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3 years ago

I need help A S A P!

Mathematics

1 answer:

SSSSS [86.1K]3 years ago

6 0

Answer:

ewfwefwfewfwfwe

Step-by-step explanation:

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What is the solution to x 4 – 12x 2 + 10 > 0?

Umnica [9.8K]

Answer: The picture contains the answer and the steps

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2 years ago

Two trains arrived at a station at 2:55 P.M., with one arriving on Track A, and the other arriving on Track B. Trains arrive on

CaHeK987 [17]

The time before they will arrive together next is given by the LCM of 16 and 18 which is 144 minutes
2:55 PM + 2 h 24 minutes = 5.19 PM

Its B

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3 years ago

Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction

Oliga [24]

The base case is the claim that

$\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}$

which reduces to

$\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86$

which is true.

Assume that the inequality holds for n = k ; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}$

We want to show if this is true, then the equality also holds for n = k + 1 ; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}$

By the induction hypothesis,

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}$

Now compare this to the upper bound we seek:

$\dfrac{2k+1}{k+1} > \dfrac{2k+2}{k+2}$

because

$(2k+1)(k+2) > (2k+2)(k+1)$

in turn because

$2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0$

6 0

2 years ago

Read 2 more answers

Please answer and no explanation needed

rewona [7]

Wait isn’t it 30 or am I reading it wrong?

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2 years ago

What is 22b+3b simplified

ss7ja [257]

it's a simple addition:

22b + 3b =

25b

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3 years ago