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Licemer1 [7]
3 years ago
10

What is the sum of all the integers between 19 and 77

Mathematics
2 answers:
victus00 [196]3 years ago
7 0
Think of it this way: Lets add numbers in pairs, starting at the very outer 2 numbers (19 and 77)  then go in by one and add the second number and the second to last (20 and 76), then (21 and 75) and so on.  The sum of all of these pairs are all the same: 96.  How many 96s will we have?  Well since we're coming from each end toward the middle adding pairs we will have half the distance between 19 and 77, that is (77-19)/2 = 29.  So we can actually just take 96*29 = 2784.  This is the sum of all numbers between 19 and 77
dedylja [7]3 years ago
6 0

Answer:

2639

Step-by-step explanation:

We are asked to find the sum of all the integers between 19 and 77.

We will use sum of arithmetic sequence formula to solve our given problem.

S_n=\frac{n}{2}(2a+(n-1)d), where,

S_n = Sum of n terms,

n = Number of terms in the sequence,

a = 1st term,

d = Common difference.

Since we are asked to take the sum of all the integers between 19 and 77, so common difference is 1.

n=77-19=58

S_n=\frac{58}{2}(2(17)+(58-1)1)

S_n=29(34+57)

S_n=29*91

S_n=2639

Therefore, the sum of all integers between 19 and 77 is 2639.

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It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

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Complete the square in the denominator:

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Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

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Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

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\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

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\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
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