Daily math problems for September 24, 2026 — Analytic Geometry, Trigonometry, Systems of Linear Equations & more

Slope $=2$. The line through $(0,1)$: $y-1=2(x-0)\Rightarrow y=2x+1$.

Area $={\displaystyle \frac{1}{2}}\cdot 6\cdot 8\cdot \sin {\displaystyle \frac{\pi }{6}}=24\cdot {\displaystyle \frac{1}{2}}=12$.

Multiplying the first by $2$: $2x+2y=2\mathrm{\ne }5$. Hence the system is inconsistent — no solution.

$S_3$ (or equivalently $D_3$) has $6$ elements and is non-abelian. All groups of order $1,2,3,4,5$ are abelian.

Din ${\log }_2(x-3)=4$ rezultă $x-3=2^4=16$, deci $x=19$. Cum $19>3$, domeniul este respectat.

$\bullet$ Add row $1$ and row $3$ component-by-component:

$$R_1+R_3=(0,0,0)$$

$\bullet$ The three rows are linearly dependent ($R_3=-R_1$), hence the determinant vanishes:

$$\det =0$$

For $x\in (0,1)$, multiplying by $x$ shrinks the value: $x^3=x\cdot x^2<x^2$ since $0<x<1$.

$f^{\mathrm{\prime }}(x)<0\iff 0<x<2$. So $f$ is decreasing on $(0,2)$. (Increasing on $(-\mathrm{\infty },0)$ and $(2,+\mathrm{\infty })$.)

$\bullet$ Simplify the exponent on the right: $i^{20}=(i^4{)}^5=1$, so $i^{20}+1=2$:

$$C_x^5=C_x^2$$

$\bullet$ Recall the combinatorial symmetry:

$$C_x^k=C_x^{x-k}$$

$\bullet$ Since $5\mathrm{\ne }2$, the symmetry forces $5+2=x$:

$$x=7$$

${\int }_0^{\pi \mathrm{/}2}{\sin }^{2}xdx={\int }_0^{\pi \mathrm{/}2}{\displaystyle \frac{1-\cos 2x}{2}}dx={\displaystyle \frac{1}{2}}{[x-\frac{\sin 2x}{2}]}_0^{\pi \mathrm{/}2}={\displaystyle \frac{\pi }{4}}$.