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natka813 [3]
3 years ago
7

Cant seem to find the rule?

Mathematics
2 answers:
LiRa [457]3 years ago
7 0
Each time you subtract 9 from x
Mamont248 [21]3 years ago
3 0
The rule is take away 9 so n=11
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I like the idea that life is for fun. Just chase your dreams~
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Solve for 3a:<br> 4a − 2 = a + 4<br> 3a =
ziro4ka [17]
4a-2=a+4
+2 to both sides
4a=a+6
-a from both sides
3a=6
8 0
3 years ago
I need help.. i really want to go sleep.. thank you so much...
Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given  \sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}

To verify that the above equality is true or false:

Now find \sum\limits_{k=0}^8\frac{1}{k+3}

Expanding the summation we get

\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3} \sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}

Now find \sum\limits_{i=3}^{11}\frac{1}{i}

Expanding the summation we get

\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}

 Comparing the two series  we get,

\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i} so the given equality is true.

2) Given \sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}

Verify the above equality is true or false

Now find \sum\limits_{k=0}^4\frac{3k+3}{k+6}

Expanding the summation we get

\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}

\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}

now find \sum\limits_{i=1}^3\frac{3i}{i+5}

Expanding the summation we get

\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}

\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}

Comparing the series we get that the given equality is false.

ie, \sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}

6 0
3 years ago
Solve for A: B = 7/8(A-9)
Darina [25.2K]
B = 7/8(A-9)

first you must distribute

B =  7/8A -7 7/8

add over the -7 7/8

B + 7 7/8 = 7/8A

now divide both sides by 7/8



B+7 7/8
-----------     =   A
    7/8
4 0
3 years ago
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The correct answer would be B
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