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1 year ago

Find K by evaluating Limit

Mathematics

1 answer:

Lubov Fominskaja [6]1 year ago

6 0

Answer:

Step-by-step explanation:

$\lim_{x \to \infty}\frac{1-cos4x}{1-cos2x}= \lim_{x \to 0} \frac{(1-cos4x)'}{(1-cos2x)'}= \lim_{x \to 0}\frac{4sin4x}{2sin2x}= \lim_{x \to 0}\frac{2*2sin2xcos2x}{sin2x}\\ = \lim_{x \to 0}4cos2x\\ =4$

{L'Hospital's rule}

<em>I hope this helps you</em>

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Solve x^2-3x-20=8 for quadratic equation

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Answer:

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Step-by-step explanation:

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2 years ago

In the triangles,BC is congruent to RS and AC is congruent to TS If RT is greater than BA, which correctly compares angles C and

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Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$

Viktor [21]

When $n=0$ , we have

$5^{3^0} + 1 = 5^1 + 1 = 6$

$3^{0 + 1} = 3^1 = 3$

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for $n=k$ , that

$3^{k + 1} \mid 5^{3^k} + 1$

Now for $n=k+1$ , we have

$5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)$

so we know the left side is at least divisible by $3^{k+1}$ by our assumption.

It remains to show that

$3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1$

which is easily done with Fermat's little theorem. It says

$a^p \equiv a \pmod p$

where $p$ is prime and $a$ is any integer. Then for any positive integer $x$ ,

$5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3$

Furthermore,

$5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3$

which goes all the way down to

$5^{3^k} \equiv 5 \pmod 3$

So, we find that

$\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3$

QED

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1 year ago

Differentiate 3x^2-5x

weqwewe [10]

Answer:

6x - 5

Step-by-step explanation:

$Let \: y = 3 {x}^{2} - 5x \\ \\ \frac{dy}{dx} = \frac{d}{dx} (3 {x}^{2} - 5x) \\ \\ = 3 \frac{d}{dx} {x}^{2} - 5 \frac{d}{dx} x \\ \\ = 3(2x) - 5(1) \\ \\ \boxed{\boxed{\huge \purple{\frac{dy}{dx}} \pink{ = } \orange{6x - 5}}}$

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2 years ago