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One morn a boy who lov'd to roam,
Along the meadows took his way, The vagrant wander'd far from home,
And playful pass'd the joyous day. As near a hedge he sauntering strolld,
He saw a Nightshade blooming gay,
And fruit that flam'd the ruby's ray.
And eat the fruit, and smelt the flowers;
The fatal berries' subtle powers. By his example warn'd, beware,
Lest you the same sad fate should meet; Pursuing pleasure, dread the snare,
That's spread to catch thy wareless feet.
ON SEEING A LAMB THAT WAS FATTENING FOR SLAUGHTER,
WANDERING through the mazy wild,
In life's gay morn, what joys are ours !
IN ALGARKIRK CHURCH, NEAR BOSTON, LINCOLNSHIRE, ON
JOHN BERIDGE, M.B. OBIIT OCT. 17, 1788.
These hallow'd stones an English heart infold,
LINES WRITTEN ON THE SEA-COAST.
Swift approaching to the main,
Flaming on the verge of hear'n,
Ah me! how little knows the human heart
The pleasing task of soft'ning others' woe; Stranger to joys that pity can impart,
And tears, sweet sympathy can teach to flow.
Pity the man who hears the moving tale
Ünmov'd; to whom the heartfelt glow's unknown, On whom the widows' plaints could ne'er prevail,
Nor made the good man's injur'd cause his own. The splendid dome, the vaulted roof to rear,
The glare and pride of pomp be, grandeur, thine : To wipe from misery's eye the falling tear,
And soothe th' oppressed orphan's woe be mine.
Be mine the blush of modest worth to spare,
To change to smiles affliction's rising sigh: The kindred warmth of charity to share,
Till joy shall sparkle from the tear-fill'd eye.
Can the loud laugh, the mirth-inspiring bowl,
The dance, or choral song, or jocund glee,
IN NO. V. ANSWERED.
1. Qu. (56) Answered by Mr. E. Webster, Armley
Mills. LET X?, 25x4, and 49x2, be the numbers required; then by the question, their common diff. 24x2 is to be a cube number, let nx he its root, then 24x2 = *3 n, and x 24
if n= 2, we find x = 3, consequently 9,225 and 763 441 are the numbers required.
, and the number required is (VS??)=
It was answered exactly in the same manner by Messrs. Gawthorp, Hirst, Rylando, and Whitley; and nearly so by Messrs. Baines, Brooke, Butterworth, Cattrall, Cummins, Dunn, Eyres, W. Harrison, Burton Pidsea, Hine, Maffett, Nesbit, Putsey, Tomlinson, and Winward. 2. Qu. (57) Answered by Messrs. Baines, Brooke, and
Whitley. Let m" denote the required number, then its square, cube, and biquadrate roots are respectively to, x*, and 7', and by the question, x + 2 = 2x; hence x3 –2x+ 1=0. One root of this equation is evidently 1, and dividing by r- 1, we have x + x-1=0, therefore X5
2 161 - 72 v5, whose square root is 9 475, its cube
and its biquadrate root 5-2. These roots form an arithmetical progression whose common
575 -11 diff. is
2 And thus nearly it was answered by Messrs. Butterworth, Cattrall, Cummins, Eyres, Gawthorp, W. Harrison, Hine, Hirst, Maffett, Nesbit, Putsey, Rylando, Tomlinson, Webster, and Winward. 3. Qu. (58) Answered by Messrs. Cattrall, Cummins,
Gawthorp, Hine, Maffett, Nesbitt, Rylando, and Winward.
Put x for the perpendicular, y the hypothenuse, and g= 16 feet: then by art. 300 Marrat's Mechanics, a body descending down y in the time t, will describe
gt's, which, in this case, is = y, therefore
y y = gtx, and the base is = ✓(gt-x --- ze). But the sum of the sides less the hypotheneuse is equal to the diameter of the inscribed circle; that is w (gtar x*)
Vigter + x = 4;-when t=1", we have (gx-x) - gr + x = 4, from whence x = 14.52, then y= 15.2816, and the base 4.7622.