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pogonyaev
3 years ago
5

Efectuati ridicarea la putere

Mathematics
1 answer:
KATRIN_1 [288]3 years ago
8 0
What power? ce <span>putere?</span>
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Mrs. Chan buys a new car for $21,000. She pays a sales tax of $0.06 for every $1. She pays a property tax of $16.80 on every $1,
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Answer:

$1612.80

Step-by-step explanation:

<u>Sales Tax</u> would be 0.06 * 21,000 = $1260

<u>Property Tax:</u>

16.80 for every 1000, so how many "1000"s are there in 21,000? Simple, 21 of them.

So the property tax would be 16.80 * 21 = $352.80

Hence, total taxes (for the year) = 1260 + 352.80 = $1612.80

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Can somebody prove this mathmatical induction?
Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

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

3 step:

Check for n=k+1 whether the statement

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

is true.

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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