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3 years ago

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which phrase translates this expression x - 8 < 23 A: a number is less than 23 B: 8-- x is at most C: a number decreased by 8

is less than 23 D: 8 decreased by another number is larger than 23

1 answer:

Elza [17]3 years ago

4 0

Answer:

I'm not smart enough to know

Step-by-step explanation:

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A group of 8 friends went to lunch and spent a total of $76, which included the food bill and a tip of $16. they decided to spli

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Answer:

A, c(excluding tip, if the tip is included then c is not the answer)

Step-by-step explanation:

I solved the Equations:

A)1/8(x+16)=76/8----60 (this is the correct answer!)

B)1/8(x+16=76)------ 592 (NOT the correct answer

c)x=60 Represents total food bill excluding tip

d) x=60 does represent each friends share of the food bill

E) not correct

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Find the slope of the line passing through the points (-2, 9) and (-10, 33).

Nina [5.8K]

Slope = rise / run = Δy / Δx = (y2 - y1) / (x2 - x1) = [33 - 9] / [ (-10 - (-2)]

Slope = 24 / ( -8) = - 3

Answer: - 3

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3 years 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

4 years ago