How would you solve this?

>>8394230
Suppose x < 9. Then 5x < 45. Then 5x + 7 < 52. Therefore if 5x + 7 >= 52, then x >= 9.

>>8394230
I'd probably use math.

>>8394237
Thanks, but I was talking about number 7

>>8394244
>number 7

Nigga we can't even see the whole problem.

>>8394244
I mean 6, my bad

>>8394247
See
>>8394248

bro this is some 6th grade level math

If both triangles are of equal height... and they share a hypoteneuse... they're both... right triangles... and this is Euclidean space... then the answer is obviously pi.

trig logs

>>8394238
Underrated.

Law of sines on triangles RSQ and STQ gives you

RQ / sin(90) = 8 / sin(<SRQ) = 8 / sin(<TRQ)

so that <SRQ = <TRQ and

<TQR = 90 - <TRQ = 90 - <SRQ = <SQR

>>8394230

Dear god I hope you don't mean number six. In that case you just got trolled hard by your textbook, lol. That problem has been proven unsolvable in 1886 by mathematician R. McGuire

>>8394230
Nigga are you dumb

File: 1474603797781.jpg (107KB, 640x640px) Image search: [Google]

107KB, 640x640px

>>8394238
LMAO

aren't they the same triangle????

>>8394346
Proof? Too lazy to Google it

43deg

>>8394472
Look at the picture and try to imagine it as a physical object and you'll realize that you're looking at the corner of a three dimensional. three-faceted pyramid. Later you'll learn that this has to do with the bond angles in its VSPER structure, but that's a subject for another class. At your level, it's acceptable to simply say that the angle is indeterminate, and the proof is trivial.

File: 1440104481132.jpg (38KB, 383x500px) Image search: [Google]

38KB, 383x500px

>OP is this retarded.

>>8394485
good bait

43°, No?

>>8394230

I was kind of being a dipshit before but now I'll be serious.


The angle is 43 degrees because they are equilateral triangles by side-angle-angle association, meaning that two sides and one angle are corresponding.

>>8394746
well yeah duh it's 43 degrees that ones labeled

it's asking about the other one, kike

Try to use rational geometry, it's superior by any means to regular geometry

>>8394762
>being so much of a brainlet you don't even know angle sum of a triangle

>>>/out/

>>8394230
The proof will depend on the knowledge you can use.
* SQ = QT implies that Q is on the perpendicular bisector of ST. (main property of a perpendicular bisector) (a)

* Pythagora is RQS : SQ^2 + RQ^2 = RS^2
* Pythagora in RQT : QT^2 + RT^2 = RS^2
* SQ = QT, so (1) and (2) make RQ = RT, that is : R is on the perpendicular bisector of ST. (b)
* (a) and (b) yields : if R and Q are on the same straight line, RQ is the perpendicular bisector of ST.

* Let's consider orthogonal symmetry % RQ.
* S(Q) = Q, S(R) = R, S(T) = S, so S(SQRT) = QRS).
* Orthogonal symmetry keeps angles and distances, ≤RQT = ≤RQS = 43°.

>>8394230
The top triangle is a copy paste of the bottom one but flipped.

Also, there is an 8 on both of them, so there should be a hint there. Also the fact that it's a right triangle.

43. You should know this.