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2 years ago

Rewrite using a single positive exponent. 7^9 x 7^-2

Mathematics

1 answer:

lidiya [134]2 years ago

6 0

Answer:

7^7

Step-by-step explanation:

When you multiply exponents with the same base, you add the exponents

Hope it helped!

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Yara saved 440$. She spent 3/4 of it on a tablet. How much money did yara spend on the tablet

TiliK225 [7]

440 times 3/4=330
$330

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3 years ago

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What is y+4=9 what is y?

amid [387]

Answer:

$y = 5$

Step-by-step explanation:

Subtract the 4 from 9

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3 years ago

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Factor 6x4 – 5x2 + 12x2 – 10 by grouping. What is the resulting expression?

lisov135 [29]

Your answer would be the last option, (6x² - 5)(x² + 2).
This is because when you expand it, you get:
6x² × x² = 6x⁴
6x² × 2 = 12x²
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-5 × 2 = -10
Which are all the correct terms.

I hope this helps!

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3 years ago

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-1/4x+2y=28 find the x and y intercept

Advocard [28]

The y intercept is (0, 14) and the x intercept is (-112,0) all you have to do is just make y zero if you are trying to solve for x and the same for the x if you are trying to solve for y.

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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

3 years ago