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3 years ago

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Mathematics

1 answer:

Nat2105 [25]3 years ago

3 0

The answer is D plane Qpr

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How is the pythogorean theorem used to determine the distance formula

Fiesta28 [93]

Answer:

The distance formula uses the coordinates of points and the Pythagorean Theorem to calculate the distance between points. If A and B form the hypotenuse of a right triangle, then the length of AB can be found using this formula, leg2 + leg2 = hypotenuse2 (or you can use A squared + B squared = C squared)

Am not a real genius but I hope that answers your question! ^w^

5 0

3 years ago

The question is "find the lowest common multiple of 4 and 6"<br> with step by step explaination

Tcecarenko [31]

Answer:

Step-by-step explanation:

4s multiples:

4,8,12,16,20,24

6s multiples:

6,12,18,24,30,36

lowest number that is a common multiple between both 4 and 6:

4 0

3 years ago

Read 2 more answers

What are the coordinates of L.

Triss [41]

Answer:

that would be b

Step-by-step explanation:

trust me just took the test

6 0

3 years ago

Read 2 more answers

Find the domain of the Bessel function of order 0 defined by [infinity]J0(x) = Σ (−1)^nx^2n/ 2^2n(n!)^2 n = 0

Snowcat [4.5K]

Answer:

Following are the given series for all x:

Step-by-step explanation:

Given equation:

$\bold{J_0(x)=\sum_{n=0}^{\infty}\frac{((-1)^{n}(x^{2n}))}{(2^{2n})(n!)^2}}\\$

Let the value a so, the value of $a_n$ and the value of $a_(n+1)$ is:

$\to a_n=\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}$

$\to a_{(n+1)}=\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}$

To calculates its series we divide the above value:

$\left | \frac{a_(n+1)}{a_n}\right |= \frac{\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}}{\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}}\\\\$

$= \left | \frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2} \cdot \frac {2^{2n}(n!)^2}{(-1)^2n x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)!^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\= \left | \frac{x^{2n}\cdot x^2}{2^{2n} \cdot 2^2(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\$

$= \frac{x^2}{2^2(n+1)^2}\longrightarrow 0$ for all x

The final value of the converges series for all x.

8 0

3 years ago

Simplify the expression.<br> 7- 5n + n + 5

kodGreya [7K]

-4n+12

Explanation-
Like terms

4 0

2 years ago