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1) From the problem, it can be seen that the master assembles 3 units every day, does a day and rests for 3 days, that is, 3 units can be configured in 4 days, that is, 3+1=4 (days) The apprentice assembles 1 unit every 3 days, and does three days and rests for 1 day. It is also 4 days to configure three sets 3+1=4 (days) and then use the apprentice's plus the master's 24 divided by (3+3)=4 4 times 4=16 (days).
2) For example, if there are 43 machines, divide 43 by 6 = 7 and leave 1 for one more day, so the number of days is 7 times 4 + 1 = 29
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The efficiency of the master is: 3 sets for 4 days.
The efficiency of the apprentice is: 1 unit for 4 days.
Time required to assemble 24 machines = 24 (3 4 + 1 4) = 24 days.
Time required to assemble 43 machines = 43 (3 4 + 1 4) = 43 days.
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Assuming that it takes at least x days to produce 24 machines, then on the last day, both master and apprentice are working, and on day x-1 the master assembles ((x-1) 4)*3 machines, the apprentice assembles ((x-1) 4)*1 machines, and on day x, the master assembles 3 machines and the apprentice assembles 1 machine.
So: ((x-1) 4)*3+((x-1) 4)*1+3+1=24
x=21
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The picture is not clear, and it is better to be troublesome.
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1.Use the number method. Suppose the distance from A to B is 120 km.
120 (120 2 30 120 2 60) = 40 (kmh) 2Solution: If he runs 1 lap in x seconds, he will run 2 laps for a total of 800 meters.
4x+5x=800
9x=800
x=800/9
He ran 400 2 5=40 (seconds) in the second half and 800 9 40=440 9 (seconds) in the first half
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4、a²=96×12=12×8×12=12^2×8=12^2×2^2×2=24^2×2
So, a = 24 2cm
5. From the Pythagorean theorem: the square of the hypotenuse = (2 3+1) 2+(2 3-1) 2
26So, hypotenuse = 26
6、x=√5-1
So: x 2+5x-6=( 5-1) 2+5( 5-1)-6=(6-2 5)+5 5-5-6
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Suppose 12 years ago, the son was x years old, and the father was 4x years old, there is (x+12)+(4x+12)=69
i.e. 5x=45, x=45, 5=9
So now the son is 9 + 12 = 21 years old, and the father is 4 * 9 + 12 = 48 years old.
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(69-12x2) (1x4).
99 ten 12 = 21 (son) years old.
69-21 = 48 years (father).
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