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:L x W x H
Step-by-step explanation:
Answer:
43.75feet
Step-by-step explanation:
Given the height of the tennis ball expressed as;
h(t) = -15t^2 + 45t + 10
At maximum height, the velocity of the ball is zero
v = dh/dt
dh/dt = -30t + 45
Since dh/dt = 0, hence;
0 = -30t + 45
30t = 45
t = 45/30
t = 3/2
t = 1.5secs
Substitute t = 1.5 into the expression given
h(1.5) = -15(1.5)^2 + 45(1.5) + 10
h(1.5) = -15(2.25)+67.5+10
h(1.5) = -33.75+77.5
h(1.5) = 43.75feet
Hence the maximum height of the ball is 43.75feet
There are no factors of 20 that add up to 7.
Factors of 20 are:
1*20
2*10
4*5
