heres what I got... I dont know if its right tho but I think it is. I dont know if I can simplify x^8 - x^2 and I (think ) that is the simplest form
https://prnt.sc/nbjsAXW2EXCV

Reply 5

username4158250

19

Original post by nrevenco

heres what I got... I dont know if its right tho but I think it is. I dont know if I can simplify x^8 - x^2 and I (think ) that is the simplest form
https://prnt.sc/nbjsAXW2EXCV

i got a very strange answer.. i got 50/12xroot3 -x^3

(edited 1 year ago)

Reply 6

nrevenco

OP

6

Original post by dahlia06

i got a very strange answer.. i got 50/12xroot3 -x^3

btw when its like 20N/5cm^2 if you want to divide the newtons by two you have to multiply the area by two and so on. basically just saying because I see you have 50N instead of 100N. This question is really hard bro idk how to it lol hopefully my teacher will tell me friday (next math lesson)

Reply 7

username4158250

19

Original post by nrevenco

btw when its like 20N/5cm^2 if you want to divide the newtons by two you have to multiply the area by two and so on. basically just saying because I see you have 50N instead of 100N. This question is really hard bro idk how to it lol hopefully my teacher will tell me friday (next math lesson)

im actually so confused
here's what i did
i worked out what 'x' is then worked out the area of circle as a whole .. wait **** i forgot to divide by two you're right give me a minute while i amend this.

Reply 8

username4158250

19

ok i got 25/24x^5-x^4... which actually looks even worse
so as i said, i worked out what x is so the side, then worked out the area of circle then took it away from the original area. and then i did pressure=force/area so then rearranged to find pressure and yh

(edited 1 year ago)

Reply 9

nrevenco

OP

6

Original post by dahlia06

ok i got 25/24x^5-x^4... which actually looks even worse
so as i said, i worked out what x is so the side, then worked out the area of circle then took it away from the original area. and then i did pressure=force/area so then rearranged to find pressure and yh

wait u dont need to rearrange it because its already in the correct format. also u dont need to find "x" it says in the question "give your answer in terms of x"

Reply 10

username4158250

19

Original post by nrevenco

wait u dont need to rearrange it because its already in the correct format. also u dont need to find "x" it says in the question "give your answer in terms of x"

yeah but then how else are you supposed to work out the area of the circle

Reply 11

username4158250

19

lemme try again... ffs

Reply 12

username4158250

19

i got 50x^2/12root3-pix^4

Reply 13

nrevenco

OP

6

Original post by dahlia06

i got 50x^2/12root3-pix^4

heres how I did it:
Area of hexagon / 6 = area of triangle
Area of hexagon / 12 = area of half triangle
Area of triangle = 1/2bc cos A
Find B and C
Find A using Pythagoras
Divide A by two and then do Pi * R^2 to get area of both BIG half circles
Then divide area AGAIN by two and do 4 * Pi * R^2 to get area of 4 smaller circles, then divide by two to get 2 total circles
then you do Big circles - small circles area and you get the area of shaded region
Then you do area of the hexagon (3absinc) - shaded area
That is the area which exerts pressure

Reply 14

username4158250

19

Original post by nrevenco

heres how I did it:
Area of hexagon / 6 = area of triangle
Area of hexagon / 12 = area of half triangle
Area of triangle = 1/2bc cos A
Find B and C
Find A using Pythagoras
Divide A by two and then do Pi * R^2 to get area of both BIG half circles
Then divide area AGAIN by two and do 4 * Pi * R^2 to get area of 4 smaller circles, then divide by two to get 2 total circles
then you do Big circles - small circles area and you get the area of shaded region
Then you do area of the hexagon (3absinc) - shaded area
That is the area which exerts pressure

seems more logical than my 'method' lmk what the answer is from ur teacher :smile:

Reply 15

mqb2766

21

Original post by nrevenco

heres how I did it:
Area of hexagon / 6 = area of triangle
Area of hexagon / 12 = area of half triangle
Area of triangle = 1/2bc cos A
Find B and C
Find A using Pythagoras
Divide A by two and then do Pi * R^2 to get area of both BIG half circles
Then divide area AGAIN by two and do 4 * Pi * R^2 to get area of 4 smaller circles, then divide by two to get 2 total circles
then you do Big circles - small circles area and you get the area of shaded region
Then you do area of the hexagon (3absinc) - shaded area
That is the area which exerts pressure

Sounds roughly right approach though I dont get your answer and some of the steps could be combined/simplified. With a bit of pattern spotting the area must have a factor like (6 - sqrt(3)x^2) as that occurs in alll the areas (hexagon and circles) so will be a factor of the total area. Your answer doesnt have that. Doing exact (surd) arithmetic gives a quadratic on the numerator and denominator which again, youd expect.

As a simpler way to derive it, assume the area of the hexagon is just A. So get the pressure as a function of A, then sub the expression for the given hexagon area (the quadratic in x.

Also thinking about the shaded area (well the r^2 part) as a scaling (by k) of the hexagon area will probably simplify your approach and you can write down most of the answer as
100x^2 / ((1 - k*pi)(6 - sqrt(3)x^2))

Must admit, I cant see any reason for specifying the original hexagon area like that. Nothing gives in the working and it simply obfuscates the answer. So 3/10 for question setting for your teacher.

(edited 1 year ago)

I need help on this question, any ideas? [Calculator P2 Higher]

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