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lesya692 [45]
3 years ago
8

you have four chains of three links each. although it is difficult to cut the links, you wish to make a single loop with all 12

links. what is the fewest numbers of cuts you must make to accomplish this task
Mathematics
1 answer:
mel-nik [20]3 years ago
3 0
You have 4 parts that you want to combine so you need to make 3 cuts
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Help for the two questions shown, please!
LekaFEV [45]

Answer:

5) 140.8m^2   6) 360in^2

Step-by-step explanation:

5.

6.4 x 6.4 = 40.96

(7.8 x 6.4) ÷ 2 = 24.96

24.96 x 4 = 99.84

99.84 + 40.96 = 140.8m^2

6.

12 x 10 = 120in

13 x 10 = 130in

5 x 10 = 50in

(12 x 5) ÷ 2 = 30in

30 x 2 = 60in

60 + 50 + 130 + 120 = 360in^2

8 0
3 years ago
Divide the difference between 2,000 and 500 by 5
mart [117]

Answer:

300

Step-by-step explanation:

2000-500= 1500

1500/5= 300

5 0
3 years ago
1. What is 1/6 plus 2/6?<br><br> 2. What is 4/8 plus 1/8?<br><br> 3. What is 2/3 plus 2/3?
Anon25 [30]
<span>1. What is 1/6 plus 2/6?
          1/6 + 2/6 = 3/6 =1/2

2. What is 4/8 plus 1/8?
           4/8 + 1/8 = 5/8

3. What is 2/3 plus 2/3?
          2/3 + 2/3 = 4/3 = 1 1/3</span>
3 0
3 years ago
Read 2 more answers
23
BaLLatris [955]

Answer:

B

Step-by-step explanation:

6 0
1 year ago
A rubber ball dropped on a hard surface takes a sequence of bounces, each one 1/6 as high as the preceding one. If this ball is
Brums [2.3K]

Answer:

\frac{259}{54}\text{ or }4.8\text{feet}

Step-by-step explanation:

GIVEN: A rubber ball dropped on a hard surface takes a sequence of bounces, each one \frac{1}{6} as high as the preceding one.

TO FIND: If this ball is dropped from a height of 12 feet, how far will it have traveled when it hits the surface the fifth time.

SOLUTION:

once the ball is dropped on hard surface it bounces \frac{1}{6} of preceding one and comes down the same distance.

When the ball is dropped from 12 feet height

after first hit =12\times\frac{1}{6}\text{ up}+12\times\frac{1}{6}\text{ down}=4

new height =12\times\frac{1}{6}=2\text{feet}

after second hit =2\times\frac{1}{6}\text{ up}+2\times\frac{1}{6}\text{ down}=\frac{2}{3}

new height =2\times\frac{1}{6}=\frac{1}{3}\text{ feet}

after third hit  =\frac{1}{3}\times\frac{1}{6}\text{ up}+\frac{1}{3}\times\frac{1}{6}\text{ down}=\frac{1}{9}

new height =\frac{1}{3}\times\frac{1}{6}=\frac{1}{18}\text{ feet}

after fourth hit =\frac{1}{18}\times\frac{1}{6}\text{ up}+\frac{1}{18}\times\frac{1}{6}\text{ down}=\frac{1}{54}

adding all distance =4+\frac{2}{3}+\frac{1}{9}+\frac{1}{54}

                                 =\frac{259}{54} feet

Hence the ball will travel   \frac{259}{54} feet before it hits the surface fifth time.

                                 

4 0
3 years ago
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