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laila [671]
3 years ago
10

Please help I was able to do the hard part but I don’t know how to crack the code

Mathematics
1 answer:
iren2701 [21]3 years ago
4 0

Step-by-step explanation:

what do you mean by "the hard part" ?

anyway,

we need to sum the numbers in every column creating 5 sums.

then we need to multiply these 5 numbers (product) and divide the result by 16 (we need 1/16 of that product).

so, we have

5 + 1 = 6

1 + 6 + 4 + 8 = 19

2 + 1 + 3 = 6

2 + 5 = 7

1 + 0 + 7 = 8

6×19×6×7×8 = 38,304

38,304 / 16 = 2394

that is your code : 2394

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3 years ago
For the equation, complete the solution. 8x + y = −7 <br> (x, y) = , 1
butalik [34]

Answer:

(x, y) = (-1, 1)

Step-by-step explanation:

8x + y = −7

for y = 1

8x + 1 = -7

subtract 1 from boh sides

8x = -8

divide bot sides by 8

x = -1

(x, y) = (-1, 1)

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=f%20-%20354%20%3D%201221" id="TexFormula1" title="f - 354 = 1221" alt="f - 354 = 1221" align="
alukav5142 [94]

f-354=1221           <em>add 354 to both sides</em>

f+354-354=1221+354\\\\\boxed{f=1,575}\to\boxed{D.}

6 0
3 years ago
Find the volume of the triangular prism.<br> 2.5 m<br> 6 m<br> 3.1 m
ahrayia [7]

Answer:

46.5

Step-by-step explanation:

5 0
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:

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>
When i = 3; <span>3(3) + 2 = 11. (OK)
</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.


3 0
3 years ago
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