Solve this problem, answers in comments pls.
in the figure ABCD is in a square. Calculate the length of side EF
https://pbs.twimg.com/media/ELNdxn3W4AIaFa1?format=jpg&name=4096x4096
Math problem IQ test
posted in
Off Topic
Solve this problem, answers in comments pls.
in the figure ABCD is in a square. Calculate the length of side EF
https://pbs.twimg.com/media/ELNdxn3W4AIaFa1?format=jpg&name=4096x4096
in the figure ABCD is in a square. Calculate the length of side EF
https://pbs.twimg.com/media/ELNdxn3W4AIaFa1?format=jpg&name=4096x4096
Don't you just use pythag theorem for AE, then pythag theorem again for EF?
edit: oh yeah lmao its an isosceles triangle and 1/2 of it is 5 so it's just 10 lmao
Don't you just use pythag theorem for AE, then pythag theorem again for EF?
edit: oh yeah lmao its an isosceles triangle and 1/2 of it is 5 so it's just 10 lmao
edit: oh yeah lmao its an isosceles triangle and 1/2 of it is 5 so it's just 10 lmao
thinly veiled homework thread for 5th grade homework
thinly veiled homework thread for 5th grade homework
Yeah, but at least now I know that I can read swedish math problems.
Now let him beräkna the längden on his own.
Yeah, but at least now I know that I can read swedish math problems.
Now let him beräkna the längden on his own.
Now let him beräkna the längden on his own.
inb4 some bot necros this thread a year from now advertising their maths homework website or some shit
inb4 some bot necros this thread a year from now advertising their maths homework website or some shit
1st triangle hypotenuse is sqrt(50) which is also side length of 2nd
So 2nd has hypotenuse sqrt100 or just 10
1st triangle hypotenuse is sqrt(50) which is also side length of 2nd
So 2nd has hypotenuse sqrt100 or just 10
So 2nd has hypotenuse sqrt100 or just 10
I good to figure this out without doing using Pythagorean identities is to look at this is with a little bit of geometry. Key things to notice are the 5 cm side lengths. AB and ED are both 5 cm, and the dotted line indicates that point D is the midpoint between CE. If one side of the midpoint is 5 cm, that means the other side is, so CD is also 5 cm. This shows that ABCD is a perfect square. Now figuring out EF is a little trickier, but visually we can see that point A is 5 cm above the line CE, and also makes a right angle with points F and E. We're given angle values, and both of these are 45 degrees. This means that if you drew a straight line from A to the middle of EF, the two triangles created are symmetrical. If one of these sides we know is the same length as AD, then the other must be as well. So we get 2 x 5cm, or 10 cm.
https://cdn.discordapp.com/attachments/629152685584809984/652987283607257098/20191207_163729.jpg
A picture to explain cutting the triangle into two symmetrical ones.
I good to figure this out without doing using Pythagorean identities is to look at this is with a little bit of geometry. Key things to notice are the 5 cm side lengths. AB and ED are both 5 cm, and the dotted line indicates that point D is the midpoint between CE. If one side of the midpoint is 5 cm, that means the other side is, so CD is also 5 cm. This shows that ABCD is a perfect square. Now figuring out EF is a little trickier, but visually we can see that point A is 5 cm above the line CE, and also makes a right angle with points F and E. We're given angle values, and both of these are 45 degrees. This means that if you drew a straight line from A to the middle of EF, the two triangles created are symmetrical. If one of these sides we know is the same length as AD, then the other must be as well. So we get 2 x 5cm, or 10 cm.
[img]https://cdn.discordapp.com/attachments/629152685584809984/652987283607257098/20191207_163729.jpg[/img]
A picture to explain cutting the triangle into two symmetrical ones.
[img]https://cdn.discordapp.com/attachments/629152685584809984/652987283607257098/20191207_163729.jpg[/img]
A picture to explain cutting the triangle into two symmetrical ones.
mirror BAEC along BA and you get a square with sides 10 cm, no calculations required
mirror BAEC along BA and you get a square with sides 10 cm, no calculations required
Arch_https://imgur.com/NGMT1Tl.png
Also I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long
[quote=Arch_][img]https://imgur.com/NGMT1Tl.png[/img][/quote]
Also I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long
Also I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long
5unnyArch_https://imgur.com/NGMT1Tl.pngAlso I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long
"smallest positive integer values"
"30 digits long"
nani the fuck?
