How long until unisex multi-stall bathrooms are common at United States universities?
>>79065924
Probably 2017. It will be one of Hillary's first initiatives.
Those are called dormitories, though
>>79066070
Really?
>>79066639
Don't they just have male floors and female floors? You mean people in dorms don't give a fuck and use the wrong bathroom, or you mean dorm bathrooms aren't even gender-designated? I've never lived in a dorm but I've heard they are gender-designated like normal, but some floors have only male and others have only female.
>>79066800
I was implying that the dorms are the bathrooms. Piss and shit on the floors, chairs and beds.
Also, it is different school-to-school and different building-to-building even within one school.
When I went, it was a mixed gender dorm, three women and three men.
>>79065924
The pic is totally irrelevant, you could easily solve the problem
>protip its 80units
>>79066991
>I was implying that the dorms are the bathrooms. Piss and shit on the floors, chairs and beds.
I'm glad I never lived in one.
>>79067485
>The pic is totally irrelevant,
So?
you could easily solve the problem
Yep
>>protip its 80units
Nope, you're a fucktard
>>79067485
The point is that the shape couldn't exist.
>>79067784
Ah but it could, it's just not drawn to scale.
>>79067485
>you could easily solve the problem
run the math again
the shape is impossible. Someone jotted down random numbers and made a shape that couldn't exist
>>79068036
But that's wrong! You're as wrong as the guy that said 80 units!
>protip: not drawn to scale
>>79067896
>>79068214
you're mom is not drawn to scale
>>79067485
>>79067634
>>79067784
>>79068036
>>79068214
SLIDE THREAD
SLIDE THREAD
SLIDE THREAD
THIS IS POSTED VERY REGULARLY
INANE QUESTION + COMMON CORE IMAGE.
>>79065924
but anon, university campuses are the epicenter of rape culture with 7 out of ever 4 women being raped!
>>79067485
The shape is 70 square units
5 x 14 = 70
if you swing the left triangle over to the right to make a rectangle it becomes a 5 x 14 rectangle
you are adding extra length to the rectangle by doing 5 x 16
>>79069361
Yes but trannies are peaceful people! They would never rape anyone!
>>79069998
No you retard, 70 and 80 are both wrong
>>79070140
what is it if not 70?
it makes a perfect rectangle 5 x 14 = 70
explain how this is not the right answer, pro tip you can't
The answer is 91.25. You're all retarded.
>>79070260
70 and 80 and two of infinity correct answers. Therefore, 70 is not "the" right answer. Prove me wrong, protip: you can't. And no it doesn't make a perfect rectangle you fucktarded fucktard, what part of NOT DRAWN TO SCALE do you not fucking understand?
The lower limit of all possible areas of the shape approaches A=sqrt(1008)=31.749015733 from above. The upper limit of all possible areas of the shape is A=40+sqrt[((20+sqrt(25+(16-sqrt39)^2))/2)(((20+sqrt(25+(16-sqrt39)^2))/2)-(sqrt(25+(16-sqrt39)^2)))(((20+sqrt(25+(16-sqrt39)^2))/2)-12)(((20+sqrt(25+(16-sqrt39)^2))/2)-8)]=82.684891941.
Prove that wrong, protip: you can't (and yes the decimal numbers are approximations not exact answers, the non-decimal numbers are exact though)
>>79070831
Nope you're even more retarded than tha 70 and 80 retards because unlike 70 and 80, 91.25 does not even fall within the range of possible areas. Alch yourself.
>>79070260
this shape cannot exist.
from the Pytaghorean theorem, CA = BD = sqrt(39) > 6
now because AB is parallel to the top edge AB = 12
therefore CD = CA + AB + BD > 16
>>79067896
>american education
>>79070967
Area of triangle = (1/2) * (bh)
5^2 + b^2 = 8^2
b=6.25
(1/2) * (6.25*5) = 15.625 (2)
Square:
b*h
5*12 = 60
60 + 15.625 + 15.625 = 91.25
>>79071277
Ok I see what you mean with the triangles
5^2 + 2^2 =/= 8^2
so really they should not have even included the 8's on the sides, the question itself is fucked up
but if you ignore the 8's then the answer is 70
>>79071400
>not realizing I'm right
>br education>>79071277
>cannot exist
>Israeli education
>>79071420
>91.25
>clapistani education
Let me lay it out REALLY FUCKING CLEARLY for you incompetent fucktards.
We have a quadrilateral with sides 8, 12, 8, and 16; with the 8 sides being opposite each other; with it being possible to draw a line segment of length 5 from one of the angles adjacent to the 12 side to some point along the 16 side.
This is ALL WE FUCKING KNOW. Not enough information for a single solution, but we can get a range of solutions.
The lower limit to all possible solutions occurs when the angle between the left 8 line and the 16 line approaches 0. In this situation, the 5-unit line segment can be ignored. We therefore essentially turn the quadrilateral into a triangle of sides 12, 8, and 8. We can solve this with Heron's formula.
A=sqrt[s(s-a)(s-b)(s-c)]
s=(a+b+c)/2
s=14
a=sqrt(14*2*6*6)=sqrt(1008)=31.749015733
Since we are making the angle approach 0 rather than be equal to 0, the area is not technically equal to sqrt(1008), rather it approaches sqrt(1008) from above.
>>79072109
The upper limit occurs when the intersection of the 12 line and the left 8 line are as far from the 16 line as possible. Therefore we have a right triangle with sides 8, 5, and sqrt39 along with a quadrilateral with sides 5, 12, 8, and 16-sqrt39 where the sides of length 5 and 16-sqrt39 form a right angle. We can split the quadrilateral into two triangles, drawing a line segment from the upper left corner to the lower right corner.
