Answer:
15 units
Step-by-step explanation:
d = 14.866
d = 15
Well 32 x 405 is 12,960, but it ask for 2 numbers added to make 12,960 so the answer is 810 + 12,150 which is 2 products that add up to the answer 12,960.
If the CAT operates with a constant speed of 53 mph (which means it beats 53 miles in one hour) and makes a trip in about 2 and 1/2 hours, we just have to multiply 53 by 2 1/2 to find out the overall distance it's beating in that time:
53 * 2 1/2 =
= 53 * 2.5 =
= 132.5
Answer: City A is about <u>132.5 miles</u> from city B.
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:
see below
Step-by-step explanation:
The graph extends to the left more or less horizontally, approaching the line y=3. The only choice that expresses that is the third one.
