When You Have Tests Tomorrow

Chat

jonah: WASH AD JEFF MAD MO AD JACK

jonah: VAN HAR TY POLK TAY FILL PIERCE

jonah: BUCH LINC JOHN GRANT HAYES GAR ART

jonah: CLEVE HAR CLEVE MC ROOS TAFT WIL

jonah: HARD COOL HOOV ROOS TRU EIS KEN

jonah: JOHN NIX FORD CART RE-A-GAN

jonah: BUSH CLINT BUSH

jonah: OOOOOOOOO!

jonah: Did I do it right?

cas: I have no idea what you did, so no.

jonah: I’m ignoring you.

jonah: 10 REDOS

jonah: ADAMS’S

jonah: J AND JQ, JOHN AND JOHN QUINCY

jonah: HARRISON’S

jonah: HARRISON, WE HAVE BAND, WH AND B

jonah: JOHNSON’S

jonah: A AND L

jonah: ROOSEVELT’S

jonah: T AND F, TRUE AND FALSE

jonah: BUSH’S

jonah: GHW AND GW

jonah: Did I do that right?

cas: Dunno.

jonah: I’m ignoring you, Cas.

jonah: LET’S GO

jonah: WASHINGTON ADAMS (J) JEFFERSON MADISON MONROE ADAMS (JQ) JACKSON

jonah: VAN BUREN HARRISON (WH) TYLER POLK TAYLOR FILLMORE PIERCE

jonah: BUCHANAN LINCOLN JOHNSON (A) GRANT HAYES GARFIELD ARTHUR

jonah: CLEVELAND HARRISON (B) CLEVELAND McKINLEY ROOSEVELT (T) TAFT WILSON

jonah: HARDING COOLIDGE HOOVER ROOSEVELT (F) TRUMAN EISENHOWER KENNEDY

jonah: JOHNSON (L) NIXON FORD CARTER REAGAN

jonah: BUSH (GHW) CLINTON BUSH (GW) OBAMA!

jonah: Did I do that right?

jonah: I think so.

crea: *applauds politely*

jonah: Thank you, thank you.

jonah: I’ll be here all year.

alex: You have a math test.

jonah: Heck yeah

jonah: CIRCUMCENTER IS PERPENDICULAR BISECTORS

jonah: BECAUSE IT’S THE ONLY WORD WHOSE LETTERS CAN BE SPLIT EVENLY SO THAT ONE SIDE SAYS “CENTER”

jonah: INCENTER IS ANGLE BISECTORS

jonah: CAUSE IT JUST IS

jonah: CENTROID IS MEDIANS

jonah: CAUSE IT’S THE ONLY ONE THAT DOESN’T END WITH CENTER AND MEDIANS ARE WEIRD

jonah: WHICH LEAVES ORTHOCENTER FOR ALTITUDES

jonah: BECAUSE YOU SHOULDN’T GET ORTHODONTICS DONE AT HIGHER ALTITUDES

jonah: Yeah

alex: How do you find them using the triangle’s coordinates? Assume A, B, and C to be the coordinates of triangle ABC.

jonah: FOR CIRCUMCENTER, YOU FIND THE MIDPOINT OF LINE SEGMENT AB. THEN YOU FIND THE SLOPE OF LINE SEGMENT AB AND USE THE PERPENDICULAR SLOPE OF IT. YOU PLUG THEM INTO THE POINT-SLOPE EQUATION AND SOLVE UNTIL YOU GET Y=MX+B. THEN YOU DO THE SAME FOR LINE SEGMENT BC. THEN YOU SET THE TWO EQUATIONS EQUAL, SO MX+B=MX+B. THEN YOU SOLVE FOR X. THEN YOU PLUG X INTO ONE OF THE ORIGINAL EQUATIONS AND SOLVE FOR Y AND THEN YOU PUT THEM IN THE COORDINATE PARENTHESIS LIKE (X,Y).

jonah: FOR THE CENTROID, YOU FIND THE MIDPOINT OF LINE SEGMENT AB. THEN YOU USE THIS COORDINATE AND THE COORDINATE OF C TO FIND THE SLOPE. THEN YOU PLUG THAT ALL INTO A POINT-SLOPE AND SOLVE TO GET Y=MX+B. THEN YOU DO THE SAME FOR MIDPOINT OF LINE SEGMENT BC AND THE COORDINATE OF A. THEN SET THEM EQUAL SO MX+B=MX+B AND SOLVE FOR X. THEN PLUG IN X AND GET Y AND PUT THEM IN THE PARENTHESIS.

jonah: FOR THE ORTHOCENTER, YOU FIND THE SLOPE OF LINE SEGMENT AB AND USE THE PERPENDICULAR SLOPE. THEN YOU PLUG THAT AND THE COORDINATES OF C INTO POINT-SLOPE, AND SOLVE FOR SLOPE-INTERCEPT. THEN YOU DO THE SAME WITH THE PERPENDICULAR SLOPE OF LINE SEGMENT BC AND THE COORDINATES OF A. THEN YOU SET THEM EQUAL SO MX+B=MX+B, SOLVE FOR X, PLUG X INTO THE EQUATION FOR Y, AND PUT THEM IN PARENTHESIS.

alex: What about the incenter?

jonah:

jonah: We never learned that.

alex: You will.

alex:  (x_I,y_I)=((ax_1+bx_2+cx_3)/(a+b+c),(ay_1+by_2+cy_3)/(a+b+c)).

alex: For Cartesian vertices and incenters.

jonah:

jonah: Okay.

alex: How do you indirect proof something?

jonah: FIRST, ASSUME THE “PROVE” IS FALSE.

jonah: THEN, SHOW THAT THE “PROVE” BEING FALSE CONTRADICTS WITH SOMETHING (THEOREM, POSTULATE, UNIVERSALLY ACCEPTED TRUTH, GIVEN, ETC., ETC.)

jonah: THEN, BECAUSE YOUR FALSE CONCLUSION IS FALSE, THEN THE CONCLUSION MUST BE TRUE SO THE PROOF IS PROVED

alex: Kay

crea: What about your Ing bomb test? Or the Judaical Branch test?

jonah: YOLO SWAG

zach: Shut up, Jonah.

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