Daily Ohio statesman. [volume] (Columbus, Ohio) 1855-1870, 1867-11-30
10 sec. Bearings N. SO deg. "a niu.. sec. W., s.
"a deg. II min. sec w. Value ,OsI 81.
No, 2sS. x4 plus x2 equals 9.
. WAsHINaron. al NOW as. ls67.

u .rl.
fear some Of contr butors from this corner are
becoming painfully prslix in our introductions to
NOTES Among these doubtless an
few words. and as well ordered as my professional
duties and the great difficulty Of the problems will
Spent much time on Sanmy's first. (No. 2s49 Find
| the same without the number in Legendre (DavieO
superior aSns Mith IN II sbe aas tSin
errs in regard to the radius Of the inscribed cir
data. radius cannot be much over feet sS inches.
lf more then the side Of square exceeds data. /
find sides Of triangle the FIRST hypo hess to be IS
isoscses: His second / think. have sovea. but
withhold solution till ! see others. may have as
No. 2ss By our new fsSd ieSd hsE. the Ter
ritry late State of Virginia. is to nv taste exact
fourth power and fourth power equal the mum,
bers Then finding the root Of eauaton BY the rule
sum of both members being equal to zero. - d
NO; 1ILSImpson's apple is funny. Boys must
No 2s3 -Otiose demands a solution of this: If K
plus ~ plus y3 xaa15. and these numbers squared
ami aided d=So. Place sum and sum of squares in
form of fraction. Suppress x in each term of BOTH.
Find root by above rule, Substitute and you have
Je~hiiikens" makes bold start with a long
name and proposition. hand the one to ''Ote."
Trigononetry. nor Il must say, of interest dis
count. GUERNSEY.
WASHINGTON. o.. Nov. 25.
DEAR PrzzLEs: Permit me to "enter my appear
ance" at that Convention. Brothers. leave for a
sluare, ana trianrles. and lets make jolly eircls
new. but pleasant--that Puzzer's Conventon--ands
worth remembering as long as We remain finite
quantities.
ANSWERS TO PROBLEMS.
No. ea6-s12sl 4Il9. --,, s.-- =
No. 2S6-l3 % S< lbs.
No. 2S7-s feet. -.,s.4si,l -.
No. 2sl-Find proportional parts in cubic inches.
Reduce A sphere subtract
diameter from diameter Of sphere when - efS out,
sSs rs 70 OF 'ho
No. 29taNot able to dIaM figure fer this problem
BY Those root Bro
G enter an equation. like the animals entered the
No. sss~Limestone SO oysters; Delta II oysters.
If that's the way the bi-vaives are be dshed out
recollect that am to be the quartcider-cup bearer
No. 297-1 mile and feet. - ,--...--.s
No. 29s-I4 equa 81; " equals 81.
DEMONSTRATIONS.
Theorem 2ss-Draw figure according to directions
of Bre Sammy. Required to prove the angle NAM
: Draw 40 parallel to Be. Then outer angle DAM
equal inner angle b; exterior angle CNA equals
Gao: Gao eaaals ps plus BAN; UAS ssuais j.
BAN; Gao plus CAN equals 9 b by adding DAM
sqpa aPSye Gao Gan plus A. squ
Hence N AM equals 5 B: This. proves the theorem;
Theorem 2sO-A B is right angled triangle.
is right angle, ID is perpendicular t. the hypote.
muse PROLONGED till it meets perpendicular erected
at A, ID being thu intersection; ID and e are
equiangular aud therefore similar, and D A B ::
b: b G. This proves the first proportion. Don't
yet see the other one. but b. pc to learn how next
Saturday. I'd, ts4t.'.-a- f'i'f H DELTA.
PROBLEM NO. 313. -4i
Hypotcnuse Of triangle is 10: radius of inscribed
circle is2. What are the sides? DELTA.
PROBLEM NO. 314.
Base of triangle is H. sum of squares of the other
sides is $50. What are they? DELTA.
PROBLEM NO. 315.
[For Limestone.1
Two garter snakes. eaci SO inches in length. un-
dertook to swallow each other at the time. tail
foremost. Suppose they progressed at the rate Of
TWO inches per minute. in what condition were they
at the end of 10 minutes?
p. s. Jealous advocates of Organic Union" be-
tween the Old and New School Presbyterian
chur hes are anxious to know for obvious reasons.
So. Brother l.. please bring up the answer: adT
PROBLEM NO. 316. 5tg5 q
Given X raised to 4nth power, minus OK raised to
3nsh power plus raised to nth power. equals (6),
six to find X. SAMMY.