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General Series
1. To find the sum of first n natural numbers. 2. To find the sum to squares of first n natural numbers. 3. To find the sum to the cubes of first n natural numbers. 4. Method of finding sum of a series whose nth term is know..
Examples:
i) 1 + 4 + 7 + 10 + ... is a series in which first term is 1, second term is 4, third term is 7 and so on. ii) 3 - 9 + 27 - 81 + ... is also a series in which the first term is 3, second term is -9, third term is 27 and so ..
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Question : Shortest distance from point (-3,6) to the line x=4?
Shortest distance from point (4,-3) and y=2?
I know... I've asked a million of these questions. But my tutor is out of town and I have a major assignment due Monday.
Sorry guys.
Answer : 1. the line x=4 is a vertical line at x=4.., so the shortest distance would be at the point where the line is across from the point and that would be at the point (4,6) (y's are equal and x stays at 4 always)... you can figure out that the distance is the distance between -3 and 4, which is 7 or use the distance formula (two points are (4,6) and (-3,6)) distance = square root of [(diff of x's)^2 + (diff of y's)^2] square root of [(-3-4)^2 + (6-6)^2] square root of [(-7)^2 + (0)^2] square root of 49 7 2. line at y= 2 is horizontal so it would be closest to the point (4,-3) when it is directly above it, or at (4,2). Again, by inspection the distance is the distance between -3 and 2 or 5. If you prefer to use the distance formula: (4,-3) and (4,2) distance = square root of [(diff of x's)^2 + (diff of y's)^2] square root of [(4-4)^2 + (-3-2)^2] square root of [(0)^2 + (-5)^2] square root of 25 5.. More from Yahoo Answers
Answer : 1. the line x=4 is a vertical line at x=4.., so the shortest distance would be at the point where the line is across from the point and that would be at the point (4,6) (y's are equal and x stays at 4 always)... you can figure out that the distance is the distance between -3 and 4, which is 7 or use the distance formula (two points are (4,6) and (-3,6)) distance = square root of [(diff of x's)^2 + (diff of y's)^2] square root of [(-3-4)^2 + (6-6)^2] square root of [(-7)^2 + (0)^2] square root of 49 7 2. line at y= 2 is horizontal so it would be closest to the point (4,-3) when it is directly above it, or at (4,2). Again, by inspection the distance is the distance between -3 and 2 or 5. If you prefer to use the distance formula: (4,-3) and (4,2) distance = square root of [(diff of x's)^2 + (diff of y's)^2] square root of [(4-4)^2 + (-3-2)^2] square root of [(0)^2 + (-5)^2] square root of 25 5.. More from Yahoo Answers
Question : There is this one question that has been giving so much trouble and my test is coming up but I won't be there to ask my teacher how to do it so please help me. The question is:
A geometric series has three terms. The sum of the three terms is 42. The third term is 3.2 times the sum of the other two. What are the terms?
I need the answer and how to do it please.
Answer : The goal here is not to over think things. You will need to do some critical thinking to find these answers, but they are not hard. First, since we know the answer is a whole number, we can start by making an assumption that the three terms are also probably whole numbers. Second, since all three terms add up to 42, the terms are probably not very big. Third, the first and second, have to add to something that when multiplied by 3.2, gives you a whole number. For my first try, I used 5 as my sum. For trials sake, i tried the first two terms as 2 and 3. This gave me the third term as (2+3) = 5 X 3.2 = 16. Now, 2 + 3 + 16 = 21 At my first test, I found exactly half of the final rule, that all parts must sum to be 42. By double my initial amounts of 2 and 3 to 4 and 6 we derive the following: (4+6) X 3.2 =32 4 + 6 + 32 = 42 and there is you series 4, 6, and 32... More from Yahoo Answers
Answer : The goal here is not to over think things. You will need to do some critical thinking to find these answers, but they are not hard. First, since we know the answer is a whole number, we can start by making an assumption that the three terms are also probably whole numbers. Second, since all three terms add up to 42, the terms are probably not very big. Third, the first and second, have to add to something that when multiplied by 3.2, gives you a whole number. For my first try, I used 5 as my sum. For trials sake, i tried the first two terms as 2 and 3. This gave me the third term as (2+3) = 5 X 3.2 = 16. Now, 2 + 3 + 16 = 21 At my first test, I found exactly half of the final rule, that all parts must sum to be 42. By double my initial amounts of 2 and 3 to 4 and 6 we derive the following: (4+6) X 3.2 =32 4 + 6 + 32 = 42 and there is you series 4, 6, and 32... More from Yahoo Answers