By combining the lower left of these two triangles with the triangle we initially separated, we have a single triangle of base 16 and height 5, giving us an area of 40.
We are now left with one triangle of area 40 and one triangle with a side length of 12 and a side length of 8. In order to solve the area of the latter triangle, we need to determine either the remaining side or an angle. We cannot solve for any of the angles without first solving for the remaining side, so we'll just solve for the remaining side. To do this, we need to take the right triangle that we created in the middle of the figure whose hypotenuse is the same as the unknown side on the triangle we still have to solve. Given that we know the other two sides are 5 and 16-sqrt39, we can use the Pythagorean Theorem to determine the length of the hypotenuse as sqrt(25+(16-sqrt39)^2). We now have all three side lengths for the triangle of unknown area and can use Heron's formula to solve.
a=sqrt(25+(16-sqrt39)^2)
b=12
c=8
A=sqrt(s(s-a)(s-b)(s-c))
s=(a+b+c)/2
s=(20+sqrt(25+(16-sqrt39)^2))/2
A=sqrt[((20+sqrt(25+(16-sqrt39)^2))/2)(((20+sqrt(25+(16-sqrt39)^2))/2)-(sqrt(25+(16-sqrt39)^2)))(((20+sqrt(25+(16-sqrt39)^2))/2)-12)(((20+sqrt(25+(16-sqrt39)^2))/2)-8)]=42.684891941
Now add to the triangle of area 40 to obtain the area of the quadrilateral.
A=40+sqrt[((20+sqrt(25+(16-sqrt39)^2))/2)(((20+sqrt(25+(16-sqrt39)^2))/2)-(sqrt(25+(16-sqrt39)^2)))(((20+sqrt(25+(16-sqrt39)^2))/2)-12)(((20+sqrt(25+(16-sqrt39)^2))/2)-8)]=82.684891941.
>>79071852
Why do you feel compelled to arbitrarily ignore the 8s when you could just as easily ignore the 5, the 12, or the 16 (and any of those would be ignoring only one measurement, not two) and to assume that the shape is a trapezium when it is not stated to be one, you incompetent fucktard?
>>79071420
yah but if the base of each triangle is 6.25 the the bottom length would have to be 12 + 6.25 + 6.25 = 24.5
but they give a length of 16, so the shape is messed up look at the jew >>79071277
he has the answer
>>79071277
>>79071852
Americans getting a better education from /pol/ than their high schools.
>>79072280
The kike has the wrong answer dumbass
>>79072395
Except the kike is wrong and is teaching the clapistani nothing.
>>79072484
Nope, the jew is correct. It's called Pythagoras theorem. That shape with those measurements cannot exist.
I'm guessing that in america you don't learn trigonometry until college.
>>79072159
Hold on are we assuming that the two triangles on the left and right form 90 degree angles at the base or not?
if they "do" then this shape can't exist with the given measurements
I'm thinking that this question is messed up
if the triangles are right triangles then the base of the object should be 24.5 units long not 16
see here>>79072280
>>79072827
>Pythagoras theorem
I think the term you're looking for is 'Pythagorean Theorem', dumbfuck. And the Pythagorean Theorem is irrelevant to this problem.
>That shape with those measurements cannot exist.
Wrong. Pic related are the limits and >>79072109 and >>79072159 for an explanation.
>I'm guessing that in america you don't learn trigonometry until college.
Some people don't. I learned it during summer between 9th and 10th grade, then calculus during 10th grade.
>>79072958
>Hold on are we assuming that the two triangles on the left and right form 90 degree angles at the base or not?
Does the problem say they form 90 degree angles? No? Then fuck yourself with your unfounded assumptions, faggot. And by the way, I only see one triangle in that picture, not two.
>>79072484
The jew is right and the triangle is wrong. If the hypotenuse is 8 the catheti can't be 2 and 5.
>>79073201
>The jew is right
Nope
>and the triangle is wrong.
Nope
>If the hypotenuse is 8 the catheti can't be 2 and 5.
What the fuck are catheti?
>>79065924
70?
>>79073201
>The jew is right
Never thought I'd type these words...
>>79073011
It's called Pythagoras' theorem as well burger boy.
>>79073423
You left out the apostrophe though you fucktard.
>>79073011
Just from looking at the shape, would the area of the right one increase if you moved the 8 line so it is vertical?
>>79073278
Catheti(plural of cathetus) are the sides of a right triangle that aren't the hypotenuse. According to Pythagoras' equation the triangle is impossible. If the hypotenuse is 8 (whatever units) and one cathetus is 5 (units) long, the other cathetus can't be 16-12/2=2 long. In fact, it should be about 6,25 if we assume the triangles in the pic are equal in size.
>>79073131
>one triangle in that picture, not two.
one triangle on the left and one on the right
OK, if the triangles do not form a right angle then what is even the point of the measurement at all? we know that the base must be 16 and the 5 line could meet at any arbitrary point along the 16 line
simple answer is they fucked up the question by including the 8's on the side without them the answer is simply 70, with them we get an impossible shape
probably some idiot test maker thought "Hey those two sides ought to have measurements, I'll just throw an 8 on there"
But they did not realize that it messes up the whole shape
>>79072827
I can honestly tell you that the SAT and ACT standardized tests have a few math problems where the figure purposely isn't drawn to scale. They still want you to solve the problem based upon the numbers provided (I took my tests in 07 and 08). Trigonometry for me was bundled with geometry and pre-calculus. I took first semester calculus (AP calculus AB) as my year long senior math class. A smaller amount of seniors in my class, (maybe 25 students) were taking AP calculus BC, second semester college calculus.
I don't wanna provide my input on solving the problem though because it looks like I have gotten really rusty on trig. Still good with calculus even though I only do biology graduate coursework.
