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inna [77]
3 years ago
12

The crew on a fishing boat caught 4 fish that weighed a total of 1,092 pound. The tarpon weighed twice as much as the amberjack

and the white marlin weighed twice as much as the tarpon. The weight of the tuna was 5 times the weight of the Amber jack. How much did each fish weigh?
Mathematics
1 answer:
Savatey [412]3 years ago
7 0

ddddd

Let T be the weight of Tarpon

Let A be the weight of amberjack

Let W be the weight of white marlin

Let TU be the weight of Tuna

The crew on a fishing boat caught 4 fish that weighed a total of 1,092 pound.

Total weight = 1092

T + A + W + TU = 1092

The tarpon weighed twice as much as the amberjack

T= 2 A

  the white marlin weighed twice as much as the tarpon.

W = 2 T

We know T is 2A  so W = 2(2A) = 4 A, W = 4A

The weight of the tuna was 5 times the weight of the Amber jack.

TU = 5A

T = 2A  , W = 4A , TU = 5A

Replace it in our equation

T + A + W + TU = 1092

2A + A + 4A + 5A = 1092

12A = 1092

Divide both sides by 12

A = 91

the weight of amberjack = 91 pounds

the weight of white marlin = 4A = 4(91) = 364 pounds

the weight of Tarpon = 2A = 2(91) = 182 pounds

the weight of Tuna= 5A = 5(91)=455 pounds

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Write 169=13^2 in logarithmic form​
e-lub [12.9K]

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\log_{13}(169)=2

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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
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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:

Sigma notation 1:

  10
<span>   Σ (2i + 3)
</span>i = 3

@ i = 3

2(3) + 3
12

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

  10
<span>   Σ (3i + 2)
</span><span>i = 3
</span>
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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)
</span>When i = 10; <span>3(10) + 2 = 32. (OK)

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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