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kolbaska11 [484]
3 years ago
8

The summation expression in the following series has an absolute value in it. Expand and evaluate the summation notation. What i

s the sum of the series?

Mathematics
1 answer:
nikitadnepr [17]3 years ago
4 0
|3i-10|=\begin{cases}3i-10&\text{for }3i-10\ge0\\-(3i-10)&\text{for }3i-10

We have that 3, which means |3i-10| reduces to 3i-10 when i\ge4, or reduces to 10-3i when i\le3.

So we can expand the summation as

\displaystyle\sum_{i=1}^6|3i-10|=\sum_{i=1}^3(10-3i)+\sum_{i=4}^6(3i-10)

Notice that the 10 contributes a total of 30 from the first sum, and -30 from the second sum, so those terms cancel, leaving us with

\displaystyle3\left(\sum_{i=4}^6i-\sum_{i=1}^3i\right)=3((4+5+6)-(1+2+3))=3(15-6)=3(9)=27
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Papessa [141]

Answer:

6) y = x^(5/3)

7) B

8) C

10) A

Step-by-step explanation:

6) The fifth root is the same as raising to the 1/5 power, so we can write this in exponent form as:

f(x) = (x^(1/5))³

f(x) = x^(3/5)

To find the inverse, switch x and y and solve for y.

x = y^(3/5)

y = x^(5/3)

7) f(x) = 2√(x − 4) + 8

Switch the x and y and solve for y:

x = 2√(y − 4) + 8

x − 8 = 2√(y − 4)

(x − 8) / 2 = √(y − 4)

(x − 8)² / 4 = y − 4

(x² − 16x + 64) / 4 = y − 4

¼x² − 4x + 16 = y − 4

y = ¼x² − 4x + 20

8) Find the inverse:

x = 5√(y + 3) − 2

x + 2 = 5√(y + 3)

(x + 2) / 5 = √(y + 3)

(x + 2)² / 25 = y + 3

y = -3 + (x + 2)² / 25

The inverse function is an upwards parabola with a vertex at (-2, -3).  The best fit is C.

desmos.com/calculator/fbabg5wc8b

10) √(4x − 31) = x − 7

Square both sides:

4x − 31 = (x − 7)²

4x − 31 = x² − 14x + 49

Combine like terms:

0 = x² − 18x + 80

Factor:

0 = (x − 8) (x − 10)

x = 8 or 10

Check for extraneous solutions.

√(4×8 − 31) = 8 − 7

1 = 1

√(4×10 − 31) = 10 − 7

3 = 3

x = 8 and x = 10 are both solutions.

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Answer:

B

Step-by-step explanation:

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Read 2 more answers
Find the sum of the series given in sigma notation.
Kryger [21]

Answer:

Option b is correct 175

Step-by-step explanation:

n = 7

k = 6

3k -2 ------1

put k = 6 in above eq. for finding first term

  a1 = 3(6) - 2  = 18 - 2 = 16

put k = 7 in above eq. for finding first term

 a2 = 3(7) - 2 = 21 - 2 = 19

 a3 = 3 (8) - 2 = 24 - 2 = 22

16, 19 , 22, ...  //Arithmetic series formation

a1 = 16 , a2 = 19

d = a2 - a1 = 19 - 16 = 3 //Difference of first two terms

Using sum forumula for arithmetic series

sum = \frac{n}{2}(2a1 + (n-1)d)

      = \frac{7}{2}(2a1 + (7-1)d)

      = \frac{7}{2}(2(16) + (6)3)

      =  \frac{7}{2}(32 + 18)    

      =  \frac{7}{2}(50)

      = 7 * 25

      = 175




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