Daily math problems for October 26, 2026 — Geometry, Fundamental Algebra, Differential Equations & more

${\displaystyle \frac{5}{\sin {30}^{\circ }}}={\displaystyle \frac{b}{\sin {90}^{\circ }}}$.

$\sin {30}^{\circ }=\frac{1}{2}$ and $\sin {90}^{\circ }=1$, so ${\displaystyle \frac{5}{1\mathrm{/}2}}={\displaystyle \frac{b}{1}}$.

$b=10$.

$49.2\approx 50$ and $5.1\approx 5$, so the estimate is $50\times 5=250$.

Verify $g$: $g^{\mathrm{\prime }}(x)=-1$ and $g+x=-x-1+x=-1$, so $g^{\mathrm{\prime }}=g+x$. ✓

The homogeneous equation $y^{\mathrm{\prime }}=y$ has general solution $C{e}^x$.

So the general solution of $y^{\mathrm{\prime }}=y+x$ is $y(x)=-x-1+C{e}^x$.

For $y=a{x}^2+bx+c$, the vertex is at $x=-{\displaystyle \frac{b}{2a}}$.

$x=-{\displaystyle \frac{-4}{2(1)}}=2$.

$y=(2{)}^2-4(2)+7=4-8+7=3$.

Vertex: $(2,3)$.

Left limit: ${\displaystyle \underset{x\to 2^{-}}{\lim }(x+1)=3}$.

Right limit: ${\displaystyle \underset{x\to 2^{+}}{\lim }(x^2-1)=3}$.

$f(2)=2^2-1=3$.

All three agree, so $f$ is continuous at $x=2$.

The coefficient of $X$ is $1$, so $x_1{x}_2+x_1{x}_3+x_2{x}_3=1$.

$\tan \text{\hspace{0.17em}⁣}(\pi +{\displaystyle \frac{\pi }{6}})=\tan {\displaystyle \frac{\pi }{6}}={\displaystyle \frac{1}{\sqrt{3}}}$.

$\bullet$ Evaluate each special entry:

$$i^2=-1,\lg 100=2,\ln 1=0$$

$$\sin \frac{\pi }{2}=1,i^{24}=1,C_4^3=4$$

$\bullet$ Assemble the numerical matrix:

$$A=\left(\begin{array}{cccc}1& 0& 2& 4\\ -1& 1& 3& 2\\ 0& 1& 3& 0\\ 1& 1& 4& 7\end{array}\right)$$

$\bullet$ Row-reduce. $R_2+=R_1$, $R_4-=R_1$:

$$\left(\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& 5& 6\\ 0& 1& 3& 0\\ 0& 1& 2& 3\end{array}\right)$$

$\bullet$ $R_3-=R_2$, $R_4-=R_2$:

$$\left(\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& 5& 6\\ 0& 0& -2& -6\\ 0& 0& -3& -3\end{array}\right)$$

$\bullet$ $R_4-=\frac{3}{2}{R}_3$ gives upper-triangular with pivots $1,1,-2,6$:

$$\det A=1\cdot 1\cdot (-2)\cdot 6=-12$$

By Vieta's relations: $r_1+r_2+r_3=6$ and $r_1{r}_2+r_1{r}_3+r_2{r}_3=11$. So $\sum r_i^2=6^2-2\cdot 11=36-22=14$.

$\bullet$ Evaluate $A(3)$ and its determinant:

$$A(3)=\left(\begin{array}{cc}4& 6\\ 3& -4\end{array}\right)$$

$$\det A(3)=4\cdot (-4)-6\cdot 3=-34$$

$\bullet$ Evaluate $A(2)$ and its determinant:

$$A(2)=\left(\begin{array}{cc}3& 4\\ 3& -5\end{array}\right)$$

$$\det A(2)=3\cdot (-5)-4\cdot 3=-27$$

$\bullet$ $A(3)$ has trace $0$, so by Cayley–Hamilton $A(3{)}^2=34I$. Inverting:

$$A^{-2}(3)=\frac{1}{34}I$$

$\bullet$ Invert $A(2)$:

$$A^{-1}(2)=\frac{1}{27}\left(\begin{array}{cc}5& 4\\ 3& -3\end{array}\right)$$

$\bullet$ Square it:

$$A^{-2}(2)=\frac{1}{729}\left(\begin{array}{cc}37& 8\\ 6& 21\end{array}\right)$$

$\bullet$ One more product:

$$A^{-3}(2)=\frac{1}{19683}\left(\begin{array}{cc}209& 124\\ 93& -39\end{array}\right)$$

$\bullet$ Add the matrices and apply $\det (sI+M)=s^2+s\mathrm{tr}M+\det M$ with $s=\frac{1}{34}$, $\mathrm{tr}M=\frac{170}{19683}$, $\det M=-\frac{1}{19683}$:

$$\det =\frac{1}{1156}+\frac{170}{34\cdot 19683}-\frac{1}{19683}$$

$$=\frac{1}{1156}+\frac{4}{19683}$$

$$=\frac{24307}{1156\cdot 19683}$$