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C-a

= 2w + 2

:: S =
b

G
C-a

b - a
:: P =

G

C-a and

bra apP + csS =(ap. + cs. i. apP + csS : bgG :: ap . (b − a) + cs (c – b) : bg, the analogy required.

924. First the question requires x, y, z, to be the integers. Then 29

y

5
X = +
36

3
.. 29x = 362 to 12y + 5

5

5
..y = 2.2

3z +
12

12
5
Let (x - 1) = w. Then
12
12w

2
-1=

+

36

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and w must be of the form 5p.
Hence

X = 1 + 12p
and
y = 2 + 29p

3z, which give the values required, z being assumed any whatever.

925. The earth being a spheroid, the latitude will be the length of the arc of the generating ellipse measured from the equator to the poles. Calling this arc s, and the distance required 3, by the equation to the ellipse we have

ba g =

1 - e'cos.28 Also if e denote the angular distance of g from the equator, and

p the I from the centre upon the tangent at the extremity of p, we have, from the equation to the ellipse,

aob? p =

;

a + 6 and the elemental triangle gives

pido

р

ds =

.

Hence we easily get

-62 do =

de e cos. O sin. 6.po

6? cos. A =

e pa

b? – (1 – eo)??, &c. sin.' 0 =

es and substituting, we have,

p2 - a2 +62 ds = a

xede. ( 6°) (a’ – p)

2

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Again, making a' – g = v* and substituting, we get

6% - u ds a

69 U?

x du

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which is integrable by the method expounded in pp. 144, Vol. II. and gives g in terms of s.

926. The diurnal path of a star is a circle inclined to the horizon of the place by an angle which measures the colatitude. Hence the orthographic projection of the circle upon the horizon is an ellipse, whose semi-axes are

a and a cos. colat. ( = a sin. lat.) a being the radius of the star's path.

927. By Wood's Optics, we easily obtain the distance of the focus of refracted rays (9) from that of incidental rays (Q).

Hence if A be the point of incidence on the refractor, the required thickness will be

Aq Qq-QA

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928. Let a be the side of the isoceles right-angled A, whose plane suppose to be inclined to the horizontal surface of the water at the angle La.

Then taking any element of the A parallel to the horizon ydor at the depth x sin. a, the pressure on that element acting I to the A is

ydx x x sin. a = (a – x) xdx sin. a, and the effort of this pressure to turn the A about its side at the surface of the water is

(a x) xdx sin. a X x. Hence the whole force from x = 0 to x = x, tending to make the A thus revolve, is

ar3

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Again, since the whole pressure on the A is

а. s (a - x) xdx sin. a = (

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sin. a

3

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:: if y

be the distance of the centre of pressure from the horizontal side of the A, the effort of the pressure all applied at this centre is

a’sin.cz
6

a* sin, a But this must equal the effort

Consequently 12

ху.

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The centre of pressure will also evidently be in the line joining the vertex and bisection of the horizontal base. Therefore the

1 distance required is

a

a

N 929. Let be the fraction in its lowest terms; then

D since every circulating decimal is equivalent to the vulgar fraction

A

10" — 1' where n is the number of digits in the period, and A the integer

N represented by those digits, if be a circulating decimal, we

D
have
N

A
D

10"-1 that is

10" - 1

Х N.
D

A =

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But since N is not divisible by D, 10% – 1 is divisible by D. Again, 10" i is prime to 10; for if not it is of the form 2 x P, 5 x Q or 10 x R, and its last digit is either even, 5 or 0, or 0, whereas the last digit is actually g. Hence that 10"– 1 may be divisible by D, D must also be prime

N to 10; that is that

D

may produce a circulating decimal, the fraction being in its lowest terms, the denominator must be prime to 10. Hence, conversely, &c.

930. The weight of the bodies are Pm and Qn in vacuo, and when immersed in water

Pm – P, Qn – Q. Hence, if x, a - x ; x', a - x' be the distances of the fulcrum from the extremities of the lever in the respective cases, we have

Pmx = Qn (a – x)

(Pm – P) x' = (Qn - Q) (a - x') which give

Qna

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Pm + Qn

Q.(n − 1) a

P (m-1)+Q (n-1) Hence the alteration required is

PQa (m-10) x - r'

(Pm+Qn) (P.m - 1+Q.n-1)

Х

931. Let l be the length of the cylinder, and M its bulk. Then by the question the weight of the part immersed is

5
M

5
M)=

M

12 the vertical tendency upwards of that part. The weight of the part not immersed is

M

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6

Hence by the property of the lever, if x be the distance of a fulcrum from the centre of gravity of the part which is out of the water, placed so as to produce an equilibrium between these forces

1

51
acting in the centres of gravity of the parts and
posite vertical directions, we have
M

5
X X = M + x)
12

12

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and in op

6

6

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Again the re-action of this fulcrum downwards is
M 5M

M

:)

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12

3

whence the magnitude direction, &c. of the force required.

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