[quote=5unny][quote=Arch_][img]https://imgur.com/NGMT1Tl.png[/img][/quote]
Also I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long[/quote]
"smallest positive integer values"
"30 digits long"
nani the fuck?
Also I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long[/quote]
"smallest positive integer values"
"30 digits long"
nani the fuck?
5unnyArch_https://imgur.com/NGMT1Tl.pngAlso I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long
90 digits solution:
Show Content
36875131794129999827197811565225474825492979968971970996283137471637224634055579/(4373612677928697257861252602371390152816537558161613618621437993378423467772036+154476802108746166441951315019919837485664325669565431700026634898253202035277992)+4373612677928697257861252602371390152816537558161613618621437993378423467772036/(36875131794129999827197811565225474825492979968971970996283137471637224634055579+154476802108746166441951315019919837485664325669565431700026634898253202035277992)+154476802108746166441951315019919837485664325669565431700026634898253202035277992/(4373612677928697257861252602371390152816537558161613618621437993378423467772036+36875131794129999827197811565225474825492979968971970996283137471637224634055579)
[quote=5unny][quote=Arch_][img]https://imgur.com/NGMT1Tl.png[/img][/quote]
Also I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long[/quote]
90 digits solution:
[spoiler]36875131794129999827197811565225474825492979968971970996283137471637224634055579/(4373612677928697257861252602371390152816537558161613618621437993378423467772036+154476802108746166441951315019919837485664325669565431700026634898253202035277992)+4373612677928697257861252602371390152816537558161613618621437993378423467772036/(36875131794129999827197811565225474825492979968971970996283137471637224634055579+154476802108746166441951315019919837485664325669565431700026634898253202035277992)+154476802108746166441951315019919837485664325669565431700026634898253202035277992/(4373612677928697257861252602371390152816537558161613618621437993378423467772036+36875131794129999827197811565225474825492979968971970996283137471637224634055579)
[/spoiler]
Also I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long[/quote]
90 digits solution:
[spoiler]36875131794129999827197811565225474825492979968971970996283137471637224634055579/(4373612677928697257861252602371390152816537558161613618621437993378423467772036+154476802108746166441951315019919837485664325669565431700026634898253202035277992)+4373612677928697257861252602371390152816537558161613618621437993378423467772036/(36875131794129999827197811565225474825492979968971970996283137471637224634055579+154476802108746166441951315019919837485664325669565431700026634898253202035277992)+154476802108746166441951315019919837485664325669565431700026634898253202035277992/(4373612677928697257861252602371390152816537558161613618621437993378423467772036+36875131794129999827197811565225474825492979968971970996283137471637224634055579)
[/spoiler]
Rahmed5unnyArch_https://imgur.com/NGMT1Tl.pngAlso I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long
"smallest positive integer values"
"30 digits long"
nani the fuck?
[quote=Rahmed][quote=5unny][quote=Arch_][img]https://imgur.com/NGMT1Tl.png[/img][/quote]
Also I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long[/quote]
"smallest positive integer values"
"30 digits long"
nani the fuck?[/quote]
https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4/answer/Alon-Amit
Also I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long[/quote]
"smallest positive integer values"
"30 digits long"
nani the fuck?[/quote]
https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4/answer/Alon-Amit
I'm not sure why you all can't count but it's 80/81/81 digits.
I don't think that's the smallest solution either, just one that was relatively easy to find.
Either way there are two ways to solve it. The proper way involves a lot of number theory, the quick and dirty way still requires you to be smart about it and probably a computer to brute force a part of it because of the numbers involved.
Essentially it is easy to get any solution, fairly easy to get one with positive numbers and incredibly easy to get one with whole numbers from one with rational numbers (just multiply all 3 numbers with all 3 denominators). Getting a positive and rational solution takes some fiddling though, either via brute force with numbers that a standard calculator won't like or a deep enough understanding of number theory.
I'm not sure why you all can't count but it's 80/81/81 digits.
I don't think that's the smallest solution either, just one that was relatively easy to find.
