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2 years ago

Who can someone answer this im having trouble

Mathematics

1 answer:

yaroslaw [1]2 years ago

4 0

Answer:

Im not 100% sure but i know 2 and 4 are correct, the rest I think are all wrong but im not sure sorry

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How do you find the product of -9over5 times 5over 3? please help.

Inessa [10]

Here look at this for my answer...

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3 years ago

Put in order from greatest to least: -5, -10, -1/2, -2.5, -0.25

Sindrei [870]

Answer:

-0.25, -1/5, -2.5, -5, -10

Step-by-step explanation:

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2 years ago

aleksley [76]

Answer:

Whelan

Step-by-step explanation:

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3 years ago

-5y - 6bStep 1 of 2: Identify the first term of the algebraic expression. Indicate whether theterm is a variable term or a const

ira [324]

Answer:

a. Variable term

b. Variable term

Explanation:

a) We were given the algebraic expression:

$-5y-6b$

The first term of the algebraic expression is:

$-5y$

The first term is a variable term

The variable is "y" and its coefficient is "-5"

b) We were given the algebraic expression:

$-5y-6b$

The second term of the algebraic expression is:

$-6b$

The second term is a variable term

The variable is "b" and the coefficient is "-6"

6 0

1 year ago

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