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3 years ago

I need help ASAP!

Mathematics

1 answer:

Readme [11.4K]3 years ago

7 0

Answer:

i) Equation can have exactly 2 zeroes.

ii) Both the zeroes will be real and distinctive.

Step-by-step explanation:

$x^{2} - 7x + 3$ is the given equation.

It is of the form of quadratic equation $a^{2} + bx + c$ and highest degree of the polynomial is 2.

Now, FUNDAMENTAL THEOREM OF ALGEBRA

If P(x) is a polynomial of degree n ≥ 1, then P(x) = 0 has exactly n roots, including multiple and complex roots.

So, the equation can have exact 2 zeroes (roots).

Also, find discriminant D = $b^{2} - 4ac = (-7)^{2} - 4(1)(3) = 49 - 12 = 37$

⇒ D = 37

Here, since D > 0, So both the roots will be real and distinctive.

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In a game, Janeesa started with 0 points. She then earned 50 points, lost 80 points and earned 10 points. Which number line show

umka2103 [35]

Answer:

-20

Step-by-step explanation:

If Jeneesa started with, 0 points and she earned 50, then she would have 50. If she lost 80 points, she would go into the negatives. She would be at -30, and when she earned 10 more points, she would be at -20.

I don't know what your options look like, but look at the one(s) with -20 at the end and figure out what one best fits my explanation (and the numbers I use)!

3 0

3 years ago

Evaluating expressions:<br> 1. x + 9 when x = 3

balu736 [363]

⇨ The numeric value of this expression = 12.

To solve this expression, just <u>replace '</u><u>x</u><u>'</u> by the <em>indicated value</em>, being the number 3, and <em>perform</em> the <em>indicated </em>math <em>operations</em>.

${\large \displaystyle \sf {1 \cdot x+9 }}$

${\large \displaystyle \sf {x=3 }}$

${\large \displaystyle \sf {1 \cdot 3+9 }}$

${\large \displaystyle \sf { 3+9 }}$

${\orange{\boxed{\boxed{\pink {\large \displaystyle \sf { \ 12 \ }}}}}}\\\\$

Therefore, we can conclude that, the correct value of this expression = 12.

${\orange{\boxed{\boxed{\pink {\large \displaystyle \sf { \ 12 \ }}}}}}\\\\$

<h3><em>Hope this helps! ❤️</em></h3>

4 0

2 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:

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

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

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

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