1797-7 Mathematical Corre[pondence, 285 
2 
+ . (peegzinfar 2:2 523 Latest &c.) 
a = (pmmgzeh 12? nsz3 12} &c.)3 
4 
$2 pte etal te 
&e. 
(eb Bh-+y+itef &c.) 
$2 (iH HE +A Be) 
FE CHEE HOd Be) 
“a = (Aft 4tst. $c.) 
&c. 
AIT. Nowas the fquare, the cube, &c. of the polinomial pmmgz—br2?—123--tatene &c. == P, 
is likewife a polinomial of the fame kind: the firft term of P, raifed to the m** power multiplied 
air : | . 
by =~» is homologous with the fecond term of P raifed to the (m#—~r)th power multi- 
me 
I m=z 
~ = or with the third term of P raifed to the (m—z)'® power multiplied by ey 
— Nim 2, 
; ~m 
plied. by 


or with the fourth term of P saifed to the (m%—3)*” power multiplied by , and foon. Cone 

a 
fequently, by putting the coefficients of the homologous terms of one fide of the equation equal 
to the coefficients of the homologous terms of the other, we will have a™4™.)m—)jm_,m 
&c. — the firft term of P raifed to the m‘t power = i 

Pag fecond term of P raifed to the 
(7—1)™ power — X third term of P raifed to the (m—z)' term +. = x fourth 
—— a—_ 
term of P raifed to the (m—3)'® term- &c. 
Thus, becaufe (A—gzr2i—m &c)?—p?—297p2+(92Larp)ar— &e. 
(p—gztrei— &c.)3 = p3—3 7722-1 &e, 
and (p—q2+tr23—  &c.)4== pt— &e. 
We have wt boty tet eat &c. =p. 
OE Bb pee Se, ao pK gaap mag 
PLOL PPO ke =P x app + xrapmsemtsr 
3-1 
Merb pitdd fe, =p —— 2 xc 399 pt x (pare) —a taxes 
Tg a a ot a 
c. 
~ 

&c. 
Aberdeen, Sept.25, 1797+ fe CYGNt. 
[To be continued. | 

QuEsTIon XXX.—An/wered by Mr. T. Hickman. 
Ke On the given bafe AB, defcribe a fegment AHCB, capable of containing the 
given vertical angle. and complete the circle; draw the diameter HG bi- 
(44 feéting the bafe in D; divide the bafe in E, fo that AE is to EB, in the 
B given ratio; then through E, drawing the line GC cutting the circumference 
in C, and joining CA, and CB ; ACB is the triangle required. 
G ‘ 
Corollary 1. If the difference, inftead of the ratio, of the fegments, AE and EB, had been 
Biven, we had only to take DE = half the given difference, and proceéd as above. 
Corollary 2. If the difference of the angles at the bafe had been given (inftead of the ratio, or 
Gifference of the fegments) we had only to draw the line GC, making the angle HGC — half 
the given difference of the angles . 
Corollary 3. If the iegments whofe ratio is given, had been thofe made by the perpendicular 
(inftead of the line bifeéting the vertical angle), then divide the bafe in I, in the given ratio, 
and ereét the perpendicular IC. All which are too evident to need formal demonftrations. 
Solutions to this Queftion were alfo received from Mr. “ames Afion, Mr. Fokn Coilins, Rev. Mr. 
Fe Evans, Mr. Geo. Haworth, Wr, ‘fokn Fohnfon, Mr, R. Simpfor, Mr. Rd. Woed, and 
rom x=ly. 
Montary Mac. XXIII, Pp 
wed ernie 
