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3 years ago

15

3х – 2y < 10 solve inequality and explain how to graph

1 answer:

katrin2010 [14]3 years ago

6 0

Go to math-way :) it will answer it

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On a sheet of paper, you have 100 statements written down.

serg [7]

Answer:

50

Step-by-step explanation:

If the first statement its true (At most 0 of the statements are true), there are not true statements in the paper. So, the first statement its false.

Now, if the first statement its false, this mean there must be at least 1 true statement in the paper.

Now, if the second statement its true ( at most 1 of the statements are true) this implies that the third statement its true (if "at most 1" its true, then "as most 2" must be true).

If any statement (besides the first) its true, then all the statement that follows it must be true.

The first non false statement, then, must be the statement made by the person 51: "At most 50 statements are true"

And the 49 statements that follows are true as well.

5 0

3 years ago

Be confident and be paitent to help people tahnky ou

Aleksandr [31]

12!!!! Hope this helped and good encouragement

4 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

Is the underlined verb in the sentence transitive or intransitive?

Firlakuza [10]

A because everything after the verb is a preposition

3 0

3 years ago

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HOw dO I rUn??????????????????????????????????????????????

Semmy [17]

How do you run???????

7 0

3 years ago

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