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3 years ago

Is my answer correct? Please see attachment and notify me.

Mathematics

2 answers:

statuscvo [17]3 years ago

6 0

Last time I checked that was about right, but it's been a few years since then.

Oksana_A [137]3 years ago

3 0

Your answer looks correct.

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HELP simple question

lawyer [7]

Answer:

I think it’s step 2 where she distributed the exponents incorrectly but she got it right so since no explanation is really needed then I guess it’s step 2.

step 1 where she divided 5 on both sides. Step 3 is also correct since the answer is 0. so step 2 had the mistake.

5 0

3 years ago

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Find the 13th term of the arithmetic sequence -3x+7, -5x+3, -7x-1, ...

oksano4ka [1.4K]

Answer:

a₁₃ = - 27x - 41

Step-by-step explanation:

The nth term of an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = - 3x + 7

d = a₂ - a₁ = - 5x + 3 - (- 3x + 7)

= - 5x + 3 + 3x - 7

= - 2x - 4

Then

a₁₃ = - 3x + 7 + 12(- 2x - 4)

= - 3x + 7 - 24x - 48

= - 27x - 41

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3 years ago

PLEASE HELP & SHOW WORK , ITS URGENT <br> it’s appreciated <3 !!

natta225 [31]

Answer:

Step-by-step explanation:

4 0

3 years ago

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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.

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3 years ago

For a quantity that changes frequently from month to month, the best kind of graph is a pie chart.

lana66690 [7]

For a quantity that changes frequently from month to month, the best kind of graph is a pie chart.a true

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3 years ago

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