Answer:
76.7 liters
Step-by-step explanation:
You have ...
C1×V1 = C2×V2
so ...
V2 = V1×(C1/C2) = (115 L)×(14/21) = 76 2/3 L
V2 ≈ 76.7 L
The angle between two vectors is:
CosФ = u - v / Magnitude(u) x magnitude(v)
Magnitude of u = SQRT(7^2 + -2^2) = SQRT(49 +4) = SQRT(53)
Magnitude of v = SQRT(-1^2 +2^2) = SQRT(1 +4) = SQRT(5)
u x v = (7 x -1) + (-2 x 2) = -7 + -4 = -11
cosФ = -11 / sqrt(53) x sqrt(5)
cosФ = -11sqrt265) / 265
Ф =cos^-1(-11sqrt265) / 265)
Ф=132.51 degrees.
This is the concept of algebra, to get the slope of the line we need to rewrite the equation in slope intercept form y=mx+c, where m=slope, c=y-intercept.
Therefore re-writing our expression in slope-intercept form we get:
7x-3y=10
-3y=-7x+10
y=7/3x-10/3
The slope=7/3
hence we conclude that the slope of the line perpendicular to this is -3/7
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.
