Answer:
It would be the third one hope i helped
Step-by-step explanation:
Equation of the line in point slope form is y + 7 = -5(x - 2)
Step-by-step explanation:
- Step 1: Given slope of the line is -5 and passes through the point (x1, y1) = (2,-7). Equation in point slope form is y - y1 = m(x - x1)
⇒ y - (-7) = -5(x - 2)
⇒ y + 7 = -5(x - 2)
Answer:
Lo's answer is correct and it is a translation because applying the rule (x+4, y+8) on the coordinates N(-2, -4) will gives us N'(2,4).
Step-by-step explanation:
i) though Raheem is mathematically correct the question asks for a translation which means that we can only use addition and/or subtraction and not multiplication. So Raheem's answer is therefore incorrect.
ii) Casey's answer is incorrect as applying the rule (x+2, y+4) on the coordinates N(-2, -4) will gives us (0,0) and not N'(2,4)
iii) Andrew's answer is also incorrect as applying the rule (x+4, y+0) on the coordinates N(-2, -4) will gives us (2,-4) and not N'(2,4).
iv) Lo's answer is correct and it is a translation because applying the rule (x+4, y+8) on the coordinates N(-2, -4) will gives us N'(2,4).
Answer: 10609
Step-by-step explanation: Raise 103 to the power of 2
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.
