Daily Ohio statesman. [volume] (Columbus, Ohio) 1855-1870, 1867-06-01
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noynt sue Spofe wan ve . sumctene Soiution, re
Answer to Problem No. 4s-l. s. 8. 1s.sas 80
'ta' . zs-,s and a.. . na ,.'. 80
t...-. ~ = - Al, Couarr1oHr.
N, r,s..o.. III CoIunBus. O.. May 2s, 1857
coming mere and more interesting; in fact. is !
marked feature in newspaperdom. IL Is q source Of
mpsa lmproyement to Young. and of Frightening ur
am glad to see the sood-natursd interest taken
last is quite interesting. shows that he can with
good feel ng correct himself when mistaken. These
corTections should only be made with view to mu-
HI Goulden's answer 70 Problem No. is correct
% Ssm "a SnS 5s ~
Answer to Problem No. 43: The parts are and le.
Answer to Problem No.As: Ny age IL my fathers
Answer 10 Problem No. so The square root of 8
minus I This rs : plain answer which nny good
ham send the WREKLY Ono SrATEsMANto cox sU7.
eAnsWer to Problem No. ~. Each face vii, bs
J. inches correct to four decimal plaaes.aa
.s :; IS. Bssss eEsss: ss;: e sss
ME. EpIroR.-My Problem No. 43 in the SArEs-
NAN Of May q, is wrongly stated. II should read
sum of their square roots viii be to AaS the cube
ToesSr their product as four times the less plus
If sny Of the ..bov., have han ..-r;nsn .n;.
crahbid radical. on insufficient data. and have not
succeeded in its solution-the fault may be mine-l
G ss tSs
the pieces of cloth Come Red breast the price is
sl2 so. not $le. But his Circle Problem. No 81 is
the radius of each to be So rods nearlyt Here re
the steps for obtaining it: 2 sif's en- as5 7
I Draw circles Of the same radius externally
tangent. .s0li e 81 ..Ii tEE faaa 8fi =
2: Connect eenters with straight lines; triang .
will pc thus formed both equilateral and squiangu
Letter the angles and the points where the esides
l. Suppose radius is Is rods then, any side of tri-
snelei 30, Now bisect any angle and let fall per-
pendeulsr, on opposite base. From this the area o.
contain 7Os Sc square rods But that part of the
sector and is onesxt6 Of the surface of this crce
since IL measured by the angle at the centre equal
BY IN ssS
piled by SOS z3 rods in all the sectors. we sub
tract from area Of triangle this area Of sectors 49,
curved triangle contained between the iunstures Of
the, circles. But thii by the hypothesi: OF the
RODS. RODS. RAD. RODS.
s7;,se;so eE; squared gives sgs1.'pd the square
Answer to Ne. 4sISon's age II. fsshs.'s4l.
/" :. s-Fhs number is Sass.
s. s. as. ss.
in1i And these nothngs are obtained in subtract-
ing and f4ctqIins in the supposed. Can ye see
Answer to Problem No. 44.-First number 8, sec-
ond 81. third %. fourth 81; a..,r ,..~zf..
would measure g. SAX inches '-''-N. ''~
These are b.,tK certainly the finest problems that
dsu ss sDE a
Root. Ss' and !. e g. s : l." for giving them, >.
fraction over 40 per cent. on his claim.
No. so cannot receive an exact answer. But Twill
pay for the WEEKLY STATESMAN. for one year to
Gou@n
send it TO another friend Dunham. on the same con-
Ihaye not yet caught up with Mattie's hoop-nor
nm Stur1a enough for No. IS. ~ =. = ~.
E the data ,as IF G6 No Sd oorreet' give it
/ .i,i.. sa5l COLUMBUS. OHIO, Maya, 1857.
Mr.. EDITOR: am glad to learn that you now
st Of The Sat and neS o. hope
TO receive answers to some Of my bems and rules
for their solutions Ibesan TO fear that you had not
them. They are Nos. O. l6 %, 81 and problem not
numbered. Immediately following No.s. ~ -
and 4. .Letxplusy=grestern number and ~ the
less. Then xSlo; Substilute tots in second squa.
know auern;sys method IF ,olviny this problem.
ed from the square root of6. ffbt ~
ANS e s0T,sS
numbers they both become then factoring. and
dividing by gives o instead of finite qusntity
Will Mr.. Earhart give his rule for solving No. 3s
least. ....--,-=e I'D s-.rr.. s,, .-.e. -,-. -.
Answer to No. ls-A in days, B in 8 days, will
Mr.. Dunham direct The WEEKLY STATESMAN to
sammy. Box 607. Columbus, Ohio, for answer to
MIR. ED1roR: ]. believe your paper is becoming
more than usSsily s PSpIIsr thTSughout the State in
Answer to No. zstEach face equals ssls7 inches.
.Answ ITO 57-Diameter is IS rods correct to is
You will please excuse my poor writing, as have
crippled my right hand and have to write with my
left hand. - ,t ,i.,,,-7 .-f... --- SAMMY.
Answer to problem No. 4s-The sons age. II the
fathers, Al.
s~ 49, 7'rgsr-t t-es- JOIN F. EARHAaT,
J aa.l ,-5suib1l3 st.ls i Columbus. Ohio.
l-.il 9al DUBLIN. o., May 25th, 1867.
MR. EDIroR.-I have not much leisure to spend
with your Puszler's Column yet find it very inter
and 49, which / see are correct think the answer
to No2z will hardly do. The question % lst nvolve
o Ss
radius of and the elevation OF the out
side rail- i only give this as my opinion.
-s : ANSWER TO PROBLEM NO. 2.
A. J. Courtwright. It seems is still in doubt as to
the size Of his turnip patch and the necessary rails
patch Of Rhoie 'islsnd. down in New England. and
will tutain 6S5,Sso acres. and req uire the same
Taking one square rod as the unit of measure,
each strip of ground rod wide across the
acres
TO number Of rails required to fence square
rod. and. therefore. must contain 81 a;res. which,
multiplied BY 100 rods. gives the leagth of the field
Il. .24o rods. Spare this number anddi vide b l6o,
and you have SIS $50 aeres 10, e gives 665
$50 rails.
81 ANSWER TO PROBLEM No. 10. = szz
i see problem No. 10 is still being discussed by
YOUR correspondents. It is simply to find the sum
Of all the series Of 4 geometric progression. The
first term .ol the number Of terms 50. the ratio2:
Raise the ratio to the 5st power. IL sss
Multiply by the FIRST Lerm.,....{ .0l
Las. term. -: Ii. ass sss. .oss 4as q.
Multiply this last term by ratio. y sds' 5fs5ts
s2,5l7. .s9s ls6 .s52 IS
Subtract the first term .......-.. - i5-' .Ol
Divide by ratio less I'D 2a 5I7. .s9s l3s, ,r59 n
Sum Of all the SERIES SIT sss. s52 a
This / believe is the true answer Your corre
spondents s. ID. K. and T. L. Rodehaven have for
correct. w ez- - LIMESTONE.
Tn y ANSWER TO PRoBLEM NO. so. n
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