Either way there are two ways to solve it. The proper way involves a lot of number theory, the quick and dirty way still requires you to be smart about it and probably a computer to brute force a part of it because of the numbers involved.
Essentially it is easy to get any solution, fairly easy to get one with positive numbers and incredibly easy to get one with whole numbers from one with rational numbers (just multiply all 3 numbers with all 3 denominators). Getting a positive [b]and[/b] rational solution takes some fiddling though, either via brute force with numbers that a standard calculator won't like or a deep enough understanding of number theory.
I don't think that's the smallest solution either, just one that was relatively easy to find.
Either way there are two ways to solve it. The proper way involves a lot of number theory, the quick and dirty way still requires you to be smart about it and probably a computer to brute force a part of it because of the numbers involved.
Essentially it is easy to get any solution, fairly easy to get one with positive numbers and incredibly easy to get one with whole numbers from one with rational numbers (just multiply all 3 numbers with all 3 denominators). Getting a positive [b]and[/b] rational solution takes some fiddling though, either via brute force with numbers that a standard calculator won't like or a deep enough understanding of number theory.
5unnyRahmedhttps://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4/answer/Alon-Amit5unnyArch_https://imgur.com/NGMT1Tl.pngAlso I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long
"smallest positive integer values"
"30 digits long"
nani the fuck?
Yeah it's a really fucked up one, the site even says under those circumstances there exist no solutions. Without the positive part, there would
[quote=5unny][quote=Rahmed][quote=5unny][quote=Arch_][img]https://imgur.com/NGMT1Tl.png[/img][/quote]
Also I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long[/quote]
"smallest positive integer values"
"30 digits long"
nani the fuck?[/quote]
https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4/answer/Alon-Amit[/quote]
Yeah it's a really fucked up one, the site even says under those circumstances there exist no solutions. Without the positive part, there would
Also I'm pretty sure I saw this somewhere and the solution that has the smallest positive integer values are like 30 digits long[/quote]
"smallest positive integer values"
"30 digits long"
nani the fuck?[/quote]
https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4/answer/Alon-Amit[/quote]
Yeah it's a really fucked up one, the site even says under those circumstances there exist no solutions. Without the positive part, there would
hey bro you should wash ur hands them nails lookin kinda dirty
hey bro you should wash ur hands them nails lookin kinda dirty
Athletickhey bro you should wash ur hands them nails lookin kinda dirty
not my nails, sorry
[quote=Athletick]hey bro you should wash ur hands them nails lookin kinda dirty[/quote]
not my nails, sorry
not my nails, sorry
Why are there so many backwards A's and E's in the definition of a convergent sequence wtf
Why are there so many backwards A's and E's in the definition of a convergent sequence wtf
toads_tfnegative area square?
negative area makes sense if you consider it as the "area under a function" (so a defined integral of a function)
and you could just handwave it anyways.
This looks like a "find the overlapping area of three circles" question. It's extremely tedious, but pretty braindead with a symbolic calculator
Here's a very rough outline of how to find A3 (not wasting any more time on this):
Show Content
let's start with an A_s (assuming it's the outer square) of 1 and multiply by the actual A_s at the very end
https://i.imgur.com/2GSGwQ1.png
I guessed that the vague curves are sections of circle with center at a corner of the large square and radius the same as the square's side length
https://i.imgur.com/qny2VWS.png
the program gives the equations for the circles easily
https://i.imgur.com/mzP36RE.png
change the relevant half of the circle into a y=f(x) format
https://i.imgur.com/y15qo1k.png
https://i.imgur.com/53bLMWX.png
find the intersections of the relevant curves, and use them as the bounds of the defined integrals..
https://i.imgur.com/VqDoIGT.png
and just smack the integrals into a calculator, and you add them all together:
+ integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) dx
- integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) + 1/2 (-1 + 2 sqrt(x - x^2))) dx
+ integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx
- integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx
+ integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(2 x - x^2)) dx
- integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(1 - x^2)) dx
and multiply the result by the side length
and the final answer for A3 (unless I made mistakes....)
