The answer to this problem would be 2.5.
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.
Answer:
x + 2 = 3
Step-by-step explanation:
x is always = to 1
Answer:
Step-by-step explanation:
The volume of a cube is given by the formula :
a³ (where a is the side length )
So now we have to cube these lengths :
Part A :
(3x²y)³ =
(3x²y)(3x²y)(3x²y) =
(9x^4y²)(3x²y) =
27x^6y³ (This is now fully simplified so our final answer for a)
Part B:
(5y²)³ =
(5y²)(5y²)(5y²) =
(25y^4)(5y²) =
125y^6 (This is now fully simplified so our final answer for b)
Hope this helped and have a good day
Answer:
4.8 x 10⁷
Step-by-step explanation:
Refer to attachment.
<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em>
