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3 years ago

Complete the following statement.

Mathematics

2 answers:

Lelechka [254]3 years ago

7 0

Bigger?? I don't really understand the question you're trying to figure out

kakasveta [241]3 years ago

3 0

Greater than? I don’t fully understand what is being stated… elaborate?

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Express 0.361 recurring as a fraction.<br> (only the 61 recurs)

Oduvanchick [21]

9514 1404 393

Answer:

179/495

Step-by-step explanation:

When the repeating decimal starts at the decimal point, the repeating digits can be turned into a fraction by putting them over the same number of 9s. That is, 0.61616161... is equivalent to 61/99.

Here, the repeating part is 1/10 that value, so is 61/990. This is added to the non-repeating part, which is 0.3 = 3/10.

Then the entire decimal is ...

0.361_61 = 3/10 + 61/990 = (297 +61)/990 = 358/990 = 179/495

7 0

3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

6 times some number equals 30

den301095 [7]

Bro the answer to that is 5

4 0

3 years ago

Read 2 more answers

Place the numbers 0 to 8, inclusive, in the magic square so that the sum of the numbers in each row, column, and diagonal is the

ehidna [41]

Answer: See attached table

Step-by-step explanation:

$\begin{table}[]\begin{tabular}{lllll} & & & & \\ & & & & \\ & & & & \\ & & & & \end{tabular}\end{table}$ +---+---+---+

| 1 | 8 | 2 |

+---+---+---+

| 6 | 4 | 2 |

+---+---+---+

| 5 | 0 | 7 |

+---+---+---+

<u>Proofs:</u>
First row: 1+6+5 = 12

Second row: 8+4+0 = 12

Third row: 2+2+7 = 12

First column: 1+8+2 = 12

Second column: 6+4+2 = 12

Third column: 5+0+7 = 12

Diagonal starting from top left to bottom right: 1+4+7 = 12

Diagonal staring from top right to bottom left: 2+4+5 = 12

6 0

2 years ago

You and your friends decide to take a survey to try to convince mall managers to add a new store. You split up to survey parts A

svetlana [45]

33.3 percent or 33 and 1/3

6 0

4 years ago

Read 2 more answers