surprise, it's a massive clusterfuck:
(-47)*(9/64 - sqrt(3)/8 - (5 π)/12 + tan^(-1)(4 + sqrt(7)) + 1/64 (-17 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)((2 sqrt(5 - sqrt(7)))/3)) + (1/64 (sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + 8 (-3 + 4 sqrt(3) - sqrt(7) + 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (5 π)/12) - (1/64 (-9 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)(2 sqrt(3 - sqrt(7))))) + (-1/64 sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + π/3 + 1/8 (-1 + sqrt(7) - 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (1/192 (3 - 24 sqrt(3) + 24 sqrt(7) + 32 π - 96 cos^(-1)(1/8 (5 - sqrt(7))))))
= ~0.036566432691702221271670 - 0.019630731925696712432468 + 0.061918112844524846486709 - 0.0150274870976263361702352 + 0.04230790314076436373846 - 0.0173897792991522791591702
= 0.0887444503545161037349656 * (-47)
= ~-4.1709891666622568755433832..
and then just do that two more times for the final answer. It's not going to get much shorter.
Of course the question is pretty vague so this could be the solution to the wrong problem.
https://i.imgur.com/2GSGwQ1.png
I guessed that the vague curves are sections of circle with center at a corner of the large square and radius the same as the square's side length
https://i.imgur.com/qny2VWS.png
the program gives the equations for the circles easily
https://i.imgur.com/mzP36RE.png
change the relevant half of the circle into a y=f(x) format
https://i.imgur.com/y15qo1k.png
https://i.imgur.com/53bLMWX.png
find the intersections of the relevant curves, and use them as the bounds of the defined integrals..
https://i.imgur.com/VqDoIGT.png
and just smack the integrals into a calculator, and you add them all together:
+ integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) dx
- integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) + 1/2 (-1 + 2 sqrt(x - x^2))) dx
+ integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx
- integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx
+ integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(2 x - x^2)) dx
- integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(1 - x^2)) dx
and multiply the result by the side length
and the final answer for A3 (unless I made mistakes....)
surprise, it's a massive clusterfuck:
(-47)*(9/64 - sqrt(3)/8 - (5 π)/12 + tan^(-1)(4 + sqrt(7)) + 1/64 (-17 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)((2 sqrt(5 - sqrt(7)))/3)) + (1/64 (sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + 8 (-3 + 4 sqrt(3) - sqrt(7) + 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (5 π)/12) - (1/64 (-9 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)(2 sqrt(3 - sqrt(7))))) + (-1/64 sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + π/3 + 1/8 (-1 + sqrt(7) - 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (1/192 (3 - 24 sqrt(3) + 24 sqrt(7) + 32 π - 96 cos^(-1)(1/8 (5 - sqrt(7))))))
= ~0.036566432691702221271670 - 0.019630731925696712432468 + 0.061918112844524846486709 - 0.0150274870976263361702352 + 0.04230790314076436373846 - 0.0173897792991522791591702
= 0.0887444503545161037349656 * (-47)
= ~-4.1709891666622568755433832..
and then just do that two more times for the final answer. It's not going to get much shorter.
Of course the question is pretty vague so this could be the solution to the wrong problem.
[quote=toads_tf]negative area square?[/quote]
negative area makes sense if you consider it as the "area under a function" (so a defined integral of a function)
and you could just handwave it anyways.
This looks like a "find the overlapping area of three circles" question. It's extremely tedious, but pretty braindead with a symbolic calculator
Here's a very rough outline of how to find A3 (not wasting any more time on this):
[spoiler]
let's start with an A_s (assuming it's the outer square) of 1 and multiply by the actual A_s at the very end
https://i.imgur.com/2GSGwQ1.png
I guessed that the vague curves are sections of circle with center at a corner of the large square and radius the same as the square's side length
https://i.imgur.com/qny2VWS.png
the program gives the equations for the circles easily
https://i.imgur.com/mzP36RE.png
change the relevant half of the circle into a y=f(x) format
https://i.imgur.com/y15qo1k.png
https://i.imgur.com/53bLMWX.png
find the intersections of the relevant curves, and use them as the bounds of the defined integrals..
