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3 years ago

8x+3>_x+10 how do i solve dis HELP

Mathematics

1 answer:

VashaNatasha [74]3 years ago

5 0

Answer:

x > 1

Step-by-step explanation:

Subtract 3 from both sides

8x + 3 - 3 > x + 10 - 3

Simplify

8x > x + 7

Subtract x from both sides

8x - x > x + 7 - x

Simplify

7x - 7

Divide both sides by 7

7x/7 > 7/7

Simplify

x > 1

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Answer:

See explanation

Step-by-step explanation:

1 step:

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So, for j=1 this statement is true

2 step:

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3 step:

Check for n=k+1 whether the statement

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Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

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4 years ago

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Answer:

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Step-by-step explanation:

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