Daily Ohio statesman. [volume] (Columbus, Ohio) 1855-1870, 1867-12-28

of the sphere You ave taken the as equals. Il
>. sids of inscribed cube. sxs= given diameter
rnen K equal II plus. think.
prner ,s msks1 1a sns brothers SEND
brother, i. e., 4. b.. sends answers" &c. t s
-,.,- So. ANSWERS TO PROBLEMS.
No. al7-Woman will finish that rum" on Tues
day of thu fourth week. s-Ef
No. 3ls-x2 plus " equal SOD.
Multiply (2) by 2; add to and subtract from (D ex-
tract So root. and by adding clear off Now square
and take out as a common factor (sx minus A.. The
minus equal o. Hence equal . and ' equal 9.
the next sas is solved BY the Intesrai Cacaius %
into which sdnctcns sdnetoruus of mathematics I've
never placed profane feet, This problem IS solved
however. in August number of Our Schoolday Wis
No. ses-4 O'Clock and 5s1l minutes .,.,.
No. 32+-3l Sill minutes after2.
No. SsS--Area ssS Area 70 1000 G rods: required side 40.
No. 3:1-R Robins rate is IL miles per day.
No. 32S-Not time yet. Hf 's
Next three are mine.
No. 332-Minf axis of ellipse is not given. and
there Is no way to determine it. Assuming it TO be
the same as the maj axis. my answer is-Nary a
pusson, Tor Slpse and circle correspond .t.-
- Conventionally yours. DELTA.
WASHINGTON, o., Dec 10, 1867
Delta and Rave had ha ard day in comic sections;
consequently. my brain is all buzz with the dying
ech.es OF orjnates abscissas. oonjuga'es tangents
zeros. majors minors, squares, primes, ints ISL
and angular relations. and equations in psrabolas,
elipses and hyperb las. &c, DC. Hence turn weary
to the column, and find the following
ANSWERS.
No. 3l7-Rum would last 22H days. That is, old
woman gets the last sip on the twenty-third day
about noon. This il as it should be. Old man had
the first drink old lady the last and last word too
No. 3ls-Peculiarl Could not solve by quad
ratics pure Involved. and found root by Robin-
cess. So ution easy thus. Both and y=2.
No. 3l91Draw diagram per data and the right
lines "A little ca' eful geometrical comparson AID
aeductlon Will show the middle one of these angles
the side opposite equal TO inches TT gonometry
gives this angle=3S degrees, 80 minutes and 30
1;ds Since. therefore the two acute angles Of this
angle equal their difference. it follows that to add
e 20, and subtract i from right angle. and take
grees of the angle required. which are: "a degr es
~ minutes and IS seem IS, and 25 degrees 40 min
ntes and ls sec Jnds. This is concise method
No sao-Deta's cloth mea sures easily. Ladies'.
yard. an B's piece ya-is At IS per yard?
No 3as-Wiy. Otiose! But. '.go in" E. 3. tn
No 32s-Jack's watch indicated the time to be
ba minutss in. s past io'cl.ck p NJ
No 324-32 IL minutes after two. (Well, friend
Jack. YOU have brought us ali to time
No. 3,sIcFinaly. as the preachers sometimes
AT first it seemed fearful But on close inspection
Isfier construction) it solved under a gr IN Look
Falling to pieces like burnt limestone submerged
in water!
z sine Sine Radius Tang equals 40.
Hence, 50X20 equals 900 area.
So endeth the lessoa. GUERNSEY.
iAlthouth the Gnimus naatbenenticus WAS fix d in
The head by Polonan hooks of steel. [ believe tn It
The toothache Is capable to oxydize those hooks.
That t.ey will snap and allow the aniusus to fly
AWAY. At least it sufficient to drive all thought
ifmathematics from my mind for the last ten days,
out Will now attempt to recite
When first glanced over the column of December
LIST. Iinwardly exclaimed, behold, Samivel
NJ" How modest in exception. and. withal.
