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- amistre64

depends on what I have to do with smth :)

- angela210793

O.o..let me attach the problem pls...

- angela210793

U have to find the area of the circle...

##### 1 Attachment

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## More answers

- angela210793

it is a rhomb ***

- amistre64

i could see this as looking for the radius; which would indicate the a triangle in Q1

- amistre64

its a 3-4-5 right triangle

- amistre64

we then find the max area of a square that fits in it...maybe; or the max part of the circle to fit it

- angela210793

w8..i think i got how to solve it....all we have to do is find the height at ODC triangle and that=r :)

- angela210793

And then S=pi*r^2

- angela210793

Right?

- amistre64

##### 1 Attachment

- angela210793

this was what u were talking abt 2? Thnx a loot :)

- amistre64

the equation of the line that we want is:
(0,3) ; (4,0) ; the slope then being 3/4 or....
y = -(3/4)x + 3 right? that is the equation for the hyp ... that line of 5

- amistre64

we want a perp line to this that goes thru the origin (0,0) if anything

- amistre64

y = (4/3)x

- amistre64

the distance R is the distance from the origin to the point where these 2 lines meet right?

- angela210793

ehe...

- angela210793

r=12/5?

- amistre64

(4/3)x = (3/4)x +3
dunno yet still figuring this out the hard way lol

- watchmath

I think it should be
-(3/4)x=(4/3)x+4
Amistre...

- angela210793

w8 is easy...S(odc)= B*H/2=3*4/2=6
S(odc) is also=OE*DC/2=6----> OE=12/DC=12/5

- amistre64

the slope of the hyp I drew is -3/4 with a y int at (0,3) right?

- angela210793

I didn't get ur way......:(:(:( why with coordinates? O.o :(

- amistre64

my way is sound to me; i understand graphs and lines better than drawings

- watchmath

Your solution is nice angela :)

- angela210793

Wht abt my way? is it wrong?

- angela210793

Oh thnx :) :)

- amistre64

4 up and 3 over is the same region as 3 up and 4 over, ...

- amistre64

i dont really know if your way is right or wrong yet ;)

- angela210793

hmmmm.....

- watchmath

Her solution based on computing the area of ODC in two ways.
First: 1/2*3*4
Second* 1/2*CD*r = 1/2*5*r

- angela210793

Ehe...

- amistre64

(4/3)x = (-3/4)x + 3 [3]
4x = (-9/4)x +9 [4]
16x = -9x + 36
25x = 36
x = 36/25 did I get that right?

- angela210793

I didn't get ur way so i can't tell it if it is right or not....
Thank u very much anyway :):):):)

- watchmath

Amistre, the y-intercept is at (0,4)

- amistre64

4 36
-- * --- = y
3 25
4 12
-- * --- = y = 48/25
1 25
R = sqrt(x^2 + y^2)
R = sqrt((36/25)^2+(48/25)^2)

- amistre64

watch; it can be; but im actually using the other side; even tho I inadvertantly flipped it; it doesnt matter if we have 4 over and 3; or 3 over and 4 up; its the same triangle just flipped on itself with the same amount of circle it it ...

- amistre64

its the point of tangency that is important; in my figuring

- angela210793

let me read it from the start once again...

- amistre64

like this

##### 1 Attachment

- watchmath

Yes, your computation gives the same r :)

- angela210793

thank u both :)

- amistre64

R = 12/5 yay!! lol

- angela210793

:D ^_^

- amistre64

so that verifies your answer angela :) I know my way is solid, which is why I went that route

- angela210793

Will u see another 1 with inflexion points? (Is this how u call it?)

- amistre64

inflection points are where the graph stays along a course in a general direction.. sure

- angela210793

^_^ ok w8 :)

- angela210793

f(x)=x^4+c*x^3+x^2 what should b X so it will have 2 inflect. points? and then 1?

- amistre64

'X' or 'c' ?

- amistre64

f'= 4x^3 +3cx^2 +2x
f'' = 12x^2 +6cx +2 = 0 for inflection points

- angela210793

it says x...

- amistre64

x, as far as I can tell, is an input and has no real effect on the graph of this thing; the constants describe the path it takes

- amistre64

how are they defining inflection?

- angela210793

they who? O.o

- amistre64

they as in the people who designed the question, the authors of the material that you are using .... those guys

- amistre64

I see the answers; but the question should be what value should 'c' be...

- angela210793

C'mon...idk the words in english...............:S

- angela210793

ok u solve it 4 c...

- amistre64

f'' = 12x^2 +6cx +2 = 0 in order for this to have 2 real roots; we have to manipulate the discriminate of the quadratic formula....

- amistre64

36c^2 -4(2)(12) > 0 for 2 inflections
36c^2 -4(2)(12) = 0 for 1 inflection

- amistre64

36c^2 -96 = 0
36c^2 = 96
c^2 = 96/36
c = +- sqrt(96)/6 for one inflection

- amistre64

and greater than that for 2 inflections

- amistre64

http://www.wolframalpha.com/input/?i=sqrt%2896%2F36%29

- angela210793

whr did tht 36c^2 come from?O.o Florida? :P

- amistre64

it comes from taking the 1st and 2nd derivatives of the given equation

- amistre64

youd have to understand a little calculus to get the idea of it; but basically I used proven methods to find when the 2nd derivative = 0

- amistre64

y = x^4+c*x^3+x^2
y' = 4x^3 +3cx^2 +2x
y'' = 12x^2 +6cx +2 ; when y'' = 0 we have a point of inflection

- amistre64

y'' is a quadratic that can have its roots figured out by the quadratic formula; the discriminant of the quad formula tells us how many roots we will have

- amistre64

b^2 -4ac
(6c)^2 - 4(2)(12)
36c^2 -96 has to be greater than 0 for it to have 2 points for inflection
36c^2 -96 has to equal 0 for it to only have 1 point of inflection

- amistre64

so you solve the equations of the discriminate for c..

- amistre64

if we know what c is ro get the discriminate to = 0; then we should assume that larger values of c will make it greater then 0 for 2 roots

- angela210793

ahaaa.....i got it....Thnx ^_^

- angela210793

Will u help me in smth else?..... if u don't mind ofc...

- amistre64

i gotta finish my international business homework; its due today..

- angela210793

Oh..Ok..I'm sorry...didn't know tht....
Thnx a lot 4 helping me and good luck with ur hw :)

- amistre64

thnx :) and good luck with yours :)

- angela210793

^_^

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