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Marizza181 [45]
3 years ago
14

How many Pythagorean triples can be created by multiplying the side lenghts in a known triple by a constant? Explain

Mathematics
1 answer:
Rudiy273 years ago
5 0
Answer: There are an infinite amount of Pythagorean Triples that could be created.

A Pythagorean Triple is a set of 3 integers that would form a right triangle. They must satisfy the equation a^2 + b^2 = c^2.

The most basic one is:  3, 4, 5

A triangle with sides of 3, 4, 5 would form a right triangle.

If we multiply each side by 2, we get 6, 8, 10. This would also be a right triangle. 

We could also multiply by 3, or 4, or 5, or 6 .......

We can multiply by any number and still have a right triangle.
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Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0
Vsevolod [243]

Answer:

\dfrac{-1}{6}

Step-by-step explanation:

Given the limit of a function expressed as \lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}, to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)

Step 2: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the function

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\

Step 3: substitute x = 0 into the resulting function

= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)

Step 4: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\

=  \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)

Step 6: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\

Step 7: substitute x = 0 into the resulting function in step 6

=  \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}

<em>Hence the limit of the function </em>\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \  is \ \dfrac{-1}{6}.

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3 years ago
The amount of Jen’s monthly phone bill is normally
vredina [299]

Answer:

P(19

And we can find this probability with this difference and using the normal standard distribution

P(-3

Step-by-step explanation:

Let X the random variable that represent the amount of Jen monthly phone of a population, and for this case we know the distribution for X is given by:

X \sim N(55,12)  

Where \mu=55 and \sigma=12

We are interested on this probability

P(19

And we can use the z score formula given by:

z=\frac{x-\mu}{\sigma}

Replacing the info we got:

P(19

And we can find this probability with this difference and using the normal standard distribution

P(-3

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