Daily math problems for July 15, 2026 — Matrices, Calculus, Algebra & more

$\bullet$ Compute each entry:

$$f(1)=2\ln 1+3=3$$

$$f^{\mathrm{\prime }}(x)=\frac{2}{x}+6x\Rightarrow f^{\mathrm{\prime }}(2)=13$$

$${\log }_2(13+3)={\log }_{2}16=4$$

$\bullet$ Also $\sin 7\pi =0$ and $\mathrm{\mid }3-4i\mathrm{\mid }=5$, giving:

$$A=\left(\begin{array}{cc}3& 4\\ 0& 5\end{array}\right)$$

$\bullet$ Use $\det (A^2)=(\det A{)}^2$:

$$\det A=15\Rightarrow \det (A^2)=225$$

$u=x^2\Rightarrow du=2xdx$. So $\int \cos udu=\sin u+C=\sin (x^2)+C$.

Factoring gives $(x-2)(x-3)=0$, so the roots are $2$ and $3$. Their sum is $5$. (Equivalently, $-b\mathrm{/}a=5\mathrm{/}1$.)

The tetrahedron with these vertices has volume $1\mathrm{/}6\mathrm{\ne }0$, so the four points are NOT coplanar — they form a non-degenerate tetrahedron.

R2 − R1: $y+2z=8$. R3 − R1: $3y+8z=30$. From the first: $y=8-2z$. Substituting: $3(8-2z)+8z=30\Rightarrow 24+2z=30\Rightarrow z=3$. Then $y=2$, $x=1$. Solution: $(1,2,3)$.

$\sigma$: $1\to 2\to 3\to 4\to 1$. So ${\sigma }^2$: $1\to 3$, $3\to 1$, $2\to 4$, $4\to 2$. That's $(13)(24)$.

Set the terms as $5-r,5,5+r$. Then $(5-r{)}^2+25+(5+r{)}^2=50+2r^2+25=83$, so $2r^2=8\Rightarrow r=2$. The largest term is $5+2=7$.

$R(\theta )R(-\theta )=I_2$, so $R(\theta {)}^{-1}=R(-\theta )$.

$f^{\mathrm{\prime }}(x)=3x^2-3$ vanishes at $x=\pm 1$. $f(-1)=-1+3+1=3>0$ and $f(1)=1-3+1=-1<0$. The local max is positive and the local min is negative, so the cubic crosses the $x$-axis three times.

The three edges from origin are $(3,0,0),(0,4,0),(0,0,5)$. Determinant: $3\cdot 4\cdot 5=60$. Volume: $60\mathrm{/}6=10$.