The formula to find the surface area of a circle is: pi x radius², so if we square the radius the answer would be 25, then we multiply that by 3.14 and the answer would be around 78.5, so the answer is 78.5.
Answer:
Area≈61.79
Step-by-step explanation:
used a calculator bruh
Answer:
Answer:18+4.5
Step-by-step explanation:
This shape is a triangle with a semicircle connected to it
and this triangle is a right triangle so side A=B
that means the other leg is 6.Knowing that we can solve
The formula for the area of a triangle is (base×height)÷2
so that means 6×6 equals 36 and if you divide that by 2 you get the 18.
Now we will deal with the semicircle. We know that both of the legs are 6 so that means the diameter is 6 and now we solve 6 divided by 2 equals 3 and we will have to square that and we will 9 and since it is a semicircle we have to divide it by 2 and that will give us 4.5 and since we have to express it in terms of pi it will be 4.5(pi) and then we add both of the areas
giving us 18+4.5(pi)
and do you go to RSM
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.
