi r^2
the dia,meter of thefigure is 10, which means the radius is 5 (not the 5 given in the picture....so you'll have Area = pi 5^2.....now don't use the calculator pi button until the end of the problem....!
You with me?
now the area of a solid jelly donut looking picture is 25pi....
No you need to find the area of the hole that is cut out of the middle.....using the same area formula A=pi r^2 the distance across the "hole" is 5, so it's radius is 2.5, so plug into the formula and A (of the hole) =pi (2.5)^2
ok can u write the steps with ur fromula so i would know how to write it all out step by step
okay ....look at this first....so now you need to subtract the 2 areas....25pi-(2.5)^2 pi
are you getting the idea of what I'm doing?
1. find the area of an entire circle or donut
2 find the area of the hole
3. subtract
Now cut the entire thing in half divid by 2
ok let me restart
my pc
ok.....Part One....A=pi r^2 (write the formula down first
I had to refresh mine a minute ago
plug in the info A= pi (5)^2 = pi 25 or 25 pi
next step or part of the problem A=pi (2.5)^2 = 6.25 pi (check my math, please! no calc here
Now, third part , subtrract the two Areas STILL withoug inputting pi 25-6.25 = 18.75 I think
NOW input pi.....so type in 18.75 and hit the pi button THEN divide by 2
you therer?
found the calculator....answer is 58.90486225.....now you need to round it to what the problem says to do....
thanx laurie i will tell u if its correct when graded
it was incorrect
my prof says to find the area of the bigger semi-circle, subtract the area of the smaller semi-circle from it to get the area of the shaded region
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