For what real values of does the quadratic 12x^2+k*x+27=0

Ответ:

amw4

14.02.2022

Mathematics

If a quadratic has nonreal roots, then the discriminant . b^2 - 4ac

Plug in the values for b, a, and c into this equation to get . k^2-4 *12*27 < 0

Rearrange to get . k^2 < *12*27

Now factorize 4*12*27 and solve to get - 36 < k<26

Step-by-step explanation:

hope this help

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Ответ:

viridianasar5158

22.10.2020

Mathematics

So the answer is DNC(Does Not Convergence)

Step-by-step explanation:

Given;

$\int\limits^\infty_5 {\frac{7}{x^{2}-9 } } \ dx$

We know,

The integral is an improper integral because the upper limit of integration approaches infinity. First, replace the infinite upper limit by the finite limit 'l' and take the limit of 'l' to approach infinity.

$\int\limits^\infty_5 {\frac{7}{x^{2}-9 } } \, dx=7 \lim_{l \to \infty} \int\limits^\infty_5 {\frac{1}{x^{2}-9 } } \, dx$

$7 \lim_{l \to \infty} \int\limits^\infty_5 {\frac{1}{x^{2}-3^{2} } } \, dx$

$7\times \lim_{l \to \infty} \left [ \frac{1}{2\times 3} \log\left |\frac{x-3}{x+3} \right | \right ]_{5}^{l}$ (∵We know; $\int {\frac{1}{x^{2}-a^{2} } } \, dx= \frac{1}{2\times a} \log\left |\frac{x-3}{x+3} \right |$ )

$\frac{7}{6} \times \lim_{l \to \infty} \left[ \log\left | \frac{l-3}{l+3} \right |-\log\left | \frac{5-3}{5+3}\right |\right]$

$\frac{7}{6} \times \lim_{l \to \infty} \left[ \log\left | \frac{l-3}{l+3} \right |-\log\left | \frac{-2}{8}\right |\right]$

$\frac{7}{6} \times \left[ \log\left | \frac{\infty-3}{\infty+3} \right |-\log( \frac{2}{8})\right]$ (∵ $\left | -\frac{2}{8} \right |=(\frac{2}{8} )$ )