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docker41 [41]
3 years ago
11

In this fulcrum, the weights are perfectly balanced. How far must the fulcrum be located from the 40 lb. weight if the bar is 11

feet long? x (to the nearest tenth) =

Mathematics
2 answers:
vivado [14]3 years ago
7 0

Answer:

x=6.11  feet

Step-by-step explanation:

Given that in a fulcrum weights are perfectly balanced.

One side 40 lb weight is there and another side 50 lb weight is given

Let x be the length of 40 lb weight from fulcrum.  Then 50 lbs is at a distance of 11-x.

Then we have since weights are perfectly balanced

40x = 50(11-x)\\90x=550\\x=6.111

Thus we get x=6.11feet

Dennis_Churaev [7]3 years ago
3 0
The complete question in the attached figure

we know that
if <span>the weights are perfectly balanced
40*(x)=50*(11-x)
40x=550-50x------------> 50x+40x=550--------------> x=6.11

x=6.1 ft

the answer is 6.1 ft</span>

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if you flip a coin and roll a 6-sided die, what is the probability that you will flip a tails and roll more than a 4
Nutka1998 [239]

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3 years ago
The sum of four consecutive odd integers is three more than five times the least of the integers. Find the integers.
Olegator [25]

Answer:

<h3>              9, 11, 13, 15</h3>

Step-by-step explanation:

{k - some integer}

2k+1  - the first odd integer (the least)

5(2k+1)  - five times the least

5(2k+1)+3 -<u> three more than five times the least</u>

2k+1+2 = 2k+3  - the odd integer consecutive to 2k+1

2k+3+2 = 2k+5  - the next odd consecutive integer (third)

2k+5+2 = 2k+7  - the last odd consecutive integer (fourth)

2k+1+2k+3+2k+5+2k+7 - <u>the sum of four odd consecutive integers</u>

2k+1 + 2k+3 + 2k+5 + 2k+7 = 5(2k+1) + 3

8k + 16 = 10k + 5 + 3

     - 10k       -10k

-2k + 16 = 8

     -16       - 16    

      -2k = -8  

    ÷(-2)    ÷(-2)  

      k = 4

2k+1 = 2•4+1 = 9

2k+3 = 2•4+3 = 11

2k+5 = 2•4+5 = 13

2k+7 = 2•4+7 = 15  

Check: 9+11+13+15 = 48;  48-3 = 45;  45:5 = 9 = 2k+1

4 0
3 years ago
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