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klasskru [66]
3 years ago
10

In a clock, a large gear completes a rotation every 45 seconds, a small gear completes a rotation every 18 seconds. The gears al

igned now, how many seconds will pass before the gears are in line again?
Mathematics
2 answers:
RUDIKE [14]3 years ago
8 0
If the big gear makes 2 circles and the small gear makes 5 circles they will both be on 90 sec so it will take 90 seconds I think. My dad helped me with this one :D
sineoko [7]3 years ago
8 0

Answer:

90 seconds will pass before the gears are in line again

Step-by-step explanation:

In a clock, a large gear completes a rotation every 45 seconds.

The small gear completes a rotation every 18 seconds.

To get the number of seconds needed so these both will pass before the gears are in line again, we will simply find the LCM of 45 and 18.

45 = 3 x 3 x 5

18 = 2 x 3 x 3

LCM = 2 x 3 x 3 x 5 = 90

Hence, 90 seconds will pass before the gears are in line again

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Read 2 more answers
The sum of the first term of an ap is 240 and the sum of the next 4 term is 220 find the first term of the ap
Sindrei [870]

Answer:

The common difference is -5/4

T(n) = T(0)  - 5n/4,

where T(0) can be any number. d = -5/4

Assuming T(0) = 0, then first term

T(1) = 0 -5/4 = -5/4

Step-by-step explanation:

T(n) = T(0) + n*d

Let

S1 = T(x) + T(x+1) + T(x+2) + T(x+3) = 4*T(0) + (x + x+1 + x+2 + x+3)d = 240

S2 = T(x+4) + T(x+5) + T(x+6) + T(x+7) = 4*T(0) + (x+5 + x+6 + x+7 + x+8)d = 220

S2 - S1

= 4*T(0) + (x+5 + x+6 + x+7 + x+8)d - (4*T(0) + (x+1 + x+2 + x+3 + x+4)d)

= (5+6+7+8 - 1 -2-3-4)d

= 4(4)d

= 16d

Since S2=220,  S1 = 240

220-240 = 16d

d = -20/16 = -5/4

Since T(0) has not been defined, it could be any number.

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