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Aloiza [94]
2 years ago
12

Help Me please . I want to know Square root of 169 and 144 and 625 . please man fast ​

Mathematics
2 answers:
irga5000 [103]2 years ago
6 0

Answer:

\sqrt{169} = 13\\\\\sqrt{144} = 12\\\\\sqrt{625} = 25

Step-by-step explanation:

\sqrt{169} = \left(13^2 \right)^{\tfrac 12} = 13^{\tfrac 22} = 13^1 = 13\\\\\sqrt{144} = \left( 12^2\right)^{\tfrac 12} = 12^{\tfrac 22} = 12^1 = 12\\\\\sqrt{625}= \left(25^2 \right)^{\tfrac 12} = 25^{\tfrac 22} = 25^1 = 25

neonofarm [45]2 years ago
5 0

Answer:

13,12,25

Step-by-step explanation:

Square root of 169 is 13.

Square root of 144 is 12

Square root of 625 is 25.

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204,000 grapefruit.

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What is 5654554 + 5544527?
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3 years ago
A skier has decided that on each trip down a slope, she will do 3 more jumps than before. On her first trip she did 5 jumps. Der
taurus [48]
Since we are already given the amount of jumps from the first trial, and how much it should be increased by on each succeeding trial, we can already solve for the amount of jumps from the first through tenth trials. Starting from 5 and adding 3 each time, we get: 5 8 (11) 14 17 20 23 26 29 32, with 11 being the third trial.

Having been provided 2 different sigma notations, which I assume are choices to the question, we can substitute the initial value to see if it does match the result of the 3rd trial which we obtained by manual adding.

Let us try it below:

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<span>   Σ (2i + 3)
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@ i = 3

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The first sigma notation does not have the same result, so we move on to the next.

  10
<span>   Σ (3i + 2)
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</span>
Since the 3rd trial is a match, we test it with the other values for the 4th through 10th trials.

When i = 4; <span>3(4) + 2 = 14. (OK)
</span>When i = 5; <span>3(5) + 2 = 17. (OK)
</span>When i = 6; <span>3(6) + 2 = 20. (OK)
</span>When i = 7; 3(7) + 2 = 23. (OK)
When i = 8; <span>3(8) + 2 = 26. (OK)
</span>When i = 9; <span>3(9) + 2 = 29. (OK)
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Adding the results from her 3rd through 10th trials: </span><span>11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 = 172.
</span>
Therefore, the total jumps she had made from her third to tenth trips is 172.


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Answer:

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