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3 years ago

How does knowing the absolute value of an integer help in determining the distance of two ordered pairs?

1 answer:

6 0

Hey there!

The absolute value helps because when you the absolute value is the the distance of a number away from 0. Meaning that if your finding distance it will never be negative.

Finding the distance between the two is subtracting one of the numbers from each ordered pair from each other. Whether the number is negative or positive, you always turn it, or leave it into a positive number because distance is always positive.

Hope this helps !

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Please help anyone if you can help thanks

-BARSIC- [3]

Answer:

t=5/10n

Step-by-step explanation:

The question wants to know how much time it takes to run one lap, not how many laps can be run in a certain amount of time.

7 0

3 years ago

Factor 6(2p q)^2 -5(2p q) - 25

kherson [118]

Treat the 2p+q as x

6x2-5x-25
trial and error
(2x-5)(3x+5)
replace x with (2p+q)
(2(2p+q)-5)(3(2p+q)+5)
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3 years ago

Please help me. Hurry, linear equations.

Lisa [10]

Answer:

Let's simplify step-by-step.

3−3/4x−1/4x−7

=3+ −3/4x+ −1/4x+ −7

Combine Like Terms:

=3+ −3/4x+ −1/4x+ −7

=(−3/4x+ −1/4x)+(3+ −7)

=−x+ −4

Answer:=−x−4

Step-by-step explanation:

answer:-4

8 0

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

3 years ago

Line q passes through points (9, 9) and (1, 2). Line r is perpendicular to q. What is the slope of line r?

Alik [6]

Answer:

y=mx+b

Step-by-step explanation:

6 0

3 years ago