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4 years ago

7(3r - 1) - (r + 5) = -52 Please step by step help

Mathematics

2 answers:

dmitriy555 [2]4 years ago

7 0

Answer:

r = -2

Step-by-step explanation:

7(3r - 1) - (r + 5) = -52 distribute (the - sign too)

21r - 7 + -r - 5 = -52 add/ subtract like terms

20r - 12 = -52 subtract -12 and added to -52

+12 +12

20r = -40 divide by 20

20r/20 = -40/20

r = -2

MrMuchimi4 years ago

4 0

7(3r-1)-(r+5)=-52

21r-7-r-5=-52

21r-r-5-7=-52

20r-12=-52

Add 12 to both sides

20r=-40

Divide by 20

r=-2

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Answer:

The given sums 11+17+23+29+35+41 can be written as in sigma notation is $\sum\limits_{i=1}^{6}5+6i$

Therefore 11+17+23+29+35+41= $\sum\limits_{i=1}^{6}5+6i$

Step-by-step explanation:

Given series is 11+17+23+29+35+41

To find the given sum expressed in sigma notation :

11+17+23+29+35+41 can be written as the given sums in sigma notation is $\sum\limits_{i=1}^{6}5+6i$

That is

11+17+23+29+35+41= $\sum\limits_{i=1}^{6}5+6i$

Now expand the sums and to verify that whether the summation is true or not :

$\sum\limits_{i=1}^{6}5+6i=(5+6(1))+(5+6(2))+(5+6(3))+(5+6(4))+(5+6(5))+(5+6(6))$

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Step-by-step explanation:

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We have a system of 2 equations and 2 unknowns

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