https://i.imgur.com/VqDoIGT.png
and just smack the integrals into a calculator, and you add them all together:
+ integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) dx
- integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) + 1/2 (-1 + 2 sqrt(x - x^2))) dx
+ integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx
- integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx
+ integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(2 x - x^2)) dx
- integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(1 - x^2)) dx
and multiply the result by the side length
and the final answer for A3 (unless I made mistakes....)
surprise, it's a massive clusterfuck:
(-47)*(9/64 - sqrt(3)/8 - (5 π)/12 + tan^(-1)(4 + sqrt(7)) + 1/64 (-17 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)((2 sqrt(5 - sqrt(7)))/3)) + (1/64 (sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + 8 (-3 + 4 sqrt(3) - sqrt(7) + 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (5 π)/12) - (1/64 (-9 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)(2 sqrt(3 - sqrt(7))))) + (-1/64 sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + π/3 + 1/8 (-1 + sqrt(7) - 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (1/192 (3 - 24 sqrt(3) + 24 sqrt(7) + 32 π - 96 cos^(-1)(1/8 (5 - sqrt(7))))))
= ~0.036566432691702221271670 - 0.019630731925696712432468 + 0.061918112844524846486709 - 0.0150274870976263361702352 + 0.04230790314076436373846 - 0.0173897792991522791591702
= 0.0887444503545161037349656 * (-47)
= ~-4.1709891666622568755433832..
and then just do that two more times for the final answer. It's not going to get much shorter.
Of course the question is pretty vague so this could be the solution to the wrong problem.
[/spoiler]
negative area makes sense if you consider it as the "area under a function" (so a defined integral of a function)
and you could just handwave it anyways.
This looks like a "find the overlapping area of three circles" question. It's extremely tedious, but pretty braindead with a symbolic calculator
Here's a very rough outline of how to find A3 (not wasting any more time on this):
[spoiler]
let's start with an A_s (assuming it's the outer square) of 1 and multiply by the actual A_s at the very end
https://i.imgur.com/2GSGwQ1.png
I guessed that the vague curves are sections of circle with center at a corner of the large square and radius the same as the square's side length
https://i.imgur.com/qny2VWS.png
the program gives the equations for the circles easily
https://i.imgur.com/mzP36RE.png
change the relevant half of the circle into a y=f(x) format
https://i.imgur.com/y15qo1k.png
https://i.imgur.com/53bLMWX.png
find the intersections of the relevant curves, and use them as the bounds of the defined integrals..
https://i.imgur.com/VqDoIGT.png
and just smack the integrals into a calculator, and you add them all together:
+ integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) dx
- integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) + 1/2 (-1 + 2 sqrt(x - x^2))) dx
+ integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx
- integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx
+ integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(2 x - x^2)) dx
- integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(1 - x^2)) dx
and multiply the result by the side length
and the final answer for A3 (unless I made mistakes....)
surprise, it's a massive clusterfuck:
(-47)*(9/64 - sqrt(3)/8 - (5 π)/12 + tan^(-1)(4 + sqrt(7)) + 1/64 (-17 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)((2 sqrt(5 - sqrt(7)))/3)) + (1/64 (sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + 8 (-3 + 4 sqrt(3) - sqrt(7) + 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (5 π)/12) - (1/64 (-9 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)(2 sqrt(3 - sqrt(7))))) + (-1/64 sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + π/3 + 1/8 (-1 + sqrt(7) - 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (1/192 (3 - 24 sqrt(3) + 24 sqrt(7) + 32 π - 96 cos^(-1)(1/8 (5 - sqrt(7))))))
= ~0.036566432691702221271670 - 0.019630731925696712432468 + 0.061918112844524846486709 - 0.0150274870976263361702352 + 0.04230790314076436373846 - 0.0173897792991522791591702
= 0.0887444503545161037349656 * (-47)
= ~-4.1709891666622568755433832..
and then just do that two more times for the final answer. It's not going to get much shorter.
Of course the question is pretty vague so this could be the solution to the wrong problem.
[/spoiler]
geogebra is taking our jobs
edit: wait i think thats desmos
geogebra is taking our jobs
edit: wait i think thats desmos
edit: wait i think thats desmos
Please
sign in through STEAM
to post a comment.