Maths

${\alpha }^2-6\alpha -2=0,{\beta }^2-6\beta -2=0$

${\alpha }^2-2=6\alpha ,{\beta }^2-2=6\beta$

$\frac{a_{10}-2a_8}{3a_9}=\frac{{\alpha }^{10}-{\beta }^{10}-2\left({\alpha }^8-{\beta }^8\right)}{3\left({\alpha }^9-{\beta }^9\right)}=\frac{{\alpha }^8\left({\alpha }^2-2\right)-{\beta }^8\left({\beta }^2-2\right)}{3\left({\alpha }^9-{\beta }^9\right)}$

$\frac{{\alpha }^8.6\alpha -{\beta }^8.6\beta }{3\left({\alpha }^9-{\beta }^9\right)}=2$

Yes. Determinants can be calculated for any square matrix of n-order, and it is done by expansion of rows and columns. Even in higher dimensions, their job is to define hyper volumes and transformations.

In such cases, the value of determinant turns out to be zero. this is because swapping the values changes the magnitude into the opposite sign (fundamental property of determinants), which results in the final answer being zero.

Yes. Determinants can be fractions and irrational numbers depending on the values of the matrix. This is because the values are calculated as sums and products of the numbers in the matrix which can turn out to be any type of integer.

$f:\mathbf{R}\to \mathbf{R}$

$$$(\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{x}\mathrm{c}\mathrm{o}\mathrm{s}?\mathrm{y})\left(f\right(2\mathrm{x}+2\mathrm{y})-f(2\mathrm{x}-2\mathrm{y}\left)\right)=(\mathrm{c}\mathrm{o}\mathrm{s}?\mathrm{x}$ Put $\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{y}\left)\right(f(2\mathrm{x}+2\mathrm{y})+f(2\mathrm{x}-2\mathrm{y}))$

$$$\mathrm{x},\mathrm{y}\in \mathbf{R}$ Put $f^{\text{'}}\left(0\right)=\frac{1}{2}$

$24f^{\text{'}\text{'}}\left(\frac{5\pi }{3}\right)$

$\mathrm{}(\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{x}\mathrm{c}\mathrm{o}\mathrm{s}?\mathrm{y})\left(\mathrm{f}\right(2\mathrm{x}+2\mathrm{y})-\mathrm{f}(2\mathrm{x}-2\mathrm{y}\left)\right)=(\mathrm{c}\mathrm{o}\mathrm{s}?\mathrm{x}\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{y})$

$\left(f\right(2x+2y)+f(2x-2y\left)\right)$

$\mathrm{f}(2\mathrm{x}+2\mathrm{y})(\mathrm{s}\mathrm{i}\mathrm{n}?(\mathrm{x}-\mathrm{y}\left)\right)=\mathrm{f}(2\mathrm{x}-2\mathrm{y})\mathrm{s}\mathrm{i}\mathrm{n}?(\mathrm{x}+\mathrm{y})$

: $?2$ adj $(3\mathrm{A}$  adj(2A))|
${=2}^3.?3\mathrm{A}$ adj(2A)| ${\left.\right|}^2$

${=2}^3\cdot {\left(3^3\right)}^2\cdot |\mathrm{A}{|}^2\cdot |\mathrm{a}\mathrm{d}\mathrm{j}\left(2\mathrm{A}\right){|}^2$  intersect the line $=2^3\cdot 3^6\cdot |\mathrm{A}{|}^2\cdot {\left(|2\mathrm{A}{|}^2\right)}^2$ at the point

$=2^3\cdot 3^6\cdot |\mathrm{A}{}^{}$

Using the standard equations of a hyperbola:

$9\left({\mathrm{e}}^2+l\right)$ and directrix $focus\mathrm{a}\mathrm{e}=\sqrt[]{10}$

By multiplying both focus and directrix, we get
$\frac{\mathrm{a}}{\mathrm{e}}=\frac{9}{\sqrt[]{10}}$ and $\Rightarrow {\mathrm{a}}^2=9$
Now $\mathrm{e}=\frac{\sqrt[]{10}}{3}$
$\mathrm{}(ae{)}^2=a^2+b^2$

We can convert 50! In terms of prime factor: $\mathrm{}{2}^{\alpha }\cdot 3^{\beta }\cdot 5^{\gamma }$ & using the greatest integer function.

n $=\left. [\frac{50}{3}\right]+\left. [\frac{50}{3^2}\right]+\left. [\frac{50}{3^3}\right]+\left. [\frac{50}{3^4}\right]$
$=16+5+1+0$ The maximum value of n is 22.

$\underset{x\to 7}{lim}\frac{18-\left[1-x\right]}{\left[x-3a\right]}$

exist &   $a\in I.$

$=\underset{x\to 7}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}$

exist

RHL =   $\underset{x\to 7^{+}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{25}{7-3a}\left[a\ne \frac{7}{3}\right]$

$LHL=\underset{x\to 7^{-}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{24}{6-3a}\left[a\ne 2\right]$

LHL = RHL

$\Rightarrow \frac{25}{7-3a}=\frac{8}{2-a}$

$\therefore a=-6$

$\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}$

$=bxdy+cydy+ady=axdx-bydx+adx$

$=\frac{c{y}^2}{2}+ay-\frac{a{x}^2}{2}-ax+bxy=k$

$a{x}^2+a{y}^2+2ax-2ay=k$

$\Rightarrow x^2+y^2+2x-2y=\lambda$

Short distance of (11,6)

= $\sqrt[]{12^2+5^2}-5$ $\text{exception 25:}$

= 13 – 5

= 8