He appeared at the first reading to have surround
ed me with inextricable difficulties wall of iron
let in posts of adamant. whist the Delta had backed
ilgated his complications and perplexities more
lo,ely. they (the wall. Delta and alb foated away
sgenty and easily as the evanescent vapors of the
Problem No. 2sI-There is no five in the data. but
'dSTe N 1ssMI multiply nn G BY and divide by
radius of sphere multiplied by 1.I5f= side of
No. 27O-Am sorry Sammy" took me for such an
ss as to 6e in earnest in my remarks concerning
No. lsa-Do you always mean circle when you
No. los--When said any boy should understand
houd be aoe to work it, for there is great differ
Lee between ur terstanding the data and working
.lformation, not being able to solve it myself.
think it can be solved from the data given.
No 2s7-Bro. Sammy's answer is not correct 8 is
No. sss-Again not correct; vide my remark to In
MS relating u this problem.
Sammy, your ans wers to my problems are right
th3ug solutions are longer than necessary.
Delta gets in a stew over Theo em 200. He says
BRO. o. reasoning appl es to that case only"
assure him it il the only wanted it to apply
2 III You would reduce oushe of wteai to poun Il.
you wouldn't ere whether or not you could reduce
yards 70 inches by the same rule.
No. 3i3-6 andS tr
Method--Given. a equal hyp ,tenuse equal 10, andl
equal radius ins. eirci equal 2. Let and equa
siues. By applcat af Algebra to geometry we OBJ
tain this formula (Which NAV ba re luced f,rther)
ory equal the ,quare of (a2 divided by plu:
or minus the square root of al divided by minus
le of a and r. and you have the answer.
o, Nc; 3u-Irhe ram drinking viii be e,ded on th'
No: sls. The siies are. sa4, IS 91s. and 90: an
giles. n degrees. 33 min. and lS deg. q min;
pothsnuss of triangle Then the following equs
tions are evident.
: A. B. /. R,binson).
After .suo tituting. eliminating, &C.. we obtai
values of x, and z; and tn n find angles b, trigO
No. 800. A yds. at dollars per yd; b 8 yds a
sdos. per yd
No 1913. 53H min. after4.
No. 324. q minutes of 3.
No s25 Area=s6o 50. rods. Distance from as
TO pl le= 4O rods. Data W IS not printed plain in In
A / have not rected for two weeks. am neces
OT1osE
Nachez, u, Dec. Gath, 1867.
PETERSBURG. VA. Dec. Is. 1857.
Mr.. EDIroR-DeGr Sir: According to your ad
vice have worked on some of the Problems whic
l. have not answered:
ing the area Of segment; when the diameter of tn
circle and the hight of the segment are given; sine
we know the area of the segment and the diamete
of the CIRCLE we can easily find the height of tn
segment; subtract the heisht of the segment froa
the radius of the circle. the remainder will be on
the radius Of the circle. and the remaining side
half of tae chord. which cuts OFF onefour tn of th
No. 3O2-56 deg, 8 min., II sec.; 81 deg, 25 min
No. 3l4-l4 03. 7.2s. JACK.
DEcEMB-R R 20. 1867.
FRIEND PUZzLERs :-Once more bid the elias
show de de-" We had good column last weel
and through It al ther was that ge9tlemanly con
answers ss has all along characterized this goriou
column Like f, n: ~ feel better and thin
let better mathematicians take my room. As on
of The c1ntributors said good while ago, it seem
as would recognize nearly any every one of tn
friends at Brother Innis New Years eve ard ais is
prved from SO doing f. ON certsin private reasoa @
unavoidable. | thank friead Delta for answern
use In your last paper Simpson reasons like
phi losopher Of old. '.Simpson against the Puzzler
Iagree with Brother Oti Sc that it i, more salts
factory to have a free and full solution to fev
than the mere answers TO twenty Can this eQua
tion be answered, and if so. answer:
3xpus ss equals ~, value Of
Answer to No. 313-Perpend ioular s, base 6, by
pot eDuse10.
No. 3l4-One side As, other side 80.
SoUB IQUET.
CoLLINsvILLE, o., Dec a. 1857
MR.. EDIToR-I received my last paper yestel
day, and write my answer to-day; SO you see al
prompt. received no papers for couple of week
AMI SO hst nothing but gossip to write, and
thought the column would te better filled wit
mathematcs expect to be on hands at the con
the ladies too. +. b. has not said anything abou
attending. am very anxious to see him. for
to scold him. besides he needs no scolding. hav
male as many little innocent mistakes as any ont