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10.

(n-1)p-(m-1)9, 1-1; first term, 11; common diff., 3; or first

n-m

n-m

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3.

- 300°, - 210°, – 120°, – 30°, 60°, 150°, 240°, and 330°.

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5. 5/3 = 8.66 lbs., 5/7 = 13.22 lbs.

6. 45.

7. 180°cos s-1" to the direction of motion of the carriage. 8.

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2u

9. W lbs.-wt. vertically upwards; when moving with uniform velocity force = W lbs.-wt.

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3. The asymptotes are parallel to AB, AC bisecting the perpendiculars from O on AB, AC.

4. (a2b3 - a3b2) (b1x – a1y) = a1(a263 - A3C2) + 61(6263-6362) and two similar

[blocks in formation]

a2-b2 1(1 - m2) b2-a2 m(1-12)
12+ m2
12+ m2
where W= wt. of sphere,

Wp
csin B'
and the perpendicular from its point of contact to side BC.

8. Tension at A = 1 wt. of board +

4m - m' 4(m + m')

5m
4(m+m')

10.

. И,

. u.

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11. 9, 15.

a

10. 3430 is 4th term G. P., 3475 is 496th term A.P.; + 2n+I

II. PART I. SECOND PAPER.

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10. (i.) ab = 1, (ii.) ab = x, (iii.) a = b2, where a, b are the given numbers.

IV. PART II. FIRST PAPER.

1. acx2+2(a+c) bx + (a+c)2 = 0. 4. 9:2. 6. 86 4307 ft.

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10. 53.76.

V.√3.

√3=4.29 f.s.

11.

12. V.

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5.

12. Ngr.

3. 6x + 7y = II, 4.

cr. (1-1), radius I; the origin and the pt. - 26, 2 6. y = (x + 1).

9.

7. (a2-c2) x2 + a2y2 = a2 (a2 – c2).

8. 6. If A, B, C be the corners at which the tensions 1, 2, 3 act respectively,

the weight must be placed at the mid. pt. of CD, where AD = 2DB.

V3 seconds.

12.

. M1Ma
m1+ m

10. W.

11.

.8.

[blocks in formation]

a

10.

3430 is 4th term G.P., 3475 is 496th term A.P.;

2n + I

[blocks in formation]

10. (i.) ab = 1, (ii.) ab = x, (iii.) a = b2, where a, b are the given numbers.

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10. 53.76.

3. 2.68 inches from A.

4. 6 lbs.

7.

V.√3.

9.

lbs.

8

11.

VJ3=4.29 f.s.

12. V.

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