Please help with 4 questions
ASAP medal+fan

- anonymous

Please help with 4 questions
ASAP medal+fan

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- schrodinger

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- anonymous

##### 1 Attachment

- anonymous

2-5

- anonymous

@mathmate @mathmale

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- anonymous

2-5

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- anonymous

Please help

- anonymous

@pooja195

- anonymous

@mathmate @mathwiz916 @mathmale @mathwizzard3 @MathematicalyExactOne

- anonymous

@mathmale

- anonymous

Can you please help

- anonymous

2-5 on this attachment

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- anonymous

please help for fan and medal

- anonymous

someone help!

- anonymous

i am so confused i have been working for 3 hours and got 1 problem done

- amy0799

ok, I think I know how to do this.
Can you tell me what the area of a circle is? and the circumference of a circle is?

- anonymous

a=pi times r

- anonymous

c=2pi times r

- amy0799

you forgot the squared for area.
A=pi(r)^2

- amy0799

for the first problem, you want to set the given circumference to the equation, 2pi(r)
352=2pi(r)

- anonymous

oh yeah

- anonymous

i got 9859

- anonymous

but for 2-5 i didn't get

- amy0799

It's basically the same set up.
For 2, 3 1/7 = pi(r)^2, solve for r, and plug it into the C=2pi(r)

- anonymous

what did you get

- amy0799

i got C=1.0272

- anonymous

Same

- anonymous

Let's do number 3

- anonymous

wait what did you get for number 1 i got 9859

- anonymous

is that what you got?

- amy0799

i got 9856

- anonymous

that's what i got the first time

- amy0799

that's due to rounding probably, i used 22/7 for pi

- anonymous

what did you get for number 3

- amy0799

did you already solve for it?

- anonymous

not yet i'm about to

- amy0799

i got my answer, tell me yours first

- anonymous

i don't get this one

- anonymous

i got something really off so i don't really want to say it

- anonymous

how do i solve this

- amy0799

the set up is exactly the same as question 2
15,400=2pi(r), solve for r and plug into C=2pi(r)

- anonymous

how do you solve for r

- amy0799

oops, that should be pi(r^2), because it's squared, you need to take the square root of it
so \[\sqrt{15,400/22/7}\]

- anonymous

i don't get this

- amy0799

first solve for 15,400/2pi, what do you get? use 22/7 for pi

- amy0799

i keep messing up, sorry, 15,400/pi

- anonymous

i got this

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- amy0799

that's because you divided it the other way around
divide 15,400 by 22/7
15,400/(22/7)

- anonymous

100

- amy0799

plug what i have typed exactly, with the parenthesis

- anonymous

4900

- amy0799

good, now you need to square root it

- anonymous

24010000

- amy0799

that's squaring it
square root is \[\sqrt{4900}\]

- anonymous

how do you square root it

- amy0799

do you have a calculator?

- anonymous

yes

- amy0799

is it Texas Instruments TI-30X?

- anonymous

yes

- amy0799

so hit the 2nd button on it, its in light blue, and then the square root symbol, which is under the ^ symbol

- anonymous

then what

- amy0799

plug in 4900 into the square root

- amy0799

did you get it?

- anonymous

1.001735585

- amy0799

for the square root of 4900?

- anonymous

yes

- anonymous

no 70

- anonymous

70

- amy0799

ok good, now plug in 70 into C=2pi(r)
r is 70

- anonymous

i got 440

- amy0799

yay! correct

- amy0799

do you understand how to solve the rest now?

- anonymous

finally

- anonymous

not really

- anonymous

but i'll try

- amy0799

do you still need my help?

- anonymous

probably sorry

- amy0799

it's fine, can you try 4, you got the correct answer on this on 1, so it's the same setup

- anonymous

can you tell me the steps one more time

- amy0799

you are given the circumference, 12(4/7),
12(4/7)=2pi(r), solve for r and plug in the value of r into A=pi(r)^2

- amy0799

tell me what you get for r first

- anonymous

bro i know this sounds wrong but can you please give me the answers i have to go to bed

- amy0799

im sorry i cant do that, it's really not that hard

- amy0799

ill give you some help

- amy0799

|dw:1449629566837:dw|

- anonymous

bro i have to go i will give you a fan and a medal im sorry

- amy0799

|dw:1449629664052:dw|

- amy0799

last one is 13690

- anonymous

Thanks bye

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