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⚜??Brainly Buddy??⚜ 6 years, 1 month ago
=>6(7)² 22/7×7/2×7/2
=>294 + 38.5 = 336.5
Yogita Ingle 6 years, 1 month ago
The hemisphere is mounted on the cubical block
The diameter of the hemisphere = the side of the cube = 7cm
TSA of the solid = TSA of cube - Area of base of hemisphere + CSA of hemisphere
⇒ TSA = 6×7² - π×(3.5)² + 2×π×(3.5)²
⇒ TSA = 6×7² + π×(3.5)²
⇒ TSA = 332. 465 cm²
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Posted by Sarvesh Pandey 6 years, 1 month ago
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Yogita Ingle 6 years, 1 month ago
Let r be radius and h be the height of cylinder.
r = {tex} 7 \over 2{/tex}cm, h = 19 - 2 {tex} 7 \over 2{/tex}= 12 cm
Volume of solid = Vol. of cylinder + Vol. of two hemispheres
=
r2h + 2[2/3 r3]
= (7/2)2× 12 +4/3 [7/2]3
= 641.66 cm3
T.S.A. = C.S.A. of cylinder + S.A. of two hemisphere.
= 2rh + 2 (2 r2) = 2r (h + 2r)
=2× 7/2[12 + 7]
= 22 
19 cm2
= 418 cm2
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Yogita Ingle 6 years, 1 month ago
Let three consecutive positive integers be, n, n + 1 and n + 2.
When a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, ⇒ n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.
If n = 3p + 2, ⇒ n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.
⇒ n (n + 1) (n + 2) is divisible by 3.
Similarly, when a number is divided 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q ⇒ n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.
If n = 2q + 1 ⇒ n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2.
⇒ n (n + 1) (n + 2) is divisible by 2.
Hence n (n + 1) (n + 2) is divisible by 2 and 3.
∴ n (n + 1) (n + 2) is divisible by 6.
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Shruti Gupta 6 years, 1 month ago
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Yogita Ingle 6 years, 1 month ago
A = 18
-3 = 18 + 3d
-3-18 = 3d
-21/3 = d
d =-7
a = A + d
a = 18 +(-7)
a = 11
b = 18 +2(-7)
b = 18-14
b = 4
a + b = 11 + 4 = 15
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Yogita Ingle 6 years, 1 month ago
Given that the smallest number when divided by 28 and 32 leaves remainder 8 and 12 respectively.
28 - 8 = 20 and 32 - 12 = 20 are divisible by the required numbers.
Therefore the required number will be 20 less than the LCM of 28 and 32.
Prime factorization of 28 = 2 × 2 × 7
Prime factorization of 32 = 2 × 2 × 2 × 2 × 2
LCM(28,32) = 2 × 2 × 2 × 2 × 2 × 7
= 224.
Therefore the required smallest number = 224 - 20
= 204.
Verification:
204/28 = 28 × 7 = 196.
= 204 - 196
= 8
204/32 = 32 × 6 = 192
= 204 - 192
= 12.
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Ajay ???? 6 years, 1 month ago
1Thank You